Contents
Chapter 4 Theory of Matrices
4.1 Introduction
We have already seen that an \(m \times n\) matrix \(A\) is an array of \(mn\) numbers \(a_{ij}\) where \(1 \leq i \leq m\), \(1 \leq j \leq n\) arranged in \(m\) rows and \(n\) columns as follows
\[ A = \begin {pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \cdots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \\ \end {pmatrix} \]
We shall denote this matrix by the symbol \(a_{ij}\). If \(m=n\), \(A\) is called a square matrix of order \(n\).
4.2 Elementary Transformations
Definition 4.2.1. Let \(A\) be an \(m \times n\) matrix over a field \(F\). An elementary row (column) operation on \(A\) is of any one of the following three types.
1. The interchange of any two rows (columns).
2. Multiplication of a row (column) by a non-zero element \(c\) in \(F\).
3. Addition of any multiple of one row (column) with any other row (column).
Example 4.2.2. Let \(A=\left (\begin {array}{rr}1 & 2 \\ 2 & 1 \\ 3 & -1\end {array}\right )\), \(A_{1}=\left (\begin
{array}{rr}3 & -1 \\ 2 & 1 \\ 1 & 2\end {array}\right )\), \(A_{2}=\left (\begin {array}{rr}2 & 2 \\ 4 & 1 \\ 6 & -1\end {array}\right )\),
\(A_{3}=\left (\begin {array}{rr}1 & 2 \\ 5 & 7 \\ 3 & -1\end {array}\right ) .\)
\(A_{1}\) is obtained from \(A\) by interchanging the first and third rows.
\(A_{2}\) is obtained from \(A\) by multiplying the first column of \(A\) by 2.
\(A_{3}\) is obtained from \(A\) by adding to the second row the multiple by 3 of the first row.
Note 4.2.3. We may use the following notations for elementary transformations.
(i) Interchange of \(i^{\text {th }}\) and \(j^{\text {th }}\) rows will be denoted by \(R_{i} \leftrightarrow R_{j}\).
(ii) Multiplication of \(i^{\text {th }}\) row by a non-zero element \(c \in F\) will be denoted by \(R_{i} \rightarrow c R_{i}\).
(iii) Addition of \(k\) times the \(j^{\text {th }}\) row to the \(i^{\text {th }}\) row will be denoted by \(R_{i} \rightarrow R_{i}+k R_{j}\).
The corresponding column operations will be denoted by writing \(C\) in the place of \(R\).
Definition 4.2.4. An \(m \times n\) matrix \(B\) is said to be row equivalent (column equivalent) to an \(m \times n\) matrix \(A\) if \(B\) can be obtained from \(A\) by a finite succession of elementary row operations (column operations).
\(A\) and \(B\) are said to be equivalent if \(B\) can be obtained from \(A\) by a finite succession of elementary row or column operations.
If \(A\) and \(B\) are equivalent. We write \(A \sim B\).
Example 4.2.6. The following \(\left (\begin {array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end {array}\right )\), \(\left (\begin {array}{lll}4 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end {array}\right )\), \(\left (\begin {array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 2 & 1\end {array}\right )\) are elementary matrices obtained from the identity matrix \(\left (\begin {array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end {array}\right )\) by applying the elementary operations \(R_{1} \leftrightarrow R_{2}\), \(R_{1} \rightarrow 4 R_{1}\), \(R_{3} \rightarrow R_{3} +2 R_2 \) respectively.
Proof : The determinant of the identity matrix of any order is 1 . Hence the determinant of an elementary matrix obtained by interchanging any two rows is \(-1\). The determinant of an elementary matrix obtained by multiplying any row by \(k \neq 0\) is \(k\). The determinant of an elementary matrix obtained by adding a multiple of one row with another row is 1. Hence any elementary matrix is non-singular. □
Proof : Let \(R_{1}, R_{2}, \ldots , R_{m}\) denote the rows of the matrix \(A\) and \(C_{1}, C_{2}, \ldots , C_{p}\) denote the columns of \(B\). By the definition of matrix multiplication
\[ A B=\left (\begin {array}{cccc} R_{1} C_{1} & R_{1} C_{2} & \ldots \ldots & R_{1} C_{p} \\ R_{2} C_{1} & R_{2} C_{2} & \ldots \ldots & R_{2} C_{p} \\ \cdot & \cdot & \ldots \ldots & \cdot \\ \cdot & \cdot & \ldots \ldots & \cdot \\ \cdot & \cdot & \ldots \ldots & \cdot \\ R_{m} C_{1} & R_{m} C_{2} & \ldots \ldots & R_{m} C_{p} \end {array}\right ) \]
It is obvious from the above representation of \(A B\) that if we apply any elementary row operation on \(A\) the matrix \(A B\) is also subjected to the same elementary row operation. Also if we apply any elementary column operation on \(B\) the matrix \(A B\) is also subjected to the same elementary column operation. □
Proof : Since \(A\) is an \(m \times n \) matrix we can write \(A=I A\) where \(I\) is the identity matrix of order \(m\). By Theorem 4.2.8 an elementary row operation on \(IA\) is equivalent to the same row operation on \(I\). But an elementary row operation on \(I\) gives an elementary matrix. Hence by pre-multiplying \(A\) by the corresponding elementary matrix we get the required row operation on \(A\). □
Proof : Since \(A\) is row equivalent to \(B\), \(A\) can be obtained from \(B\) by applying successive elementary row operations.
Hence \(A=E_{1} E_{2} \ldots E_{n} B\) where each \(E_{i}\) is an elementary matrix. Since each \(E_{i}\) is non-singular, \(A=P B\) where \(P=E_{1} E_{2} \ldots E_{n}\) and \(P\) is non-singular. □
Proof : Let \(E\) be an elementary matrix obtained from \(I\) by applying some elementary operations. If we apply the reverse operation on \(E\), then \(E\) is carried back to \(I\). Let \(E^{*}\) be the elementary matrix
corresponding to the reverse operation.
Then \(E^{*} E=E E^{*}=I\). Hence \(E^{*}=E^{-1}\).
Hence \(E^{-1}\) is also an elementary matrix. □
Canonical form of a matrix
We now use elementary tow and column operations to reduce any matrix to a simple form, called the canonical form of a matrix.
Theorem 4.2.15. By successive applications of elementary row and column operations, any non-zero \(m \times n\) matrix
\(A\) can be reduced to a diagonal matrix \(D\) in Which the diagonal entries are either 0 or 1 and all the 1’s proceeding all the zeros on the diagonal.
In other Words, any non-zero \(m \times n\) matrix is equivalent to a matrix of the form \(\left [\begin {array}{cc}I_{r} & 0 \\ 0 & 0\end {array}\right ]\) where \(I_{r}\) is the \(r \times r\) identity matrix and \(0\) is the zero matrix.
Proof : We shall prove the theorem by induction on the number of rows of \(A\). Suppose \(A\) has just one row.
Let \(A=\left (a_{11} a_{12} \ldots a_{1 n}\right )\).
Since \(A \neq 0\), by interchanging columns, if necessary, we can bring a non-zero entry \(c\) to the position \(a_{11}\).
Multiplying \(A\) by \(c^{-1}\) we get 1 as the first entry.
Other entries in \(A\) can be made zero by adding - suitable multiples of 1 . Thus the result is true when \(m=1\).
Now, suppose that the result is true for any non-zero matrix with \(m-1\) rows.
Let \(A\) be a non-zero \(m \times n\) matrix. By permuting rows and columns we can bring some non-zero entry \(c\) to the position \(a_{11}\).
Multiplying the first row by \(c^{-1}\) we get 1 as the first entry.
All other entries in the first column can be made zero by adding suitable multiples of the first row to each other row.
Similarly all the other entries in the first row can be made zero:
This reduces \(A\) to a matrix of the form
\[ B=\left (\begin {array}{ll} I_{1} & O \\ 0 & C \end {array}\right ) \]
where \(C \) is an \((m-1) \times (n-1)\) matrix.
Now by induction hypothesis \(C\) can be reduced to the desired form by elementary row and column operations.
Hence \(A\) is equivalent to a matrix of the required form. □
Proof : By Corollary 4.2.16, \(P A Q=\left (\begin {array}{ll}I_{r} & O \\ O & 0\end {array}\right )\).
Since \(P, A, Q\) are all non-singular \(\left (\begin {array}{ll}I_{r} & O \\ O & O\end {array}\right )\) is non-singular. This is possible if and only if \(\left (\begin {array}{cc}I_{r} & O \\ O & O\end {array}\right )=I_{n}\). □
Note 4.2.19. The inverse of a non-singular matrix \(A\) can be computed by using elementary transformations. Let \(A\) be a non-singular matrix of
order \(n\). Then \(A A^{-1}=\) \(A^{-1} A=I\). Now, the non-singular matrix \(A^{-1}\) can be expressed as the product of elementary matrices.
Let \(A^{-1}=E_{1} E_{2} \ldots E_{n}\).
Then \(I=A^{-1} A=E_{1} E_{2} \ldots E_{n} A\).
Thus every non-singular matrix \(A\) can be reduced to \(I\) by pre-multiplying \(A\) by elementary matrices.
Hence \(A\) can be reduced to the identity matrix by applying successive elementary row operations.
Now, \(A=I A\). Reduce the matrix \(A\) in the left hand side to \(I\) by applying successive elementary row operations and apply the same elementary row operations to the factor \(I\) in the right hand side. Then we get \(I=B A\) so that \(B=A^{-1}\).
Solved Problems
Solution :
\(\seteqnumber{0}{4.}{0}\)\begin{align*} A & =\begin{pNiceArray}{ccc}1 & 2 & -1 \\ 1 & 1 & 2 \\ 2 & 4 & -2\end {pNiceArray}\\ &\sim \begin{pNiceArray}[last-col]{ccc} 1 & 2 & -1 \\ 0 & -1 & 3 & R_{2} \rightarrow R_{2}-R_{1}\\ 0 & 0 & 0 & R_{3} \rightarrow R_{3}-2 R_{1} \end {pNiceArray}\\ &\sim \begin{pNiceArray}[last-col]{ccc}1 & 0 & 0 \\ 0 & -1 & 3 & C_{2} \rightarrow C_{2}-2 C_{1}\\ 0 & 0 & 0 & C_{3} \rightarrow C_{3}+C_{1} \end {pNiceArray}\\ &\sim \begin{pNiceArray}[last-col]{ccc}1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 & C_{3} \rightarrow C_{3}+3 C_{2} \end {pNiceArray}\\ &\sim \begin{pNiceArray}[last-col]{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 & R_{2} \rightarrow -R_{2}\\ 0 & 0 & 0 \end {pNiceArray} \end{align*}
Solution :
\(\seteqnumber{0}{4.}{0}\)\begin{align*} \begin{pNiceArray}{ccc} 1 & 0 & 2 \\ 3 & 1 & -1 \\ -2 & 1 & 3\end {pNiceArray} & = A \begin{pNiceArray}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end {pNiceArray} \\ \Rightarrow \begin{pNiceArray}{ccc} 1 & 0 & 2 \\ 0 & 1 & -7 \\ 0 & 1 & 7\end {pNiceArray}&=A \begin{pNiceArray}[last-col]{ccc} 1 & 0 & 0 \\ -3 & 1 & 0 & R_{2} \rightarrow R_{2}-3 R_{1} \\ 2 & 0 & 1 & R_{3} \rightarrow R_{3}+2 R_{1} \end {pNiceArray} \\ \Rightarrow \begin{pNiceArray}{ccc} 1 & 0 & 2 \\ 0 & 1 & -7 \\ 0 & 0 & 14 \end {pNiceArray}&=A \begin{pNiceArray}[last-col]{ccc} 1 & 0 & 0 \\ -3 & 1 & 0 \\ 5 & -1 & 1 & R_{3} \rightarrow R_{3}-R_{2} \end {pNiceArray} \\ \Rightarrow \begin{pNiceArray}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end {pNiceArray} & =A \begin{pNiceArray}[last-col]{ccc} \frac {2}{7} & \frac {1}{7} & -\frac {1}{7} & R_{1} \rightarrow R_{1}-\frac {1}{7} R_{3} \\ -\frac {1}{2} & \frac {1}{2} & \frac {1}{2} & R_{2} \rightarrow R_{2}+\frac {1}{2} R_{3} \\ \frac {5}{14} & -\frac {1}{14} & \frac {1}{14}& R_{3} \rightarrow \frac {1}{14} R_{3} \end {pNiceArray}\\ \Rightarrow A^{-1}&=\begin{pNiceArray}{ccc} \frac {2}{7} & \frac {1}{7} & -\frac {1}{7} \\ -\frac {1}{2} & \frac {1}{2} & \frac {1}{2} \\ \frac {5}{14} & -\frac {1}{14} & \frac {1}{14}\end {pNiceArray} \end{align*}
Solved Problems
Solution : Let \(S\) be the set of all \(n \times n\) matrices.
