Contents
Chapter 3 Inner Product Space
Upto this point we have dealt with the algebraic properties of a vector space and these properties are consequences of the basic operations, namely, vector addition and scalar multiplication defined in the vector space. We know that in the usual three dimensional vector space \(V_{3}(\mathbb {R})\) it is possible to talk about the length of a vector and angle between two vectors. These concepts of length and angle can be defined in terms of the usual “ dot product” or “ scalar product” of two vectors. The dot product of \(u=(a_{1},\ b_{1},\ c_{1})\) and \(v=(a_{2},\ b_{2},\ c_{2})\) is defined by
\[ u\cdot v=a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} \]
We note that the length of \(u\) is given by \(\sqrt {uu}\) and the angle \(\theta \) between \(u\) and \(v\) is determined by \(\displaystyle \cos \theta =\frac {uv}{\sqrt {uu}\sqrt {vv}}\). Hence \(u\) and \(v\) are perpendicular or orthogonal if and only if \(u\cdot v=0.\)
An inner product on a vector space is a generalisation of the dot product and in terms of such an inner product we can define the length of a vector and angle between two vectors. Our study about angle will be restricted to the concept of perpendicularity of two vectors.
Throughtout this section we shall deal only with vector spaces over the field \(F\) of real or complex numbers.
3.1 Definition and Examples
Definition 3.1.1. Let \(V\) be a vector space over \(F\). An inner product on \(V\) is a function which assigns to each ordered pair of vectors \(u, v\) in \(V\) a scalar in \(F\) denoted by \(\langle u, v\rangle \) satisfying the following conditions.
(i) \(\langle u+v, w\rangle =\langle u, w\rangle +\langle v, w\rangle \)
(ii) \(\langle \alpha u, v\rangle =\alpha \langle u, v\rangle \)
(iii) \(\langle u, v\rangle =\langle v, u\rangle \), where \(\langle v, u\rangle \) is the complex conjugate of \(\langle u, v\rangle .\)
(iv) \(\langle u, u\rangle \geq 0\) and \(\langle u, u\rangle =0\) if and only if \(u=0.\)
A vector space with an inner product defined on it is called an inner product space.
An inner product space is called an Euclidean space or unitary space according as \(F\) is the field of real numbers or complex numbers.
1 \(V_{n}(\mathbb {R})\) is a real inner product space with inner product defined by \(\langle x, y\rangle =x_{1}y_{1}+x_{2}y_{2}+\cdots +x_{n}y_{n}\) where \(x=(x_{1},\ x_{2},\ \ldots ,\ x_{n})\) and \(y=(y_{1},\ y_{2},\ \ldots ,\ y_{n})\) . This is called the standard inner product on \(V_{n}(\mathbb {R})\) .
Proof : Let \(x, y, z\in V_{n}(\mathbb {R})\) and \(\alpha \in \mathbb {R}.\)
(i) \(\langle x+y, z\rangle =(x_{1}+y_{1})z_{1}+(x_{2}+y_{2})z_{2}+\cdots +(x_{n}+y_{n})z_{n}\)
\[ =(x_{1}z_{1}+x_{2}z_{2}+\cdots +x_{n}z_{n})+(y_{1}z_{1}+y_{2}z_{2}+\cdots +y_{n}z_{n})=\langle x,\ z\rangle +\langle y,\ z\rangle . \]
(ii) \(\langle \alpha x, y\rangle =\alpha x_{1}y_{1}+\alpha x_{2}y_{2}+\cdots +\alpha x_{n}y_{n}=\alpha (x_{1}y_{1}+x_{2}y_{2}+\cdots +x_{n}y_{n})=\alpha \langle x, y\rangle .\)
(iii) \(\langle x, y\rangle =x_{1}y_{1}+x_{2}y_{2}+\cdots +x_{n}y_{n}=y_{1}x_{1}+y_{2}x_{2}+\cdots +y_{n}x_{n}=\langle y, x\rangle .\)
(iv) \(\langle x, x\rangle =x_{1}^{2}+x_{2}^{2}+\cdots +x_{n}^{2}\geq 0\) and \(\langle x, x\rangle =0\) if and only if \(x_{1}=x_{2}=\cdots =x_{n}=0 \langle x, x\rangle =0\) if and only if \(x=0\)
□
2 \(V_{n}(\mathbb {C})\) is a complex inner product space with inner product defined by \(\langle x, y\rangle =x_{1}y_{1}^{-}+x_{2}y_{2}^{-}+\cdots +x_{n}y_{n}^{-}\) where \(x=(x_{1},\ x_{2},\ \ldots ,\ x_{n})\) and \(y=(y_{1},\ y_{2},\ \ldots ,\ y_{n})\). This is called the standard inner product on \(V_{n}(\mathbb {R})\).