Let \(A \in S\).
Since \(A=I^{-1} A I\) and \(I\) is non-singular, \(A\) is similar to \(A\).
Hence similarity of matrices is reflexive.
Now, let \(A, B \in S\) and let \(A\) be similar to \(B\) matrix.
\[ A=P^{-1} B P\]
where \(P \in S\) is a non-singular. Now,
\(\seteqnumber{0}{4.}{0}\)
\begin{align*}
P^{-1} B P=A & \Rightarrow P P^{-1} B P P^{-1}=P A P^{-1} \\ & \Rightarrow B=P A P^{-1} \\ & \Rightarrow B=\left (P^{-1}\right )^{-1} A\left (P^{-1}\right )
\end{align*}
Since \(P\) is non-singular \(P^{-1} \in S\) is also non-singular.
\(B\) is similar to \(A\). Hence similarity of matrices is symmetric.
Now, let \(A, B, C \in S\).
Let \(A\) be similar to \(B\) to \(B\) be similar to \(C\). Hence there exist non-singular matrices \(P, Q \in S\) such that
\[ A=P^{-1} B P \text { and } B=Q^{-1} C Q \text {. } \]
Now,
\(\seteqnumber{0}{4.}{0}\)
\begin{align*}
A& =P^{-1} B P\\ &=P^{-1}\left (Q^{-1} C Q\right ) P \\ &=\left (P^{-1} Q^{-1}\right ) C Q P \\ &=(Q P)^{-1} C(Q P) .
\end{align*}
Since \(P, Q \in S\) are non-singular, \(Q P \in S\) is also non-singular.
Hence \(A\) is similar to \(C\). \(\therefore \) Similarity of matrices is transitive.
Hence similarity of matrices is an equivalence relation.
Solution : Let \(A\) and \(B\) be two similar matrices.
\(\therefore \) There exists a non-singular matrix \(P\) such that \(B=P^{-1} A P\). Now,
\begin{align*} |B| &=\left |P^{-1} A P\right | \\ &=\left |P^{-1} \| A\right ||P| \\ &=|A| \quad \left (\text { since }\left |P^{-1}\right |=\frac {1}{|P|}\right ) \end{align*} Hence the result.
4.3 Rank of a Matrix
We now proceed to introduce the concept of the rank of a matrix.
Definition 4.3.1. Let \(A=\left (a_{i j}\right )\) be an \(m \times n\) matrix. The rows \(R_{i}=\left (a_{i 1}, a_{i 2}, \ldots \ldots \ldots , \ldots a_{i n}\right )\) of \(A\) can be thought of as elements of \(F^{A}\). The subspace of \(F^{n}\) generated by the \(m\) rows of \(A\) is called the row space of \(A\).
Similarly, the subspace of \(F^{m}\) generated by the \(n\) columns of \(A\) is called the column space of \(A\).
The dimension of the row space (column space) of \(A\) is called the row rank (column rank) of \(A\).
Proof : Let \(A\) be an \(m \times n\) matrix.
It is enough if we prove that the row space of \(A\) is not altered by any elementary row operation.
Obviously the row space of \(A\) is not altered by an elementary row operation of the type \(R_{i} \leftrightarrow R_{j}\).
Now, consider the elementary row operation \(R_{i} \rightarrow c R_{i}\) where \(c \in F-\{0\} .\)
Since \(L\left (\left \{R_{1}, R_{2}, \ldots \ldots , R_{i}, \ldots \ldots \ldots , R_{n}\right \}\right )=\) \(L\left (\left \{R_{1}, R_{2}, \ldots \ldots \ldots , c R_{i}, \ldots \ldots , R_{n}\right \}\right )\) the row space of \(A\) is not
altered by this type of elementary row operation.
Similarly we can easily prove that the row space of \(A\) is not altered by an elementary row operation of the type \(R_{i} \rightarrow R_{i}+c R_{i}\).
Hence row equivalent matrices have the same row space and hence the same row rank. □
Similarly we can prove the following theorem.
Proof : Let \(A=\left (a_{i j}\right )\) be an \(m \times n\) matrix.
Let \(R_{1}, R_{2}, \ldots \ldots \ldots , R_{m}\) denote the rows of \(A\).
Hence \(R_{i}=\left (a_{i 1}, a_{i 2}, \ldots \ldots \ldots , a_{i n}\right )\).
Suppose the row rank of \(A\) is \(r\).
Then the dimension of the row space is \(r\).
Let \(v_{1}=\left (b_{11}, \ldots , b_{1 n}\right ), v_{2}=\left (b_{21}, \ldots , b_{2 n}\right ), \ldots \) \(\ldots , v_{r}=\left (b_{r 1}, \ldots \ldots \ldots , b_{r n}\right )\) be a basis for the row space of \(A\).
Then each row is a linear combination of the vectors \(v_{1}, v_{2}, \ldots \ldots \ldots , v_{r}\). Let,
\begin{align*} R_{1} & =k_{11} v_{1}+k_{12} v_{2}+\cdots \cdots +k_{1 r} v_{r}\\ R_{2} & =k_{21} v_{1}+k_{22} v_{2}+\cdots \cdots +k_{2 r} v_{r}\\ R_{m}=& k_{m 1} v_{1}+k_{m 2} v_{2}+\cdots \cdots +k_{m r} v_{r} && \text { where } k_{i j} \in F \end{align*} Equating the \(i^{\text {th }}\) component of each of the above equations, we get
\(\seteqnumber{0}{4.}{0}\)\begin{align*} a_{1 i}=& k_{11} b_{1 i}+k_{12} b_{2 i}+\cdots \cdots +k_{1 r} b_{r i} \\ a_{2 i}=& k_{21} b_{1 i}+k_{22} b_{2 i}+\cdots \cdots +k_{2 r} b_{r i} \\ \cdots & \cdots \cdots \cdots \cdots \\ a_{m i}=& k_{m 1} b_{1 i}+k_{m 2} b_{2 i}+ \cdots \cdots + k_{m r} b_{r i} \end{align*} Hence
\[ \left (\begin {array}{c} a_{1 i} \\ \cdot \\ \cdot \\ a_{m i} \end {array}\right )=b_{1 i}\left (\begin {array}{c} k_{11} \\ \cdot \\ \cdot \\ k_{m 1} \end {array}\right )+b_{2 i}\left (\begin {array}{c} k_{12} \\ \cdot \\ \cdot \\ k_{m 2} \end {array}\right )+\cdots +b_{r i}\left (\begin {array}{c} k_{1 r} \\ \cdot \\ \cdot \\ k_{m r} \end {array}\right ) \]
Thus each column of \(A\) is a linear combination of vectors.
Hence the dimension of the column space \(\leq r\).
\(\therefore \quad \) Column rank of \(A \leq r=\) row rank of \(A\).
Similarly, row rank of \(A \leq \) column rank of \(A\).
Hence the row rank and the column rank of \(A\) are equal. □
Note 4.3.6. Since the row rank and the column rank of matrix are unaltered by elementary row and column operations, equivalent matrices have the same rank In particular if a matrix \(A\) is reduced to its canonical form \(\left (\begin {array}{ll}I_{r} & O \\ O & O\end {array}\right )\), then rank of \(A=r\).
Thus to find the rank of a matrix \(A\), we reduce to the canonical form and find the number of non-zero entries in the diagonal.
Note that in the canonical form of the matrix there exists an \(r \times r\) sub-matrix, namely, \(I_{r}\), whose determinant is not zero.
Further every \((r+1) \times (r+1)\) sub-matrix contains a row of zeros and hence its determinant is zero.
Also under any elementary row or column operation the value of a determinant is either unaltered or multiplied by a non-zero constant.
Hence the matrix \(A\) is also such that
(i) there exists an \(r \times r\) sub-matrix whose determinant is nonzero.
(ii) the determinant of every \((r+1) \times (r+1)\) sub-matrix is zero.
Hence one can also define the rank of a matrix A to be \(r\) if A satisfies (i) and (ii).
Solved Problems
Solution :
\(\seteqnumber{0}{4.}{0}\)\begin{align*} A &=\begin{pNiceArray}{cccc} 4 & 2 & 1 & 3 \\ 6 & 3 & 4 & 7 \\ 2 & 1 & 0 & 7 \end {pNiceArray} \\ & \sim \begin{pNiceArray}[last-col]{cccc} 1 & 2 & 4 & 3 & C_{1} \leftrightarrow C_{3} \\ 4 & 3 & 6 & 7 \\ 0 & 1 & 2 & 7 \end {pNiceArray} \\ & \sim \begin{pNiceArray}[last-col]{cccc} 1 & 0 & 0 & 0 & C_{1} \rightarrow C_2 - 2 C_1 \\ 4 & -5 & -10 & -5 & C_{3} \rightarrow C_{3}-4 C_{1} \\ 0 & 1 & 2 & 7 & C_{4} \rightarrow C_{4}-3 C_{1} \end {pNiceArray} \\ & \sim \begin{pNiceArray}[last-col]{cccc} 1 & 0 & 0 & 0 \\ 0 & -5 & -10 & -5 & R_{2} \rightarrow R_{2}-4 R_{1}\\ 0 & 1 & 2 & 7 \end {pNiceArray} \\ & \sim \begin{pNiceArray}[last-col]{cccc} 1 & 0 & 0 & 0 & C_{3} \rightarrow C_{3}-2 C_{2} \\ 0 & -5 & 0 & 0 & C_4 \to C_4 - C_2 \\ 0 & 1 & 0 & 6 \end {pNiceArray} \\ & \sim \begin{pNiceArray}[last-col]{cccc} 1 & 0 & 0 & 0 \\ 0 & -5 & 0 & 0 \\ 0 & 0 & 0 & 6 & R_{3} \rightarrow R_{3}+\frac {1}{5} R_{2} \end {pNiceArray} \\ & \sim \begin{pNiceArray}[last-col]{cccc} 1 & 0 & 0 & 0 \\ 0 & -5 & 0 & 0 & C_{2} \leftrightarrow C_{3} \\ 0 & 0 & 6 & 0 \end {pNiceArray} \\ & \sim \begin{pNiceArray}[last-col]{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & R_{2} \rightarrow -\frac {1}{5} R_{2} \\ 0 & 0 & 1 & 0 & R_{3} \rightarrow \frac {1}{6} R_{3} \end {pNiceArray} \end{align*}
\[\text { Rank of } A=3.\]
Solution :
\(\seteqnumber{0}{4.}{0}\)\begin{align*} \left |\begin{array}{lll} 1 & 1 & 1 \\ 4 & 1 & 0 \\ 0 & 3 & 4 \end {array}\right | & =0=\left |\begin{array}{lll} 1 & 1 & 1 \\ 1 & 0 & 2 \\ 3 & 4 & 2 \end {array}\right | \\ \left |\begin{array}{lll} 1 & 1 & 1 \\ 4 & 1 & 2 \\ 0 & 3 & 2 \end {array}\right |& =0=\left |\begin{array}{lll} 1 & 1 & 1 \\ 4 & 0 & 2 \\ 0 & 4 & 2 \end {array}\right | \end{align*} Every \(3 \times 3\) submatrix of \(A\) has determinant zero. Also,
\(\seteqnumber{0}{4.}{0}\)\begin{align*} \left |\begin{array}{ll} 1 & 1 \\ 4 & 1 \end {array}\right | & =-3 \neq 0 \\ \text { Rank of } A& =2 \end{align*}
4.4 Simultaneous Linear Equations
In this section we shall apply the theory of matrices developed in the preceding sections to study the existence of simultaneous linear equations.
Matrix form of a set of linear equations
Consider a system of \(m\) linear equations in \(n\) unknowns \(x_{1}, x_{2}, \ldots , x_{n}\) given by
\(\seteqnumber{0}{4.}{0}\)\begin{align*} a_{11}x_1 + a_{12}x_2 + \cdots + a_{1n}x_n & = b_1 \\ a_{21}x_1 + a_{22}x_2 + \cdots + a_{2n}x_n & = b_2 \\ \cdots \cdots \cdots \cdots & = b_1 \\ a_{m1}x_1 + a_{m2}x_2 + \cdots + a_{mn}x_n & = b_m \\ \end{align*} Using the concept of matrix multiplication and equality of matrices this system can be written as \(A X=B\) where,
\[ A = \begin {pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \\ \end {pmatrix} , \quad X = \begin {pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end {pmatrix} , \quad B= \begin {pmatrix} b_1 \\ b_2 \\ \vdots \\ b_m \end {pmatrix} \]
Then \(m \times n \) matrix \(A\) is called the coefficient matrix.
Definition 4.4.1. A set of values of \(x_1, x_2, \ldots , x_n\) which satisfy the above system of equations is called a solution of the system. The system of equations is said to be consistent if it has at least one solution. Otherwise the system is said to be inconsistent.
The \(m \times (n+1)\) matrix given by
\[A = \begin {pmatrix} a_{11} & a_{12} & \cdots & a_{1n} & b_1 \\ a_{21} & a_{22} & \cdots & a_{2n} & b_2 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn}& b_m \end {pmatrix}\]
Thus the augmented matrix \((A,B)\) is obtained by annexing to \(A \) the column matrix \(B\), which becomes the \((n+1)^{th}\) column in \((A,B)\).
Proof : Let the system be consistent.
Let \(u_{1}, u_{2}, \ldots \ldots , u_{n}\) be a solution of the system.
Then \(B=u_{1} C_{1}+u_{2} C_{2}+\cdots \cdots +u_{n} C_{n}\) where \(C_{1}, C_{2}, \cdots , C_{n}\) denote the columns of \(A\).
Hence the column space of the augmented matrix \((A, B)\), namely \(\left \langle C_{1}, C_{2}, \ldots \ldots , C_{n}, B\right \rangle \) is the same as the column space \(\left \langle C_{1}, C_{2}, \ldots \ldots , C_{n}\right \rangle \) of
\(A\).
Hence the rank of \(A=\operatorname {rank}\) of \((A, B)\).
Conversely let rank of \(A=\operatorname {rank}\) of \((A, B)\).
Then the column rank of \(A=\) column rank of \((A, B)\).
\[\operatorname {dim}\left \langle C_{1}, C_{2}, \ldots , C_{n}\right \rangle =\operatorname {dim}\left \langle C_{1}, C_{2}, \ldots , C_{n}, B\right \rangle . \]
But \(\left \langle C_{1}, C_{2}, \ldots \ldots , C_{n}\right \rangle \) is a subspace of \(\left \langle C_{1}, C_{2}, \ldots \ldots , C_{n}, B\right \rangle \).
\(B\) is a linear combination of \(C_{1}, C_{2}, \ldots \ldots , C_{n}\).
If \(B=u_{1} C_{1}+\ldots +u_{n} C_{n}\) then \(u_{1}, u_{2}, \ldots \ldots , u_{n}\) is a solution of the system. This completes the proof. □
Solved Problems
Solution : The given system of equations can be put in the matrix form
\(\seteqnumber{0}{4.}{0}\)\begin{align*} A X & =\left (\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 7 \end {array}\right )\left (\begin{array}{l} x \\ y \\ z \end {array}\right )=\left (\begin{array}{c} 6 \\ 14 \\ 30 \end {array}\right )=B \end{align*} The augmented matrix is given by
\(\seteqnumber{0}{4.}{0}\)
\begin{align*}
(A, B) &=\left (\begin{array}{cccc} 1 & 1 & 1 & 6 \\ 1 & 2 & 3 & 14 \\ 1 & 4 & 7 & 30 \end {array}\right ) \\ & \sim \begin{pNiceArray}[last-col]{cccc} 1 & 1 & 1 & 6 \\ 0 & 1
& 2 & 8 & R_{2} \rightarrow R_{2}-R_{1} \\ 0 & 3 & 6 & 24 & R_{3} \rightarrow R_{3}-R_{1} \end {pNiceArray}\\ & \sim \begin{pNiceArray}[last-col]{cccc} 1 & 1 & 1 & 6 \\ 0 & 1 & 2 & 8
\\ 0 & 0 & 0 & 0 & R_{3} \rightarrow R_{3}-3 R_{2} \end {pNiceArray}
\end{align*}
Hence rank of \(A=\operatorname {rank}\) of \((A, B)=2\).
Hence the given system is consistent.
Also the given system of equations reduces to
\begin{align*} \left (\begin{array}{lll} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end {array}\right ) \left (\begin{array}{l} x \\ y \\ z \end {array}\right ) & =\left (\begin{array}{l} 6 \\ 8 \\ 0 \end {array}\right ) \\ \therefore \quad x+y+z & =6 \\ y+2 z & =8 \end{align*} Putting \(z=c\) we obtain the general solution of the system as \(x=c-2, y=8-2 c, z=c\).
Solution : The matrix form of the system is given by
\(\seteqnumber{0}{4.}{0}\)\begin{align*} \left (\begin{array}{rrr} 1 & -4 & -3 \\ 4 & -1 & 6 \\ 2 & 7 & 12 \\ 5 & -5 & 3 \end {array}\right )\left (\begin{array}{l} x \\ y \\ z \end {array}\right ) & =\left (\begin{array}{r} -16 \\ 16 \\ 48 \\ 0 \end {array}\right ) \end{align*} The augmented matrix is given by
\(\seteqnumber{0}{4.}{0}\)\begin{align*} (A, B)& =\left (\begin{array}{rrrr} 1 & -4 & -3 & -16 \\ 4 & -1 & 6 & 16 \\ 2 & 7 & 12 & 48 \\ 5 & -5 & 3 & 0 \end {array}\right )\\ & \sim \begin{pNiceArray}[last-col]{cccc} 1 & -4 & -3 & -16 \\ 0 & 15 & 18 & 80 & R_{2} \rightarrow R_{2}-4 R_{1} \\ 0 & 15 & 18 & 80 & R_{3} \rightarrow R_{3}-2 R_{1} \\ 0 & 15 & 18 & 80 & R_{4} \rightarrow R_{4}-5 R_{1} \end {pNiceArray} \\ & \sim \begin{pNiceArray}[last-col]{cccc} 1 & -4 & -3 & -16 \\ 0 & 15 & 18 & 80 \\ 0 & 0 & 0 & 0 & R_{3} \rightarrow R_{3}-R_{2} \\ 0 & 0 & 0 & 0 & R_{4} \rightarrow R_{4}-R_{2} \end {pNiceArray} \end{align*} Rank of \(A=\operatorname {Rank}\) of \((A, B)=2\) and hence the system is consistent. Also the system of equations reduces to
\[ \left (\begin {array}{ccc} 1 & -4 & -3 \\ 0 & 15 & 18 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end {array}\right )\left (\begin {array}{l} x \\ y \\ z \end {array}\right )=\left (\begin {array}{r} -16 \\ 80 \\ 0 \\ 0 \end {array}\right )\]
\(x-4 y-3 z=-16 \) and \(15 y+18 z=80 .\)
Putting \(z=c\) we obtain the general solution of the systems as
\[x=-(9 c / 5)+(16 / 3),\ y=-(6 c / 5)+(16 / 3) , z=c .\]
Solution : The matrix form of the system is given by
\(\seteqnumber{0}{4.}{0}\)\begin{align*} \left (\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 4 & 10 \end {array}\right )\left (\begin{array}{l} x \\ y \\ z \end {array}\right ) & =\left (\begin{array}{c} 1 \\ \eta \\ \eta ^{2} \end {array}\right ) \end{align*} The augmented matrix is given by
\(\seteqnumber{0}{4.}{0}\)\begin{align*} (A, B) &=\left (\begin{array}{cccc} 1 & 1 & 1 & 1 \\ 1 & 2 & 4 & \eta \\ 1 & 4 & 10 & \eta ^{2} \end {array}\right ) \\ & \sim \begin{pNiceArray}[last-col]{cccc} 1 & 1 & 1 & 1 \\ 0 & 1 & 3 & \eta -1 & R_{2} \rightarrow R_{2}-R_{1} \\ 0 & 3 & 9 & \eta ^{2}-1 & R_3 \to R_3 - R_1 \end {pNiceArray}\\ & \sim \begin{pNiceArray}[last-col]{cccc} 1 & 1 & 1 & 1 \\ 0 & 1 & 3 & \eta -1 \\ 0 & 0 & 0 & \eta ^{2}-3 \eta +2 & R_3 \to R_3 - 3 R_2 \end {pNiceArray} \end{align*} The given system is consistent if and only if \(\eta ^{2}-3 \eta +2=0\) \(\therefore \quad \eta =2\) or \(1\).
Solution : The matrix form of the system is given by
\[ \left (\begin {array}{lll} 1 & 2 & 1 \\ 4 & 6 & 5 \\ 2 & 2 & 3 \end {array}\right )\left (\begin {array}{l} x \\ y \\ z \end {array}\right )=\left (\begin {array}{c} 11 \\ 8 \\ 19 \end {array}\right ) \]
\(\therefore \) The augmented matrix is given by
\(\seteqnumber{0}{4.}{0}\)\begin{align*} (A, B) & =\left (\begin{array}{cccc} 1 & 2 & 1 & 11 \\ 4 & 6 & 5 & 8 \\ 2 & 2 & 3 & 19 \end {array}\right ) \\ & \sim \begin{pNiceArray}[last-col]{cccc} 1 & 2 & 1 & 11 \\ 0 & -2 & 1 & -36 &R_{2} \rightarrow R_{2}-4 R_{1} \\ 0 & -2 & 1 & -3&R_{3} \rightarrow R_{3}-2 R_{1} \end {pNiceArray}\\ & \sim \begin{pNiceArray}[last-col]{cccc} 1 & 2 & 1 & 11 \\ 0 & -2 & 1 & -36 \\ 0 & 0 & 0 & 33 & R_{3} \rightarrow -R_{3}-R_{2} \end {pNiceArray} \end{align*} Rank of \(A=2 \) and rank of \((A, B)=3 \). The given system is inconsistent.
4.5 Characteristic Equation And Cayley Hamilton Theorem
Example 4.5.2. If \(\left (\begin {array}{ll} 1 & 2 \\ 0 & 3 \end {array}\right )+\left (\begin {array}{ll} 1 & 1 \\ 2 & 1 \end {array}\right ) x+ \left (\begin {array}{ll} 2 & 0 \\ 3 & 1 \end {array}\right ) x^{2}\) is a matrix polynomial of degree 2 and it is simply the matrix \(\left (\begin {array}{cc}1+x+2 x^{2} & 2+x \\ 2 x+3 x^{2} & 3+x+x^{2}\end {array}\right )\).
Definition 4.5.3. Let \(A\) be any square matrix of order \(n\) and let \(I\) be the identity matrix of order \(n\). Then the matrix polynomial given by \(A-x I\) is called the characteristic matrix of \(A\).
The determinant \(|A-x I|\) which is an ordinary polynomial in \(x\) of degree \(n\) is called the characteristic polynomial of \(A\).
The equation \(|A-x I|=0\) is called the characteristic equation of \(A\).