Proof : Let \(x, y, z\in V_{n}(\mathbb {C})\) and \(\alpha \in \mathbb {C}\)
(i) \(\langle x+y, z\rangle =(x_{1}+y_{1})z_{1}^{-}+(x_{2}+y_{2})z_{2}^{-}+\cdots +(x_{n}+y_{n})z_{n}^{-}\)
\[ =(x_{1}z_{1}^{-}+x_{2}z_{2}^{-}+\cdots +x_{n}z_{n}^{-})+(y_{1}z_{1}^{-}+y_{2}z_{2}^{-}+\cdots +y_{n}z_{n}^{-})=\langle x,\ z\rangle +\langle y,\ z\rangle . \]
(ii) \(\langle \alpha x, y\rangle =\alpha x_{1}y_{1}^{-}+\alpha x_{2}y_{2}^{-}+\cdots +\alpha x_{n}y_{n}^{-}=\alpha (x_{1}y_{1}^{-}+x_{2}y_{2}^{-}+\cdots +x_{n}y_{n}^{-})=\alpha \langle x, y\rangle .\)
(iii) \(\langle y, x\rangle =\overline {y_{1}x_{1}^{-}+y_{2}x_{2}^{-}+\cdots +y_{n}x_{n}^{-}}=y_{1}^{-}x_{1}+y_{2}^{-}x_{2}+\cdots +y_{n}^{-}x_{n}=\langle x, y\rangle .\)
(iv) \(\langle x, x\rangle =x_{1}x_{1}^{-}+x_{2}x_{2}^{-}+\cdots +x_{n}x_{n}^{-}=|x_{1}|^{2}+|x_{2}|^{2}+\cdots +|x_{n}|^{2}\geq 0\) \(\langle x, x\rangle =0\) if and only if \(x=0\)
□
3 Let \(V\) be the set of all continuous real valued functions defined on the closed interval \([0,1]. V\) is a real inner product space with inner product defined by \(\langle f, g\rangle = \displaystyle \int _{0}^{1}f(t)g(t)dt.\)
Proof : Let \(f, g, h\in V\) and \(\alpha \in \mathbb {R}.\)
(i) \(\langle f+g, h\displaystyle \rangle =\int _{0}^{1}[f(t)+g(t)]h(t)dt=\int _{0}^{1}f(t)h(t)dt+\int _{0}^{1}g(t)h(t)dt=\langle f, h\rangle +\langle g, h\rangle .\)
(ii) \(\langle \alpha f, g\displaystyle \rangle =\int _{0}^{1}\alpha f(t)g(t)dt=\alpha \int _{0}^{1}f(t)g(t)dt=\alpha \langle \alpha f, g\rangle .\)
(iii) \(\langle f, g\displaystyle \rangle =\int _{0}^{1}f(t)g(t)dt\int _{0}^{1}g(t)f(t)dt=\langle g, f\rangle .\)
(iv) \(\langle f, f\displaystyle \rangle =\int _{0}^{1}|f(t)|^{2}dt\geq 0\) and \(\langle f, f\rangle =0\) if and only if \(f=0\)
□
Solved Problems
Solution :
(i) \(\langle f, g\displaystyle \rangle =\int _{0}^{1}f(t)g(t)dt=\int _{0}^{1}(t+2)(t^{2}-2t-3)dt\)
\(=\displaystyle \int _{0}^{1}(t^{3}-7t-6)dt=[\frac {t^{4}}{4}-\frac {7t^{2}}{2}-6t]_{0}^{1}=\frac {1}{4}-\frac {7}{2}-6=-\frac {37}{4}.\)
(ii) \(\Vert f\Vert ^{2}=\langle f, f\displaystyle \rangle =\int _{0}^{1}[f(t)]^{2}dt=\int _{0}^{1}(t+2)^{2}dt=\int _{0}^{1}(t^{2}+4+4)dt\)
\[ =[\frac {t^{3}}{3}+2t^{2}+4t]_{0}^{1}=\frac {1}{3}+2+4=\frac {19}{3} \]
\[ \Vert f\Vert =\frac {\sqrt {19}}{\sqrt {3}} \]
Theorem 3.1.9. The norm defined in an inner product space \(V\) has the following properties.