Solution : Then the characteristic matrix of \(A\) is \(A-x l\) given by
\(\seteqnumber{0}{4.}{0}\)\begin{align*} A-x l &=\left (\begin{array}{ll} 1 & 2 \\ 3 & 4 \end {array}\right )-x\left (\begin{array}{ll} 1 & 0 \\ 0 & 1 \end {array}\right ) \\ &=\left (\begin{array}{cc} 1-x & 2 \\ 3 & 4-x \end {array}\right ) \end{align*} The characteristic polynomial of \(A\) is
\(\seteqnumber{0}{4.}{0}\)\begin{align*} |A-xI | & = \begin{vmatrix} 1-x & 2 \\ 3 & 4-x \end {vmatrix}\\ &=(1-x)(4-x)-6 \\ &=x^{2}-5 x-2 \end{align*} \(\therefore \) The characteristic equation of \(A\) is \(|A-x I|=0\). \(\therefore x^{2}-5 x-2=0\) is the characteristic equation of \(A\),
Solution : The characteristic matrix of \(A\) is \(A-x I\) given by
\[ A-x I=\left (\begin {array}{ccc} 1-x & 0 & 2 \\ 0 & 1-x & 2 \\ 1 & 2 & -x \end {array}\right ) \]
The characteristic polynomial of \(A\) is
\(\seteqnumber{0}{4.}{0}\)\begin{align*} |A-x I| &=\left |\begin{array}{ccc} 1-x & 0 & 2 \\ 0 & 1-x & 2 \\ 1 & 2 & -x \end {array}\right | \\ &=(1-x)[(1-x)(-x)-4]-2(1-x)] \\ &=-x(1-x)^{2}-4(1-x)-2+2 x \\ &=-x^{3}+2 x^{2}-x-4+4 x-2+2 x \\ &=-x^{3}+2 x^{2}+5 x-6 \end{align*} The characteristic equation of \(A\) is
\(\seteqnumber{0}{4.}{0}\)\begin{align*} -x^{3}+2 x^{2}+5 x-6&=0 \\ \text { (i.e) } x^{3}-2 x^{2}-5 x+6&=0 \end{align*}
Proof : Let \(A\) be a square matrix of order \(n\).
\(\seteqnumber{0}{4.}{0}\)\begin{equation} \label {p726eq1} \text { Let }|A-x l|=a_{0}+a_{1} x+a_{2} x^{2}+\cdots +a_{n} x^{n} \end{equation}
be the characteristic polynomial of \(A\).
Now, \(\operatorname {adj}(A-x I)\) is a matrix polynomial of degree \(n-1\) since each entry of the matrix \(adj(A-x I)\) is a cofactor of \(A-x I\) and hence is a polynomial of degree \(\leq n-1\).
\(\seteqnumber{0}{4.}{1}\)\begin{equation} \text { Let } \operatorname {adj}(A-x I)=B_{0}+B_{1} x+B_{2} x^{2}+\cdots +B_{n-1} x^{n-1} \label {p726eq2} \end{equation}
Now,
\(\seteqnumber{0}{4.}{2}\)\begin{align*} (A-x I) \operatorname {adj}(A-x I)&=|A-x I| I . \quad (\because \quad (\operatorname {adj} A) A=A(\operatorname {adj} A)=|A| I) \\ \therefore \quad (A-x I)\left (B_{0}+B_{1} x+\cdots +B_{n-1} x^{n-1}\right ) &=\left (a_{0}+a_{1} x+\cdots +a_{n} x^{n}\right ) I \text { using } \eqref {p726eq1} \text { and }\eqref {p726eq2} . \end{align*} Equating the coefficients of the corresponding powers of \(x\) we get
\(\seteqnumber{0}{4.}{2}\)\begin{align*} A B_{0}=& a_{0} I \\ A B_{1}-B_{0}=& a_{1} I \\ A B_{2}-B_{1}=& a_{2} I \\ \ldots & \ldots . . . . \\ \ldots & \ldots . \\ AB_{n-1}-B_{n-2}=& a_{n-1} I \\ -B_{n-1}=& a_{n} I \end{align*} Pre-multiplying the above equations by \(I, A, A^{2}, \ldots \) \(A^{n}\) respectively and adding we get
\[ a_{0} I+a_{1} A+a_{2} A^{2}+\cdots +a_{n} A^{n}=0 \]
□
Note 4.5.7. The inverse of a non-singular matrix can be calculated by using the Cayley Hamilton theorem as follows.
Let \(a_{0}+a_{1} x+a_{2} x^{2}+\cdots +a_{n} x^{n}\) be the characteristic polynomial of \(A\)
Then by Theorem 4.5.6, we have
\(\seteqnumber{0}{4.}{2}\)\begin{equation} a_{0} I+a_{1} A+a_{2} A^{2}+\cdots +a_{n} A^{n}=0 \label {p726eq3} \end{equation}
Since \(|A-xI|=a_{0}+a_{1} x+a_{2} x^{2}+\cdots +a_{n} x^{n}\) we get \(a_{0}=|A|\) (by putting \(x=0\) ).
\(\seteqnumber{0}{4.}{3}\)\begin{align*} \therefore \quad a_{0} & \neq 0 (\because A \text { is a non singular matrix. }) \\ \therefore \quad I & =-\frac {1}{a_{0}}\left [a_{1} A+a_{2} A^{2}+\cdots +a_{n} A^{n}\right ] \text { by \eqref {p726eq3} }\\ {A^{-1}&=-\frac {1}{a_{0}}\left [a_{1} I+a_{2} A+\cdots +a_{n} A^{n-1}\right ] \end{align*}
Solved Problems
Solution : The characteristic equation of the matrix \(A\) is given by \(|A- \lambda I|=0\).
\(\seteqnumber{0}{4.}{3}\)\begin{align*} \begin{vmatrix}8-\lambda &-6&2\\ -6&7-\lambda &-4\\ 2&-4&3-\lambda \end {vmatrix} &=0 \\ \left (8-\lambda \right ) \begin{vmatrix}7-\lambda &-4\\ -4&3-\lambda \end {vmatrix}-\left (-6\right ) \begin{vmatrix}-6&-4\\ 2&3-\lambda \end {vmatrix}+2\cdot \begin{vmatrix}-6&7-\lambda \\ 2&-4\end {vmatrix} & =0 \\ \left (8-\lambda \right )[\left (7-\lambda \right )\left (3-\lambda \right )-\left (-4\right )\left (-4\right )] -\left (-6\right )[\left (-6\right )\left (3-\lambda \right )-\left (-4\right )\cdot 2] & \\ \quad +2[ \left (-6\right )\left (-4\right )-\left (7-\lambda \right )\cdot 2] &= 0 \\ \left (8-\lambda \right )\left (\lambda ^2-10\lambda +5\right )-\left (-6\right )\left (6\lambda -10\right )+2\left (2\lambda +10\right ) &= 0 \\ -\lambda ^3+18\lambda ^2-85\lambda +40+36\lambda -60+4\lambda +20 &= 0 \\ -\lambda ^3+18\lambda ^2-45\lambda &= 0 \\ \lambda ^3-18\lambda ^2+45\lambda &= 0 \end{align*} ie., \(\lambda ^3-18\lambda ^2+45\lambda = 0\), which represents the characteristic equation of \(A\).
Solution : The characteristic polynomial of \(A\) is
\(\seteqnumber{0}{4.}{3}\)\begin{align*} |A- \lambda I | & = \begin{vmatrix} 1-x & 2 \\ 3 & 1-x \end {vmatrix} \\ & =\left (1-\lambda \right )\left (1-\lambda \right )-2\cdot 3 \\ &=\lambda ^2-2\lambda -5 \end{align*} By Cayley-Hamiltonian theorem
\(\seteqnumber{0}{4.}{3}\)\begin{align*} A^2 -2A -5I&=0\\ I & = \dfrac {1}{5} (A^2 -2A) \\ A^{-1} & = \dfrac {1}{5} (A -2I)\\ & = \dfrac {1}{5} \left [\begin{pmatrix} 1& 2 \\ 3 & 1 \end {pmatrix} - 2 \begin{pmatrix} 1& 0 \\ 0 & 1 \end {pmatrix} \right ] \\ A^{-1}& = \dfrac {1}{5} \begin{pmatrix} -1& 2 \\ 3 & -1 \end {pmatrix} \end{align*}
Solution : The characteristic polynomial of \(A\) is
\(\seteqnumber{0}{4.}{3}\)\begin{align*} |A-\lambda I|&=0\\ \begin{vmatrix}2-\lambda &-3&1\\ 3&1-\lambda &3\\ -5&2&-4-\lambda \end {vmatrix}&=0 \\ \left (2-\lambda \right ) \begin{vmatrix}1-\lambda &3\\ 2&-4-\lambda \end {vmatrix}-\left (-3\right ) \begin{vmatrix}3&3\\ -5&-4-\lambda \end {vmatrix}+1\cdot \begin{vmatrix}3&1-\lambda \\ -5&2\end {vmatrix}&=0 \\ \left (2-\lambda \right )\left (\lambda ^2+3\lambda -10\right )-\left (-3\right )\left (-3\lambda +3\right )+1\cdot \left (-5\lambda +11\right )&=0\\ -\lambda ^3-\lambda ^2+16\lambda -20-9\lambda +9-5\lambda +11&=0 \\ -\lambda ^3-\lambda ^2+2\lambda &=0 \end{align*} By Cayley-Hamilton Theorem
\(\seteqnumber{0}{4.}{3}\)\begin{align*} -A^3 -A^2 +2A&=0\\ A^3 +A^2 -2A&=0\\ A(A^2 +A-2I)&=0\\ A(A+2I)(A-I)&=0 \end{align*}
Solution : Let \(A =\left [\begin {array}{rrr} 7 & 2 & -2 \\ -6 & -1 & 2 \\ 6 & 2 & -1 \end {array}\right ] \).
The characteristic polynomial of \(A=|A-xI|\)
\(\seteqnumber{0}{4.}{3}\)\begin{align*} |A-xI|=&\left |\begin{array}{ccc} 7-x & 2 & -2 \\ -6 & -1-x & 2 \\ 6 & 2 & -1-x \end {array}\right | \\ &=(7-x)\left [(1+x)^{2}-4\right ]-2[6(1+x)-12] -2[-12+6(1+x)]\\ &=(7-x)\left (x^{2}+2 x-3\right )-12(x-1)-12(x-1)\\ &= 7 x^{2}+14 x-21-x^{3}-2 x^{2}+3 x-12 x+12 -12 x+12 \\ &=-x^{3}+5 x^{2}-7 x+3 \end{align*} By Cayley-Hamilton theorem
\(\seteqnumber{0}{4.}{3}\)\begin{align*} -A^{3}+5 A^{2}-7 A+3 I_{3}&={0} \\ A^{3}-5 A^{2}+7 A-3 I_{3}&=0 \\ 3 I_{3}&=A^{3}-5 A^{2}+7 A \\ I_{3}&=\frac {1}{3}\left (A^{3}-5 A^{2}+7 A\right ) \end{align*} Pre (or post) multiplying by \(A^{-1}\) on both sides we get
\(\seteqnumber{0}{4.}{3}\)\begin{equation} \label {p728eq1} A^{-1}=\frac {1}{3}\left [A^{2}-5 A+7 I_{3}\right ] \end{equation}
Now,
\(\seteqnumber{0}{4.}{4}\)\begin{align*} A^{2}&=\left (\begin{array}{rrr} 7 & 2 & -2 \\ -6 & -1 & 2 \\ 6 & 2 & -1 \end {array}\right )\left (\begin{array}{rrr} 7 & 2 & -2 \\ -6 & -1 & 2 \\ 6 & 2 & -1 \end {array}\right ) \\ & =\begin{pmatrix}7\cdot \:7+2\left (-6\right )+\left (-2\right )\cdot \:6&7\cdot \:2+2\left (-1\right )+\left (-2\right )\cdot \:2&7\left (-2\right )+2\cdot \:2+\left (-2\right )\left (-1\right )\\ \left (-6\right )\cdot \:7+\left (-1\right )\left (-6\right )+2\cdot \:6&\left (-6\right )\cdot \:2+\left (-1\right )\left (-1\right )+2\cdot \:2&\left (-6\right )\left (-2\right )+\left (-1\right )\cdot \:2+2\left (-1\right )\\ 6\cdot \:7+2\left (-6\right )+\left (-1\right )\cdot \:6&6\cdot \:2+2\left (-1\right )+\left (-1\right )\cdot \:2&6\left (-2\right )+2\cdot \:2+\left (-1\right )\left (-1\right )\end {pmatrix} \\ &=\left (\begin{array}{rrr} 25 & 8 & -8 \\ -24 & -7 & 8 \\ 24 & 8 & -7 \end {array}\right ) \end{align*} From (4.4)
\(\seteqnumber{0}{4.}{4}\)\begin{align*} A^{-1}&=\frac {1}{3}\left [\left (\begin{array}{rrr} 25 & 8 & -8 \\ -24 & -7 & 8 \\ 24 & 8 & -7 \end {array}\right ) -\left [\begin{array}{rrr} 35 & 10 & -10 \\ -30 & -5 & 10 \\ 30 & 10 & -5 \end {array}\right ] +\left [\begin{array}{rrr} 7 & 0 & 0 \\ 0 & 7 & 0 \\ 0 & 0 & 7 \end {array}\right ]\right ] \\ A^{-1}&=\frac {1}{3}\left (\begin{array}{rrr} -3 & -2 & 2 \\ 6 & 5 & -2 \\ -6 & -2 & 5 \end {array}\right ) \end{align*}
Solution : The characteristic polynomial of \(A\) is
\(\seteqnumber{0}{4.}{4}\)\begin{align*} |A-x I| &=\begin{vmatrix} 3-x & 3 & 4 \\ 2 & -3-x & 4 \\ 0 & -1 & 1-x \end {vmatrix}\\ &= \left (3-x\right )\left (x^2+2x+1\right )-3\cdot \:2\left (-x+1\right )+4\left (-2\right )\\ &= -x^3+x^2+5x+6x+3-6-8\\ &=-x^{3}+x^{2}+11 x-11 \end{align*} By Cayley-Hamilton theorem
\(\seteqnumber{0}{4.}{4}\)\begin{align*} A^{3}+A^{2}+11 A-11 I_{3}&=0 . \\ \therefore \quad A^{3}-A^{2}-11 A+11 I_{3}&=0 . \\ 11 I_{3}&=-\left (A^{3}-A^{2}-11 A\right ) . \\ \therefore \quad I_{3}&=-\frac {1}{11}\left [A^{3}-A^{2}-11 A\right ] . \end{align*} Pre (post) multiplying by \(A^{-1}\) on both sides we get
\(\seteqnumber{0}{4.}{4}\)\begin{align*} A^{-1} & =-\frac {1}{11}\left [A^{2}-A-11 I_{3}\right ] \\ & =-\frac {1}{11} \left [ \begin{pmatrix} 15 & -4 & 28 \\ 0 & 11 & 0 \\ -2 & 2 & -3 \end {pmatrix} - \begin{pmatrix} 3 & 3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end {pmatrix} -11 \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end {pmatrix} \right ] \\ &=\begin{pmatrix} -\frac {1}{11} & \frac {7}{11} & -\frac {24}{11} \\ \frac {2}{11} & -\frac {3}{11} & \frac {4}{11} \\ \frac {2}{11} & -\frac {3}{11} & \frac {15}{11} \end {pmatrix} \end{align*}
Solution : The characteristic equation of \(A\) is
\(\seteqnumber{0}{4.}{4}\)\begin{align*} |A-\lambda l|&=0 . \\ \therefore \left |\begin{array}{cc} 1-\lambda & 2 \\ 4 & 3-\lambda \end {array}\right |&=0 \\ \therefore \quad (1-\lambda )(3-\lambda )-8 &=0\\ \lambda ^{2}-4 \lambda -5 &=0 \end{align*} By Cayley Hamilton’s theorem \(A\) satisfies its characteristic equation. We have
\(\seteqnumber{0}{4.}{4}\)\begin{align*} A^2 - 4A - 5I &=0 \\ A^2 &= \begin{pmatrix} 1&2 \\ 4 & 3 \end {pmatrix}\begin{pmatrix} 1&2 \\ 4 & 3 \end {pmatrix} = \begin{pmatrix} 9&8\\16&17 \end {pmatrix}\\ 4A&= 4 \begin{pmatrix} 1&2 \\ 4 & 3 \end {pmatrix} = \begin{pmatrix} 4 &8 \\ 16& 12 \end {pmatrix} \\ 5I & = 5 \begin{pmatrix} 1&0 \\0&1 \end {pmatrix} =\begin{pmatrix} 5&0 \\0&5 \end {pmatrix}\\ A^2 - 4A - 5I & = \begin{pmatrix} 9&8\\16&17 \end {pmatrix} - \begin{pmatrix} 4 &8 \\ 16& 12 \end {pmatrix} - \begin{pmatrix} 5&0 \\0&5 \end {pmatrix} \\ & = \begin{pmatrix} 0&0 \\0 &0 \end {pmatrix} =0 \end{align*} Thus Cayley-Hamilton theorem is verified.
Solution : The characteristic equation of \(A\) is
\(\seteqnumber{0}{4.}{4}\)\begin{align*} |A-\lambda I | &=0 \\ \begin{vmatrix} 1 -\lambda & 0 & -2 \\ 2 & 2 -\lambda & 4 \\ 0 & 0 & 2 - \lambda \end {vmatrix}&=0\\ (2 -\lambda ) (2 - \lambda ) (1-\lambda )&=0 \\ \lambda ^{3}- 5 \lambda ^{2}+8 \lambda -4 &=0 \end{align*} By Cayley-Hamilton’s theorem
\(\seteqnumber{0}{4.}{4}\)\begin{align} A^{3}-5 A^{2}+8 A-4 I&=0 \label {p729eq1} \\ 4I &= A^3 -5A^2 +8A \notag \end{align} (i) To find \(A^{-1}\) pre multiplying by \(A^{-1}\) we get
\(\seteqnumber{0}{4.}{5}\)\begin{align} 4 A^{-1} &=A^{-1} A^{3}-5 A^{-1} A^{2}+8 A^{-1} A\notag \\ &=A^{2}-5 A+8 I \notag \\ \therefore A^{-1} &=\frac {1}{4}\left [A^{2}-5 A+8 I\right ] \label {p729eq2} \end{align} Now,
\(\seteqnumber{0}{4.}{6}\)\begin{align*} A^{2} &=\left (\begin{array}{rrr} 1 & 0 & -2 \\ 2 & 2 & 4 \\ 0 & 0 & 2 \end {array}\right )\left (\begin{array}{rrr} 1 & 0 & -2 \\ 2 & 2 & 4 \\ 0 & 0 & 2 \end {array}\right ) \\ &=\left (\begin{array}{rrr} 1 & 0 & -6 \\ 6 & 4 & 12 \\ 0 & 0 & 4 \end {array}\right ) \end{align*} From (4.6)
\(\seteqnumber{0}{4.}{6}\)\begin{align*} A^{-1}&= \frac {1}{4}\left [\left (\begin{array}{rrr} 1 & 0 & -6 \\ 6 & 4 & 12 \\ 0 & 0 & 4 \end {array}\right ) -\left (\begin{array}{rrr} 5 & 0 & -10 \\ 10 & 10 & 20 \\ 0 & 0 & 10 \end {array}\right )+\left (\begin{array}{lll} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end {array}\right )\right ]\\ &=\frac {1}{4}\left (\begin{array}{rrr} 4 & 0 & 4 \\ 4 & 2 & -8 \\ 0 & 0 & 2 \end {array}\right ) \\ A^{-1}&=\left (\begin{array}{rrr} 1 & 0 & 1 \\ 1 & \frac {1}{2} & -2 \\ 0 & 0 & -\frac {1}{2} \end {array}\right ) \end{align*} (ii) To find \(A^{4}\). From (4.5)
\(\seteqnumber{0}{4.}{6}\)\begin{align*} A^{3} &=5 A^{2}-8 A+4 I \\ \therefore A^{4}&=5 A^{3}-8 A^{2}+4 A \\ &=5\left [5 A^{2}-8 A+4 I\right ]-8 A^{2}+4 A \text { by } \eqref {p729eq1} \\ &=17 A^{2}-36 A+20 I\\ & = 17 \left (\begin{array}{rrr} 1 & 0 & -6 \\ 6 & 4 & 12 \\ 0 & 0 & 4 \end {array}\right ) - 36 \begin{pmatrix} 1 & 0 & -2 \\ 2 & 2 & 4 \\ 0 & 0 & 2 \end {pmatrix} +20 \begin{pmatrix} 1 & 0 & 0 \\0& 1 & 0 \\0& 0 & 1 \end {pmatrix} \\ &=\left (\begin{array}{rrr} 17 & 0 & -102 \\ 102 & 68 & 204 \\ 0 & 0 & 68 \end {array}\right ) - \left (\begin{array}{rrr} 36 & 0 & -72 \\ 72 & 72 & 144 \\ 0 & 0 & 72 \end {array}\right ) +\left (\begin{array}{rrr} 20 & 0 & 0 \\ 0 & 20 & 0 \\ 0 & 0 & 20 \end {array}\right ) \\ A^{4}&=\left (\begin{array}{rrr} 1 & 0 & -30 \\ 30 & 16 & 60 \\ 0 & 0 & 16 \end {array}\right ) \end{align*}
4.6 Eigen Values And Eigen Vectors
Definition 4.6.1. Let \(A\) be an \(n \times n\) matrix. A number \(\lambda \) is called an eigen value of \(A\) if there exists a nonzero vector \(X=\left (\begin {array}{c}x_{1} \\ x_{2} \\ \vdots \\ x_{n}\end {array}\right )\) such that \(A X=\lambda X\) and \(X\) is called an eigen vector corresponding to the eigen value \(\lambda \).
Remark 4.6.3. Let \(X\) be an eigen vector corresponding to the eigen value \(\lambda \) of \(A\). Then \(A X=\lambda X\) so that \((A-\lambda I) X=0\). Thus \(X\) is a non-trivial solution of the system of homogeneous linear equations \((A-\lambda I) X=0\). Hence \(|A-\lambda I|=0\), which is the characteristic polynomial of \(A \) . Let \(|A-\lambda I|=a_{0} \lambda ^{n}+a_{1} \lambda ^{n-1}+\cdots +a_{n}\). The roots of this polynomial give the eigen values of \(A\). Hence eigen values are also called characteristic roots.
4.6.1 Properties of Eigen Values
Proof : By definition \(X \neq 0, A X=\lambda _{1} X\) and \(A X=\lambda _{2} X\).
\(\seteqnumber{0}{4.}{6}\)\begin{align*} \lambda _{1} X & =\lambda _{2} X \\ \left (\lambda _{1}-\lambda _{2}\right ) X&=0 \end{align*} Since \(X \neq 0, \quad \lambda _{1}=\lambda _{2}\). □
Proof : (i) Let \(A=\left (\begin {array}{cccc}a_{11} & a_{12} & \cdots & a_{1 n} \\ a_{21} & a_{22} & \cdots & a_{2 n} \\ \vdots & \vdots & \vdots & \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n n}\end {array}\right )\). The eigenvalues of \(A\) are the roots of the characteristic equation
\(\seteqnumber{0}{4.}{6}\)\begin{align} |A-\lambda I| & =\left |\begin{array}{cccc} a_{11}-\lambda & a_{12} & \cdots & a_{1 n} \\ a_{21} & a_{22}-\lambda & \cdots & a_{2 n} \\ \vdots & \vdots & \vdots & \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n n}-\lambda \end {array}\right |=0 \label {p731eq1} \end{align} Let,
\(\seteqnumber{0}{4.}{7}\)\begin{align} |A-\lambda I|& =a_{0} \lambda ^{n}+a_{1} \lambda ^{n-1}+\cdots +a_{n} \ldots \label {p731eq2} \end{align} From (4.7) and (4.8) we get
\(\seteqnumber{0}{4.}{8}\)\begin{equation} a_{0}=(-1)^{n} ; a_{1}=(-1)^{n-1}\left (a_{11}+a_{22}+\cdots +a_{n n}\right ) \label {p731eq3} \end{equation}
Also by putting \(\lambda =0\) in (4.8), we get \(a_{n}=|A|\). Now let \(\lambda _{1}, \lambda _{2}, \ldots , \lambda _{n}\) be the eigen values of \(A\). \(\therefore \lambda _{1}, \lambda _{2}, \ldots , \lambda _{n}\) are the roots of (4.8).