(i) \(\Vert x\Vert \geq 0\) and \(\Vert x\Vert =0\) if and only if \(x=0.\)
(ii) \(\Vert \alpha x\Vert =|\alpha |\Vert x\Vert .\)
(iii) \(|\langle x, y\rangle |\leq \Vert x\Vert \Vert y\Vert \) (Schwartz’s inequality)
(iv) \(\Vert x+y\Vert \leq \Vert x\Vert +\Vert y\Vert \) (Triangle inequality)
Proof :
(i) \(\Vert x\Vert =\sqrt {\langle x,x\rangle }\geq 0\) and \(\Vert x\Vert =0\) if and only if \(x=0.\)
(ii) \(\Vert \alpha x\Vert ^{2}=\langle \alpha x, \alpha x\rangle =\alpha \langle x, \alpha x\rangle =\alpha \overline {\alpha }\langle x, x\rangle =|\alpha |^{2}\Vert x\Vert ^{2}\)
Hence \(\Vert \alpha x\Vert =|\alpha |\Vert x\Vert .\)
(iii) The inequality is trivially true when \(x=0\) or \(y=0\). Hence let \(x\neq 0\) and \(y\neq 0.\) Consider \(z=y-\displaystyle \frac {\langle y,x\rangle }{\Vert x\Vert ^{2}}x\). Then
\(\seteqnumber{0}{3.}{0}\)\begin{align*} 0 & \leq \langle z, z\displaystyle \rangle \\ & =\langle y-\frac {\langle y,x\rangle }{\Vert x\Vert ^{2}}x, y-\displaystyle \frac {\langle y,x\rangle }{\Vert x\Vert ^{2}}x\rangle \\ & =\langle y, y\rangle -\overline {\frac {\langle y,x\rangle }{\Vert x\Vert ^{2}}}\langle y, x\displaystyle \rangle -\frac {\langle y,x\rangle }{\Vert x\Vert ^{2}}\langle x, y\displaystyle \rangle +\frac {\langle y,x\rangle \overline {\langle y,x\rangle }}{\Vert x\Vert ^{2}\Vert x||^{2}}\langle x, x\rangle \\ &=\Vert y^{2}\Vert -\frac {\overline {\langle y,x\rangle }\langle y,x\rangle }{\Vert x\Vert ^{2}}-\frac {\langle y,x\rangle \langle x,y\rangle }{\Vert x\Vert ^{2}}+\frac {\langle y,x\rangle \overline {\langle y,x\rangle }}{\Vert x\Vert ^{2}}\\ &=\Vert y^{2}\Vert -\frac {\overline {\langle x,y\rangle }\langle x,y\rangle }{||x\Vert ^{2}}\\ 0 & \leq \Vert x\Vert ^{2}\Vert y\Vert ^{2}-|\langle x,\ y\rangle |^{2}\\ |\langle x,\ y\rangle |^{2}&\leq \Vert x\Vert ^{2}\Vert y\Vert ^{2} \end{align*}
(iv) \(\begin {aligned}[t] \Vert x+y\Vert ^{2} & =\langle x+y, x+y\rangle \\ &=\langle x, x\rangle +\langle x, y\rangle +\langle y, x\rangle +\langle y, y\rangle \\ &=\Vert x\Vert ^{2}+\langle x,\ y\rangle +\overline {\langle x,y\rangle }+\Vert y\Vert ^{2}\\ &=\Vert x\Vert ^{2}+2Re\langle x,\ y\rangle +\Vert y\Vert ^{2}\\ &\leq \Vert x\Vert ^{2}+2|\langle x, y\rangle |+\Vert y\Vert ^{2}\\ & \leq \Vert x\Vert ^{2}+2\Vert x\Vert \Vert y\Vert +\Vert y\Vert ^{2} \text { (by (iii))}\\ &\leq (\Vert x\Vert +\Vert y\Vert )^{2} \\ \Vert x+y\Vert \leq \Vert x\Vert +\Vert y\Vert . \end {aligned} \)
□
3.2 Orthogonality
Note 3.2.2. \(x\) is orthogonal to \(y\)