\(\seteqnumber{0}{4.}{9}\)\begin{align*} \therefore \quad \lambda _{1}+\lambda _{2}+\cdots +\lambda _{n} & =-\frac {a_{1}}{a_{0}} \\ &=a_{11}+a_{22}+\cdots +a_{n n}(\text { using } \eqref {p731eq3} ) \\ \text {Sum of the eigen values} & =\text { trace.of } A \end{align*} (ii) Product of the eigen values = product of the roots
\(\seteqnumber{0}{4.}{9}\)\begin{align*} \text {Product of the eigen values} &=\lambda _{1} \lambda _{2} \cdots \lambda _{n} \\ &=(-1)^{n} \frac {a_{n}}{a_{0}}=\frac {(-1)^{n} a_{n}}{(-1)^{n}} \\ &=a_{n} \\ &=|A| . \end{align*} □
Proof : It is enough if we prove that \(A\) and \(A^{T}\) have the same characteristic polynomial. Since for any square matrix \(M,|M|=\left |M^{T}\right |\) we have,
\(\seteqnumber{0}{4.}{9}\)\begin{align*} |A-\lambda I|=\left |(A-\lambda I)^{T}\right | &=\left |A^{T}-(\lambda I)^{T}\right | \\ &=\left |A^{T}-\lambda I\right | . \end{align*} Hence the result. □
Proof : Let \(X\) be an eigen vector corresponding to \(\lambda \).
Then \(A X=\lambda X .\) Since \(A\) is non singular \(A^{-1}\) exists.
\begin{align*} \therefore \quad A^{-1}(A X) &=A^{-1}(\lambda X) \\ I X &=\lambda A^{-1} X\\ \therefore \quad A^{-1} X &=\left (\frac {1}{\lambda }\right ) X \end{align*} \(\therefore \quad \frac {1}{\lambda }\) is an eigen value of \(A^{-1}\) □
Proof : Let \(X\) be an eigen vector corresponding to \(\lambda \). Then
\(\seteqnumber{0}{4.}{9}\)\begin{equation} A X=\lambda X \label {prop5eq1} \end{equation}
Now,
\(\seteqnumber{0}{4.}{10}\)\begin{align*} (k A) X= & k(A X)\\ &=k(\lambda X) \quad (\text { by } \eqref {prop5eq1}) \\ &=(k \lambda ) X . \end{align*} \(\therefore \quad k \lambda \) is an eigen value of \(k A\). □
Proof : Let \(X\) be an eigen vector corresponding to \(\lambda \). Then
\(\seteqnumber{0}{4.}{10}\)\begin{equation} A X=\lambda X \label {prop6eq1} \end{equation}
Now,
\(\seteqnumber{0}{4.}{11}\)
\begin{align*}
A^{2} X &=(A A) X=A(A X) \\ &=A(\lambda X) \quad (\text { by } \eqref {prop6eq1} ) \\ &=\lambda (A X) \\ &=\lambda (\lambda X) \quad \text { (by } \eqref {prop6eq1} ) \\ &=\lambda ^{2} X .
\end{align*}
\(\lambda ^{2} \) is an eigen value of \(A^{2}\).
Proceeding like this we can prove that \(\lambda ^{k}\) is an eigen value of \(A^{k}\) for any positive integer. □
Proof : Let \(\lambda _{1}, \lambda _{2}, \ldots , \lambda _{k}\) be distinct eigen values of a matrix and let \(X_{i}\) be the eigen vector corresponding to \(\lambda _{i}\). Hence
\(\seteqnumber{0}{4.}{11}\)\begin{equation} \label {prop7eq1} A X_{i}=\lambda _{i} X_{i} \quad (i=1,2, \ldots , k) \end{equation}
Now, suppose \(X_{1}, X_{2}, \ldots , X_{k}\) are linearly dependent. Then there exist real numbers \(\alpha _{1}, \alpha _{2}, \ldots , \alpha _{k}\), not all zero, such that \(\alpha _{1} X_{1}+\alpha _{2} X_{2}+\cdots +\alpha _{k} X_{k}=0\). Among all such relations, we choose one of shortest length, say \(j\).
By rearranging the vectors \(X_{1}, X_{2}, \ldots , X_{k}\) we may assume that
\(\seteqnumber{0}{4.}{12}\)\begin{equation} \label {prop7eq2} \alpha _{1} X_{1}+\alpha _{2} X_{2}+\cdots +\alpha _{j} X_{j}=0 \end{equation}
\(\seteqnumber{0}{4.}{13}\)\begin{align} A\left (\alpha _{1} X_{1}\right )+A\left (\alpha _{2} X_{2}\right )+\cdots +A\left (\alpha _{j} X_{j}\right ) & =0 \notag \\ \alpha _{1}\left (A X_{1}\right )+\alpha _{2}\left (A X_{2}\right )+\cdots +\alpha _{j}\left (A X_{j}\right )&=0\notag \\ \alpha _{1} \lambda _{1} X_{1}+\alpha _{2} \lambda _{2} X_{2}+\cdots +\alpha _{j} \lambda _{j} X_{j} & =0 \label {prop7eq3} \end{align} Multiplying (4.13) by \(\lambda _{1}\) and subtracting from (4.14), we get
\(\seteqnumber{0}{4.}{14}\)\begin{align} \alpha _{2}\left (\lambda _{1}-\lambda _{2}\right ) X_{2}+& \alpha _{3}\left (\lambda _{1}-\lambda _{3}\right ) X_{3}+\cdots +\alpha _{j}\left (\lambda _{1}-\lambda _{j}\right ) X_{j}=0 \label {prop7eq4} \end{align} and since \(\lambda _{1}, \lambda _{2}, \ldots , \lambda _{j}\) are distinct and \(\alpha _{2}, \ldots , \alpha _{j}\) are non-zero we have
\[ \alpha _{i}\left (\lambda _{1}-\lambda _{i}\right ) \neq 0 ; \quad i=2,3, \ldots , j . \]
Thus (4.15) gives a relation whose length is \(j-1\), giving a contradiction. Hence \(X_{1}, X_{2}, \ldots , X_{k}\) are linearly independent. □
Proof : Let \(A\) be a Hermitian matrix. Hence
\(\seteqnumber{0}{4.}{15}\)\begin{equation} A=\overline {A}^{T} \label {prop8eq1} \end{equation}
Let \(\lambda \) be a characteristic root of \(A\) and let \(X\) be a characteristic vector corresponding to \(\lambda \).
\(\seteqnumber{0}{4.}{16}\)\begin{equation} A X=\lambda X \label {prop8eq2} \end{equation}
Now,
\(\seteqnumber{0}{4.}{17}\)\begin{align*} A X &= \lambda X \\ \Rightarrow \overline {X}^{T} A X &= \lambda \overline {X}^{T} X \\ \Rightarrow \left (\overline {X}^{T} A X\right )^{T}&= \lambda \overline {X}^{T} X \quad \left (\text { since } X^{T} A X\right . \text { is a $1 \times 1$ matrix)}\\ \Rightarrow X^{T} A^{T}\left (\overline {X}^{T}\right )^{T}&= \lambda \overline {X}^{T} X \\ \Rightarrow X^{T} A^{T} \overline {X}&= \lambda \overline {X}^{T} X \\ \Rightarrow \overline {X^TA^T} \overline {X} &= \overline {\lambda X^T X } \\ \Rightarrow \overline {X}^{T} \overline {A}^{T} X&= \overline {\lambda X^{T} X} \\ \Rightarrow \overline {X}^{T} A X&= \overline {\lambda } X^{T} \overline {X} \quad \text { (using \eqref {prop8eq1} ) } \\ \Rightarrow \overline {X}^{T} \lambda X&= \overline {\lambda } X^{T} \overline {X} \quad \text { (using \eqref {prop8eq2} ) } \end{align*}
\(\seteqnumber{0}{4.}{17}\)\begin{equation} \Rightarrow \lambda \left (\overline {X}^{T} X\right )= \overline {\lambda }\left (X^{T} \overline {X}\right ) \label {prop8eq3} \end{equation}
Now,
\(\seteqnumber{0}{4.}{18}\)\begin{align*} \overline {X}^{T} X &=X^{T} \overline {X}=\overline {x_{1}} x_{1}+\overline {x_{2}} x_{2}+\ldots \ldots +\overline {x_{n}} x_{n} \\ &=\left |x_{1}\right |^{2}+\left |x_{2}\right |^{2}+\ldots \ldots +\left |x_{n}\right |^{2} \\ &\neq 0 \end{align*} From (4.18) we get \(\lambda =\overline {\lambda }\). Hence \(\lambda \) is real. □
Proof : We know that any real symmetric matrix is Hermitian. Hence the result follows from the above property. □
Proof : Let \(A\) be a skew Hermitian matrix and \(\lambda \) be a characteristic root of \(A\).
\(\seteqnumber{0}{4.}{18}\)\begin{align*} \therefore |A-\lambda I| & =0 \\ &\therefore |i A-i \lambda I|& =0 \end{align*} \(\therefore i \lambda \) is a characteristic root of \(i A .\)
Since \(A\) is skew Hermitian \(i A\) is Hermitian. \(\therefore \, i \lambda \) is real. Hence \(\lambda \) is purely imaginary or zero. □
Proof : We know that any real skew symmetric matrix is skew Hermitian. Hence the result follows from the above property. □
Proof : Let \(\lambda \) be a characteristic root of an unitary matrix \(A\) and \(X\) be a characteristic vector corresponding to \(\lambda \),
\(\seteqnumber{0}{4.}{18}\)\begin{equation} AX = \lambda X \label {prop10eq1} \end{equation}
Taking conjugate and transpose in (4.19) we get \((\overline {A X})^{T}=(\overline {\lambda X})^{T}\).
\(\seteqnumber{0}{4.}{19}\)\begin{equation} \therefore \overline {X}^{T} \overline {A}^{T}=\overline {\lambda } \overline {X}^{T} \label {prop10eq2} \end{equation}
Multiplying (4.19) and (4.20), we get
\(\seteqnumber{0}{4.}{20}\)
\begin{align*}
\left (\overline {X^{T} A^{T}}\right )(A X)&=\left (\overline {\lambda } \overline {X}^{T}\right )(\lambda X) \\ \therefore \overline {X}^{T}\left (\overline {A}^{T} A\right ) X &=\overline {\lambda } \lambda \left (\overline
{X}^{T} X\right )
\end{align*}
Now, since \(A\) is an unitary matrix \(\overline {A}^{T} A=I\).
Hence \(\overline {X}^{T} X=(\overline {\lambda } \lambda ) \overline {X}^{T} X\).
Since \(X\) is non-zero vector \(\overline {X}^{T}\) is also non-zero vector and \(\overline {X}^{T} X=\left |x_{1}\right |^{2}+\left |x_{2}\right |^{2}+\ldots \ldots +\left |x_{n}\right |^{2} \neq 0\) we get \(\lambda \overline {\lambda }=1\).
Hence \(|\lambda |^{2}=1\). Hence \(|\lambda |=1\). □
Since any orthogonal matrix-is unitary the result follows from Property 10.
Proof : The eigen values of \(A\) are the roots of the characteristic equation \(|A-\lambda I|=0 .\) Now, 0 is an eigen value of \(A \Leftrightarrow |A-0 I|=0\)
\(\seteqnumber{0}{4.}{20}\)\begin{align*} \Leftrightarrow |A| =0 \\ \end{align*} \(\Leftrightarrow A\) is a singular matrix □
Proof : Let \(\lambda \) be an eigen value of \(A B\) and \(X\) be an eigen vector corresponding to \(\lambda \).
\(\seteqnumber{0}{4.}{20}\)
\begin{align*}
(A B) X & =\lambda X \\ B(A B) X& =B(\lambda X)=\lambda (B X) \\ (B A)(B X)&=\lambda (B X) \\ (B A) Y&=\lambda Y \text { where } Y=B X .
\end{align*}
Hence \(\lambda \) is an eigen value of \(B A\).
Also \(B X\) is the corresponding eigen vector. □
Proof : Let \(B=P^{-1} A P\).
To prove \(A\) and \(B\) have same eigen values, it is enough to prove that the characteristic polynomials of \(A\) and \(B\) are the same. Now
\begin{align*} |B-\lambda I| &=\left |P^{-1} A P-\lambda I\right | \\ &=\left |P^{-1} A P-P^{-1}(\lambda I) P\right | \\ &=\left |P^{-1}(A-\lambda I) P\right | \\ &=\left |P^{-1}\right ||A-\lambda I||P| \\ &=\left |P^{-1}\right ||P||A-\lambda I| \\ &=\left |P^{-1} P \| A-\lambda I\right | \\ &=|I||A-\lambda I| \\ &=|A-I \lambda | . \end{align*} The characteristic equations of \(A\) and \(P^{-1} A P\) are the same. □
Proof : Let \(f(x)=a_{0} x^{n}+a_{1} x^{n-1}+\ldots + a_{n-1} x+a_{n}\) where \(a_{0} \neq 0\) and \(a_{1}, a_{2}, \ldots , a_{n}\) are all real numbers.
\[ f(A)=a_{0} A^{n}+a_{1} A^{n-1}+\ldots +a_{n-1} A+a_{n} I . \]
Since \(\lambda \) is a characteristic root of \(A\), \(\lambda ^{n}\) is a characteristic root of \(A^{n}\) for any positive integer \(n\) (By Property 6)
\(\seteqnumber{0}{4.}{20}\)\begin{align*} A^n X & = \lambda ^n X\\ A^{n-1}X & = \lambda ^{n-1}X\\ \cdots & \cdots \cdots \\ \cdots & \cdots \cdots \\ AX & = \lambda X\\ a_0 A^n X & = a_0 \lambda ^ n X \\ a_1 A^{n-1}X & = a_1 \lambda ^{n-1}X \\ \cdots & \cdots \cdots \\ \cdots & \cdots \cdots \\ a_{n-1} A X & =a_{n-1} \lambda X \end{align*} Adding the above equations we have
\(\seteqnumber{0}{4.}{20}\)\begin{align*} a_{0} A^{n} X+a_{1} A^{n-1} X+\ldots +a_{n-1} A X& =a_{0} \lambda ^{n} X+a_{1} \lambda ^{n-1} X+\ldots +a_{n-1} \lambda X\\ \left (a_{0} A^{n}+a_{1} A^{n-1}+\ldots +a_{n-1} A\right ) X&=\left (a_{0} \lambda ^{n}+a_{1} \lambda ^{n-1}+\ldots +a_{n-1} \lambda \right ) X \\ \left (a_{0} A^{n}+a_{1} A^{n-1}+\ldots +a_{n-1} A+a_{n} I\right ) X & =\left (a_{0} \lambda ^{n}+a_{1} \lambda ^{n-1}+\ldots + a_{n-1} \lambda +a_{n}\right ) X\\ f(A) X&=f(\lambda ) X \end{align*} Hence \(f(\lambda )\) is a characteristic root of \(f(A)\). □
Solved Problems
Solution : Since \(X_{1}\) and \(X_{2}\) are given vectors corresponding to \(\lambda \), we have
\[ A X_{1}=\lambda X_{1} \text { and } A X_{2}=\lambda X_{2}. \]
Hence \(A\left (a X_{1}\right )=\lambda \left (a X_{1}\right )\) and \(A\left (b X_{2}\right )=\lambda \left (b X_{2}\right )\)
\[ A\left (a X_{1}+b X_{2}\right )=\lambda \left (a X_{1}+b X_{2}\right ) \text {. } \]
\(a X_{1}+b X_{2}\) is an eigen vector corresponding to \(\lambda \).
Solution : Since 0 is not an eigen value of \(A\), \(A\) is non singular matrix and hence \(A^{-1}\) exists.
Eigen values of \(A^{-1}\) are \(,\dfrac {1}{2},\dfrac {1}{2},\dfrac {1}{3}\) and eigen values of \(A^2\) are \(2^2, 2^2, 3^2 \).
Solution : The characteristic equation of \(A\) is obviously
\[(3-\lambda )(4-\lambda )(1-\lambda )=0.\]
Hence the eigen values of \(A\) are \(3,4,1\).
The eigen values of \(A^{5}\) are \(3^{5}, 4^{5}, 1^{5}\).
Solution : Let \(A=\left (\begin {array}{lll}3 & -4 & 4 \\ 1 & -2 & 4 \\ 1 & -1 & 3\end {array}\right )\)
Sum of the eigen values \(=\) trace of \(A=3+(-2)+3=4\).
Product of the eigen values \(=|A| .\) Now,
\begin{align*} |A| &=\left |\begin{array}{rrr}3 & -4 & 4 \\ 1 & -2 & 4 \\ 1 & -1 & 3\end {array}\right | \\ &=3(-6+4)+4(3-4)+4(-1+2) \\ &=-6-4+4=-6 \end{align*} Product of the eigen values \(=-6\).
Solution : Let \(A=\left (\begin {array}{rr}\cos \theta & -\sin \theta \\ -\sin \theta & \cos \theta \end {array}\right )\)
The characteristic equation of \(A\) is given by \(|A-\lambda I |=0\).
\begin{align*} \begin{vmatrix} \cos \theta -\lambda & -\sin \theta \\ -\sin \theta & \cos \theta -\lambda \end {vmatrix}&=0 \\ (\cos \theta -\lambda )^{2}-\sin ^{2} \theta &=0\\ (\cos \theta -\lambda -\sin \theta )(\cos \theta -\lambda +\sin \theta )&=0\\ [\lambda -(\cos \theta -\sin \theta )][\lambda -(\cos \theta +\sin \theta )]&=0 \end{align*} The two characteristic roots, (the two eigen values) of the matrix are \((\cos \theta -\sin \theta )\) and \((\cos \theta +\sin \theta )\).
Solution : The characteristic equation of \(A\) is given by \(|A-\lambda I|=0\).
\(\seteqnumber{0}{4.}{20}\)\begin{align*} \begin{vmatrix} \cos \theta -\lambda & -\sin \theta \\ -\sin \theta & -\cos \theta -\lambda \end {vmatrix} & =0 \\ -\left (\cos ^{2} \theta -\lambda ^{2}\right )-\sin ^{2} \theta &=0 \\ \therefore \quad \lambda ^{2}-\left (\cos ^{2} \theta +\sin ^{2} \theta \right ) &=0 \\ \therefore \quad \lambda ^{2}-1&=0 . \end{align*} The characteristic roots are 1 and \(-1\).
Solution : Sum of the eigen values of
\[ A=\text { trace of } A=a_{11}+a_{22} \]
Product of the eigen values of
\[ A= |A|=a_{11} a_{22}-a_{12} a_{21} . \]
Solution : The characteristic equation of \(A\) is \(|A- \lambda I|=0\).
\(\seteqnumber{0}{4.}{20}\)
\begin{align*}
\begin{vmatrix}-2-\lambda &2&-3\\ 2&1-\lambda &-6\\ -1&-2&-\lambda \end {vmatrix} &=0\\ \left (-2-\lambda \right ) \begin{vmatrix}1-\lambda &-6\\ -2&-\lambda \end {vmatrix}-2\cdot
\begin{vmatrix}2&-6\\ -1&-\lambda \end {vmatrix}-3\cdot \begin{vmatrix}2&1-\lambda \\ -1&-2\end {vmatrix} &=0\\ \left (-2-\lambda \right )\left (-\lambda +\lambda ^2-12\right )-2\left (-2\lambda -6\right )-3\left
(-\lambda -3\right ) &=0\\ -\lambda ^3-\lambda ^2+14\lambda +24+4\lambda +12+3\lambda +9 &=0\\ -\lambda ^3-\lambda ^2+21\lambda +45 &=0 \\ \lambda ^3+\lambda ^2-21\lambda -45 &=0
\end{align*}
This is a cubic equation in \(\lambda \) and hence it has 3 roots and the three roots are the three eigen values of the matrix.
The sum of the eigen values = \(- \left (\dfrac {\text {coefficient of } \lambda ^2 }{\text {coefficient of } \lambda ^3} \right )=-1\)
The sum of the elements on the diagonal of the matrix \(A = -2+1+0=-1 \). Hence the result.
Solution : Let \(\lambda _{1}, \lambda _{2}, \lambda _{3}\) be the eigen values of \(A\).
Given, product of 2 eigen values (say) \(\lambda _{1}, \lambda _{2}\) is \(16.\)
\[ \therefore \lambda _{1} \lambda _{2}=16 \]
We know that the product of the eigen values is \(|A|\).
\(\seteqnumber{0}{4.}{20}\)
\begin{align*}
\lambda _{1} \lambda _{2} \lambda _{3} & =\left |\begin{array}{rrr}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end {array}\right |\\ 16 \lambda _{3} &=6(9-1)+2(-6+2)+2(2-6) \\ &=48-8-8 \\ &=32 \\
\therefore \lambda _{3} &=2
\end{align*}
\(\therefore \) The third eigen value is 2.
Also we know that the sum of the eigen values of
\[ A=\operatorname {trace} \text { of } A=6+3+3=12 \]
Solution : Let \(\lambda _{1}, \lambda _{2}, \lambda _{1}\) be the eigen values of \(A\).
Given product of 2 eigen values, say, \(\lambda _{1}\) and \(\lambda _{1}\) is \(-12\).
\begin{equation} \lambda _{1} \lambda _{2}=-12 \label {prob10eq1} \end{equation}
We know that the product of the eigen values is \(|A|\) We know that the product of the eigen values is \(|A|\).
\(\seteqnumber{0}{4.}{21}\)\begin{align} \therefore \quad \lambda _{1} \lambda _{2} \lambda _{3} & =\left |\begin{array}{rrr} 2 & 2 & -7 \\ 2 & 1 & 2 \\ 0 & 1 & -3 \end {array}\right | \notag \\ 12 \lambda _{3}& =-12 \notag \\ \therefore \quad \lambda _{3} & =1 \label {prob10eq2} \end{align} Also we know sum of the eigen values \(=\) Trace of \(A\).
\(\seteqnumber{0}{4.}{22}\)\begin{align} \lambda _{1}+\lambda _{2}+\lambda _{3}& =2+1-3=0\notag \\ \lambda _{1}+\lambda _{2} & =-1 \quad \text { (using \eqref {prob10eq2} ) } \label {prob10eq3} \end{align} Using (4.23) in (4.21) we get
\(\seteqnumber{0}{4.}{23}\)
\begin{align*}
\lambda _{1}\left (-1-\lambda _{1}\right )&=-12\\ \lambda _{1}^{2}+\lambda _{1}-12 &=0 \\ \left (\lambda _{1}+4\right )\left (\lambda _{1}-3\right )&=0 \\ \therefore \lambda _{1}&=3 \text { or }-4
\end{align*}
Putting \(\lambda _{1}=3\) in (4.21) we get \(\lambda _{2}=-4\). Or putting \(\lambda _{1}=-4\) in (4.21) we get \(\lambda
_{2}=3\).
Thus the three eigen values are \(3,-4,1\).
Solution : Let \(\lambda _{1}, \lambda _{2}, \lambda _{3}\) be the eigen values of \(A\).
We know that \(\lambda _{1}^{2}, \lambda _{2}^{2}, \lambda _{3}^{2}\) are the eigen values of \(A^{2}\).
\begin{align*} A^{2} &=\left (\begin{array}{lll} 3 & 1 & 4 \\ 0 & 2 & 6 \\ 0 & 0 & 5 \end {array}\right ) \cdot \left (\begin{array}{lll} 3 & 1 & 4 \\ 0 & 2 & 6 \\ 0 & 0 & 5 \end {array}\right ) \\ &=\left (\begin{array}{lll} 9 & 5 & 38 \\ 0 & 4 & 42 \\ 0 & 0 & 25 \end {array}\right ) \end{align*} Sum of the eigen values of \(A^{2}=\) Trace of \(A^{2}=9+4+25\)
\[i.e., \lambda _{1}^{2}+\lambda _{2}^{2}+\lambda _{3}^{2}=38 \]
Sum of the squares of the eigen values of \(A=38\).