\(\seteqnumber{0}{3.}{0}\)\begin{align*} & \Rightarrow , \langle x, y\rangle =0 \\ &\Rightarrow \overline {\langle x,y\rangle } =\overline {0}\\ &\Rightarrow \langle y, x\rangle =0\\ &\Rightarrow y \text { is orthogonal to $x$.} \end{align*} Thus \(x\) and \(y\) are orthogonal if and only if \(\langle x, y\rangle =0.\)
Proof : Let \(\alpha _{1}v_{1}+\alpha _{2}v_{2}+\cdots +\alpha _{n}v_{n}=0\). Then
\(\seteqnumber{0}{3.}{0}\)\begin{align*} \langle \alpha _{1}v_{1}+\alpha _{2}v_{2}+\cdots +\alpha _{n}v_{n}, v_{1}\rangle & =\langle 0, v_{1}\rangle =0 \\ \alpha _{1}\langle v_{1},\ v_{1}\rangle +\alpha _{2}\langle v_{2},\ v_{1}\rangle +\cdots +\alpha _{n}\langle v_{n},\ v_{1}\rangle & =0\\ \alpha _{1}\langle v_{1}, v_{1}\rangle & =0 \ \text {(since $S$ is orthogonal)} \\ \alpha _{1} &=0\ \text { (since $v_{1}\neq 0$)} \end{align*} Similarly \(\alpha _{2}=\alpha _{3}=\cdots =\alpha _{n}=0\). Hence \(S\) is linearly independent. □
Proof :
\(\seteqnumber{0}{3.}{0}\)\begin{align*} \langle v, v_{k}\rangle & =\langle \alpha _{1}v_{1}+\alpha _{2}v_{2}+\cdots +\alpha _{n}v_{n}, v_{k}\rangle \\ &=\alpha _{1}\langle v_{1},\ v_{k}\rangle +\alpha _{2}\langle v_{2},\ v_{k}\rangle +\cdots +\alpha _{k}\langle v_{k},\ v_{k}\rangle +\cdots +\alpha _{n}\langle v_{n},\ v_{k}\rangle \\ &=\alpha _{k}\langle v_{k}, v_{k}\rangle \text { (since $S$ is orthogonal) } \\ &=\alpha _{k}\Vert v_{k}\Vert ^{2}\\ \alpha _{k}&=\frac {\langle v,v_{k}\rangle }{\Vert v_{k}||^{2}} \end{align*} □
Proof : Let \(V\) be a finite dimensional inner product space.
Let \(\{v_{1},\ v_{2},\ \ldots ,\ v_{n}\}\) be a basis for \(V\).
From this basis we shall construct an orthonormal basis \(\{w_{1},\ w_{2},\ \ldots ,\ w_{n}\}\) by means of a construction known as \(Gram-\)Schmidt orthogonalisation process.
First we take \(w_{1}=v_{1}\). Let
\[w_{2}=v_{2}-\displaystyle \frac {\langle v_{2},w_{1}\rangle }{||w_{1}\Vert ^{2}}w_{1}.\]
We claim that \(w_{2}\neq 0\).
For, if \(w_{2}=0\) then \(v_{2}\) is a scalar multiple of \(w_{1}\) and hence of \(v_{1}\) which is a contradiction since \(v_{1}, v_{2}\) are linearly independent. Also,
\begin{align*} \langle w_{2}, w_{1}\displaystyle \rangle & =\langle v_{2}-\frac {\langle v_{2},w_{1}\rangle }{||w_{1}\Vert ^{2}}w_{1}, w_{1}\displaystyle \rangle \\ &=\langle v_{2}-\frac {\langle v_{2},v_{1}\rangle }{||v_{1}\Vert ^{2}}v_{1}, v_{1}\rangle && (\because w_{1}=v_{1})\\ &=\langle v_{2},\ v_{1}\rangle -\frac {\langle v_{2},v_{1}\rangle }{||v_{1}\Vert ^{2}}\langle v_{1},\ v_{1}\rangle \\ &=\langle v_{2},\ v_{1}\rangle -\langle v_{1},\ v_{1}\rangle =0. \end{align*} Now, suppose that we have constructed non-zero orthogonal vectors \(w_{1}, w_{2}, \ldots , w_{k}.\) Then put
\[ w_{k+1}=v_{k+1}-\sum _{j=1}^{k}\frac {\langle v_{k+1},w_{j}\rangle }{\Vert w_{j}||^{2}}w_{j} \]
We claim that \(w_{k+1}\neq 0\).