Solution : The characteristic equation of \(A\) is
\(\seteqnumber{0}{4.}{23}\)\begin{align*} |A-\lambda I| & =0 \\ \therefore \left |\begin{array}{ccc} 1-\lambda & 1 & 3 \\ 1 & 5-\lambda & 1 \\ 3 & 1 & 1-\lambda \end {array}\right | & =0 \\ \therefore (1-\lambda )[(5-\lambda )(1-\lambda )-1]-[(1-\lambda )-3]+3[1-3(5-\lambda )] & =0 \\ (1-\lambda )\left (\lambda ^{2}-6 \lambda +4\right )+(\lambda +2)+3(3 \lambda -14) & =0 \\ \lambda ^{2}-6 \lambda +4-\lambda ^{3}+6 \lambda ^{2}-4 \lambda +\lambda +2+9 \lambda -42 & =0 \\ -\lambda ^{3}+7 \lambda ^{2}-36& =0 \\ \lambda ^{3}-7 \lambda ^{2}+36&=0 \\ (\lambda +2)\left (\lambda ^{2}-9 \lambda +18\right )&=0 \\ (\lambda +2)(\lambda -6)(\lambda -3)&=0 \end{align*} \(\lambda =-2,3,6\) are the three eigen values.
Case (i) Eigen vector corresponding to \(\lambda =-2\).
Let \(X=\left (\begin {array}{l} x_{1} \\ x_{2} \\ x_{3} \end {array}\right ) \) be an eigen vector corresponding to \(\lambda =-2\)
Hence \(A X=-2 X\).
\begin{align*} \left (\begin{array}{lll} 1 & 1 & 3 \\ 1 & 5 & 1 \\ 3 & 1 & 1 \end {array}\right )\left (\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end {array}\right )&=\left (\begin{array}{l} -2 x_{1} \\ -2 x_{2} \\ -2 x_{3} \end {array}\right ) \\ \therefore x_{1}+x_{2}+3 x_{3}&=-2 x_{1} \\ x_{1}+5 x_{2}+x_{3}&=-2 x_{2} \\ 3 x_{1}+x_{2}+x_{3}&=-2 x_{3} \end{align*}
\(\seteqnumber{0}{4.}{23}\)\begin{align} 3 x_{1}+x_{2}+3 x_{3}& =0 \label {prob12eq1} \\ x_{1}+7 x_{2}+x_{3}&=0 \label {prob12eq2}\\ 3 x_{1}+x_{2}+3 x_{3}&=0 \label {prob12eq3} \end{align} Clearly this system of three equations reduces to two equations only. From (4.24) and (4.25) we get
\[ \because \quad x_{1}=-2 k ; x_{2}=0 ; x_{3}=2 k. \]
\(\therefore \) It has only one independent solution and can be obtained by giving any value to \(k\) say \(k=1\).
\(\therefore \quad (-2,0,2)\) is an eigen vector corresponding to \(\lambda =-2\).
Case (ii) Eigen vector corresponding to \(\lambda =3\).
Then \(AX = 3 X\)
\begin{align*} -2 x_{1}+x_{2}+3 x_{3}&= 0 \\ x_{1}+2 x_{2}+x_{3} &=0 \\ 3 x_{1}+x_{2}-2 x_{3}&= 0 \end{align*} Taking the first 2 equations we get
\(\seteqnumber{0}{4.}{26}\)\begin{align*} \frac {x_{1}}{-5} & =\frac {x_{2}}{5}=\frac {x_{3}}{-5}=k \text { (say). } \\ x_{1}&=-k ; x_{2}=k ; x_{3}=-k . \end{align*} Taking \(k=1\) (say) \((-1,1,-1)\) is an eigen vector corresponding to \(\lambda =3\).
Case (iii) Eigen vector corresponding to \(\lambda =6\),
We have \(A X=6 X\).
\begin{align*} -5 x_{1}+x_{2}+3 x_{3} &=0 \\ x_{1}-x_{2}+x_{3} &=0 \\ 3 x_{1}+x_{2}-5 x_{3} &=0 \end{align*} Taking the first two equations we get
\[ \frac {x_{1}}{4}=\frac {x_{2}}{8}=\frac {x_{3}}{4}=k \]
\(\therefore \quad x_{1}=k ; x_{2}=2 k ; x_{3}=k\). It satisfies the third equation also.
Taking \(k=1\) (say) \((1,2,1)\) is an eigen vector corresponding to \(\lambda =6\).
Therefore \((-2,0,2), (-1,1,-1)\) and \((1,2,1)\) are the eigen vectors of the eigen values \(-2,3,6\) respectively.
Solution : The characteristic equation of \(A\) is \(|A-\lambda I|=0\).
\(\seteqnumber{0}{4.}{26}\)
\begin{align*}
\left |\begin{array}{ccc} 6-\lambda & -2 & 2 \\ -2 & 3-\lambda & -1 \\ 2 & -1 & 3-\lambda \end {array}\right | & =0 . \\ \therefore \quad (6-\lambda )\left [(3-\lambda )^{2}-1\right ]+2[(2 \lambda -6)+2]
+2(2-6+2 \lambda )&=0\\ \therefore \quad (6-\lambda )\left (8+\lambda ^{2}-6 \lambda \right )+4 \lambda -8+4 \lambda -8&=0 \\ 48+6 \lambda ^{2}-36 \lambda -8 \lambda -\lambda ^{3}+6 \lambda ^{2}+8 \lambda -16&=0 \\ -\lambda
^{3}+12 \lambda ^{2}-36 \lambda +32&=0\\ \lambda ^{3}-12 \lambda ^{2}+36 \lambda -32&=0 \\ (\lambda -2)(\lambda -2)(\lambda -8)&=0
\end{align*}
The eigen values are \(2,2,8\). We now find the eigen vectors.
Case (i) Eigen vector corresponding to \(\lambda =2\).
The eigen vector \(X=\left (\begin {array}{l}x_{1} \\ x_{2} \\ x_{3}\end {array}\right )\) is got from \(A X=2 X\)
\begin{align*}
6 x_{1}-2 x_{2}+2 x_{3}&=2 x_{1} \\ -2 x_{1}+3 x_{2}-x_{3}&=2 x_{2} \\ 2 x_{1}-x_{2}+3 x_{3}&=2 x_{3} \\ 4 x_{1}-2 x_{2}+2 x_{3}&=0 \\ -2 x_{1}+x_{2}-x_{3}&=0\\ 2 x_{1}-x_{2}+x_{3}&=0
\end{align*}
The above three equations are equivalent to the single equation \(2 x_{1}-x_{2}+x_{3}=0\).
The independent eigen vectors can be obtained by giving arbitrary values to any two of the unknowns \(x_{1}, x_{2}, x_{3}\). Let \(x_{1}=1 ; x_{2}=2\) we get \(x_{3}=0\).
Let \(x_{1}=3 ; x_{2}=4 \) we get \(x_{3}=-2\).
\(\therefore \quad \) Two independent vectors corresponding to \(\lambda = 2\) are \((1,2,0)\) and \((3,4,-2)\).
Case (ii) Eigen vector corresponding to \(\lambda =8\).
The eigen vector \(X= \begin {pmatrix} x_{1} \\ x_{2} \\ x_{3} \end {pmatrix}\) is got from \(X=8 X\).
\begin{align} \therefore \quad -2 x_{1}-2 x_{2}+2 x_{3}&=0 \label {prob13eq1} \\ -2 x_{1}-5 x_{2}-x_{3}&=0\label {prob13eq2} \\ 2 x_{1}-x_{2}-5 x_{3}&=0\label {prob13eq3} \end{align} From (4.27) and (4.28) we get
\(\seteqnumber{0}{4.}{29}\)
\begin{align*}
\frac {x_{1}}{12} &=\frac {x_{2}}{-6} &=\frac {x_{3}}{6}=k \text { (say). } \\ \therefore \quad x_{1} &=2 k ; x_{2}=-k ; x_{3}=k .
\end{align*}
Giving \(k=1\) we get an eigen vector corresponding to 8 as \((2,-1,1)\).
Therefore \((1,2,0), (3,4,-2)\) and \((2,-1,1)\) are the eigen vectors of the eigen values \(2,2,8\) respectively.
Solution : The characteristic equation of \(A\) is \(|A-\lambda l|=0\).
\(\seteqnumber{0}{4.}{29}\)
\begin{align*}
\left |\begin{array}{ccc}2-\lambda & -2 & 2 \\ 1 & 1-\lambda & 1 \\ 1 & 3 & -1-\lambda \end {array}\right |&=0 \\ (2-\lambda )[-(1-\lambda )(1+\lambda )-3] +2[-(1+\lambda )-1]+2[3-(1-\lambda )]&=0\\
(2-\lambda )\left (\lambda ^{2}-4\right )-2(2+\lambda )+2(2+\lambda )&=0\\ 2 \lambda ^{2}-8-\lambda ^{3}+4 \lambda -4-2 \lambda +4+2 \lambda &=0\\ -\lambda ^{3}+2 \lambda ^{2}+4 \lambda -8&=0\\ \lambda ^{3}-2 \lambda ^{2}-4
\lambda +8&=0 \\ (\lambda -2)\left (\lambda ^{2}-4\right )&=0\\ (\lambda -2)(\lambda -2)(\lambda +2)&=0
\end{align*}
\(\lambda =2,2,-2 \) are the three eigen values.
Case (i) Eigen vector corresponding to \(\lambda =2\).
Let \(X=\left (x_{1}, x_{2}, x_{3}\right )\) be an eigen vector corresponding to \(\lambda =2, X\) is got from \(A X=2 X\).
\[\left (\begin {array}{rrr} 2 & -2 & 2 \\ 1 & 1 & 1 \\ 1 & 3 & -1 \end {array}\right )\left (\begin {array}{l} x_{1} \\ x_{2} \\ x_{3} \end {array}\right )=\left (\begin {array}{l} 2 x_{1} \\ 2 x_{2} \\ 2 x_{3} \end {array}\right ) \]
The eigen vector corresponding to \(\lambda =2\) is given by the equations
\(\seteqnumber{0}{4.}{29}\)\begin{align*} 2 x_{1}-2 x_{2}+2 x_{3} &=2 x_{1} \\ x_{1}+x_{2}+x_{3} &=2 x_{2} \\ x_{1}+3 x_{2}-x_{3} &=2 x_{3} \end{align*}
\(\seteqnumber{0}{4.}{29}\)\begin{align} \text { (i.e.) }-x_{2}+x_{3} &=0\label {prob14eq1} \\ x_{1}-x_{2}+x_{3} &=0 \label {prob14eq2}\\ x_{1}+3 x_{2}-3 x_{3} &=0 \label {prob14eq3} \end{align} Taking (4.30) and (4.31) we get
\(\seteqnumber{0}{4.}{32}\)
\begin{align*}
\frac {x_{1}}{0} &=\frac {x_{2}}{1}=\frac {x_{3}}{1}=k \text { (say)}.\\ \therefore \quad x_{1}&=0 ; x_{2}=k ; x_{3}=k
\end{align*}
Taking \(k=1\), we get \((0,1,1)\) as an eigen vector corresponding to \(\lambda =2\).
Case (ii) Eigen vector corresponding to \(\lambda =-2\).
Corresponding to \(\lambda =-2\) we have \(A X=-2 X\).
\begin{align*} 2 x_{1}-2 x_{2}+2 x_{3} &=-2 x_{1} \\ x_{1}+x_{2}+x_{3} &=-2 x_{2} \\ x_{1}+3 x_{2}-x_{3} &=-2 x_{3} \end{align*}
\(\seteqnumber{0}{4.}{32}\)\begin{align} 2 x_{1}-x_{2}+x_{3} &=0 \label {prob14eq4} \\ x_{1}+3 x_{2}+x_{3} &=0 \label {prob14eq5}\\ x_{1}+3 x_{2}+x_{3} &=0 \label {prob14eq6} \end{align} Taking (4.33) and (4.34) we get,
\(\seteqnumber{0}{4.}{35}\)\begin{align*} \frac {x_{1}}{-4}=\frac {x_{2}}{-1}=\frac {x_{3}}{7}&=k \text { (say). } \\ x_{1}&=-4 k ; x_{2}=-k ; x_{3}=7 k. \end{align*} Taking \(k=1\) we get \((-4,-1,7)\) as an eigen vector corresponding to the eigen value \(\lambda =-2\).
0 Comments