For, if \(w_{k+1}=0\), then \(v_{k+1}\) is a linear combination of \(w_{1}, w_{2}, \ldots , w_{k}\) and hence is a linear combination of \(v_{1}, v_{2}, \ldots , v_{k}\) which is a contradiction since \(v_{1}, v_{2}, \ldots , v_{k+1}\) are linearly independent.
Also,
\begin{align*}
\langle w_{k+1},\ w_{i}\rangle & =\langle v_{k+1},\ w_{i}\rangle -\sum _{j=1}^{k}\frac {\langle v_{k+1},w_{j}\rangle }{\Vert w_{j}||^{2}}\langle w_{j},\ w_{i}\rangle \\ &=\langle v_{k+1},\ w_{i}\rangle -\frac {\langle
v_{k+1},w_{i}\rangle }{\Vert w_{i}||^{2}}\langle w_{i},\ w_{i}\rangle \\ &=\langle v_{k+1},\ w_{i}\rangle -\langle w_{i},\ w_{i}\rangle \\ &=0.
\end{align*}
Thus, continuing in this way we ultimately obtain a non-zero orthogonal set \(\{w_{1}, w_{2}, \cdots , w_{n}\}\). By Theorem 3.2.9 this set is linearly independent and hence a basis.
To obtain an orthonormal basis we replace each \(w_{i}\) by \(\displaystyle \frac {w_{i}}{\Vert w_{i}\Vert }.\) □
Solved Problems
Solution : Take \(w_{1}=v_{1}=(1,0,1)\). Then
\(\seteqnumber{0}{3.}{0}\)\begin{align*} \Vert w_{1}\Vert ^{2} =\langle w_{1}, w_{1}\rangle & =1^{2}+0^{2}+1^{2}=2 \end{align*} and
\(\seteqnumber{0}{3.}{0}\)\begin{align*} \langle w_{1}, v_{2}\rangle &=1+0+1=2 \end{align*} Put
\(\seteqnumber{0}{3.}{0}\)
\begin{align*}
w_{2} &=v_{2}-\displaystyle \frac {\langle v_{2},w_{1}\rangle }{||w_{1}\Vert ^{2}}w_{1}\\ &=(1,3,1)-(1,0,1)\\ &=(0,3,0)\\ \Vert w_{2}\Vert ^{2}&=9.
\end{align*}
Also \(\langle w_{2}, v_{3}\rangle =0+6+0=6\) and \(\langle w_{1}, v_{3}\rangle =3+0+1=4\)
Now,
\begin{align*}
w_{3} & =v_{3}-\displaystyle \frac {\{v_{3},w_{1}\rangle }{||w_{1}\Vert ^{2}}w_{1}-\frac {\{v_{3},w_{2}\rangle }{||w_{2}\Vert ^{2}}w_{2}\\ &=(3,2,1)-\frac {4}{2}(1,0,1)-\frac {6}{9}(0,3,0)\\ &=(3,2,1)-2(1,0,1)-\frac
{2}{3}(0,3,0)\\ &=(1,0,\ -1)\\ \Vert w_{3}\Vert ^{2}&=2.
\end{align*}
The orthogonal basis is \(\{(1,0,1),\ (0,3,0),\ (1,0,\ -1)\}.\)
Hence the orthonormal basis is
\[ \left \{ \left (\frac {1}{\sqrt {2}},0,\ \frac {1}{\sqrt {2}}\right ), (0,1,0), \left ( \frac {1}{\sqrt {2}},0,\ -\frac {1}{\sqrt {2}}\right ) \right \}.\]
Problem 3.2.13. Let \(V\) be the set of all polynomials of degree \(\leq \) 2 together with the zero polynomial. \(V\) is a real inner product space with inner product defined by \(\langle f, g\displaystyle \rangle =\int _{-1}^{1}f(x)g(x)dx\). Starting with the basis \(\{1,\ x,\ x^{2}\}\), obtain an orthogonal basis for \(V.\)
Solution : Let \(v_{1}=1;v_{2}=x\) and \(v_{3}=x^{2}\).
Let \(w_{1}=v_{1}.\) Then \(\Vert w_{1}\Vert ^{2}=\langle w_{1}, w_{1}\displaystyle \rangle =\int _{-1}^{1}1dx=2\)
Hence \(\Vert w_{1}\Vert =\sqrt {2}\)
\begin{align*} w_{2} & =v_{2}-\displaystyle \frac {\langle v_{2},w_{1}\rangle }{||w_{1}\Vert ^{2}} w_{1}\\ &=x-\frac {1}{2}\int _{-1}^{1}xdx\\ &=x \\ \Vert w_{2}\Vert ^{2}&=\langle w_{2},\ w_{2}\rangle \\ &=\int _{-1}^{1}x^{2}dx \\ &=\frac {2}{3}. \end{align*} Now,
\(\seteqnumber{0}{3.}{0}\)
\begin{align*}
w_{3} & =v_{3}-\displaystyle \frac {\langle v_{3},w_{1}\rangle }{||w_{1}\Vert ^{2}}w_{1}-\frac {\langle v_{3},w_{2}\rangle }{||w_{2}\Vert ^{2}}w_{2}\\ &=x^{2}-\frac {1}{2}\int _{-1}^{1}x^{2}dx-(\frac {3x}{2})\int
_{-1}^{1}x^{3}dx\\ &=x^{2}-\frac {1}{3}\\ \Vert w_{3}\Vert ^{2} &=\langle w_{3},\ w_{3}\rangle \\ &=\int _{-1}^{1}(x^{2}-\frac {1}{3})dx\\ &=\frac {8}{45}.
\end{align*}
Hence the orthogonal basis is \(\displaystyle \left \{1,\ x,\ x^{2}-\frac {1}{3} \right \}.\)
The required orthonormal basis is \(\displaystyle \left \{\frac {1}{\sqrt {2}}, \displaystyle \frac {\sqrt {3}}{2}x, \displaystyle \frac {\sqrt {10}}{4}(3x^{2}-1)\right \}.\)
Solution : Let \(x=(x_{1},\ x_{2},\ x_{3})\) be any vector orthogonal to \((1,\ 3,\ 4)\) . Then \(x_{1}+3x_{2}+ 4x_{3}=0\). Any solution of this equation gives a vector orthogonal to \((1,\ 3,\ 4)\) . For example \(x=(1,1,\ -1)\) is orthogonal to \((1,\ 3,\ 4)\) . Also \(\Vert x\Vert =\sqrt {3}\). Hence a unit vector orthogonal to \((1,\ 2,\ 3)\) is given by \((\displaystyle \frac {1}{\sqrt {3}},\ \frac {1}{\sqrt {3}},\ -\frac {1}{\sqrt {3}})\)
Solution : \((1,\ 1,\ -1)\) is a vector orthogonal to \((1,\ 3,\ 4)\) (By Problem 3.2.14).
Now, let \(y=(y_{1},\ y_{2},\ y_{3})\) be a vector orthogonal to both \((1,\ 3,\ 4)\) and \((1,\ 1,\ -1)\). Then
\begin{align*}
y_{1}+3y_{2}+4y_{3}&=0\\ y_{1}+y_{2}-y_{3}&=0
\end{align*}
Any solution of this system of equations gives a vector orthogonal to \((1,\ 3,\ 4)\) and \((1,\ 1,\ -1)\).
For example \((7,\ -5,2)\) is one such vector. (by cross multiplication method). Hence \(\{(1,3,4),\ (1,1,\ -1),\ (7,\ -5,2)\}\) is an orthogonal basis containing \((1,\ 3,\ 4)\).
3.3 Orthogonal Complement
Definition 3.3.1. Let \(V\) be an inner product space. Let \(S\) be a subset of \(V\). The orthogonal complement of \(S\), denoted by \(S^{\perp }\), is the set of all vectors in \(V\) which are orthogonal to every vector of \(S\).
\[ \text { (i.e) } S^{\perp }=\{x / x \in V \text { and }\langle x, u\rangle =0 \text { for all } u \in S\}. \]
1. \(V^{\perp }=\{0\}\) and \(\{0\}^{\perp }=V\) since 0 is the only vector which is orthogonal to every vector.
2. Let \(S=\{(x, 0,0) / x \in \mathbb {R}\} \subseteq V_{3}(\mathbb {R})\) with standard inner product. Then \(S^{\perp }=\{(0, y, z) / y, z \in \mathbf {R}\} \) (i.e) The orthogonal complement of the \(x\)-axis is the \(y z\) plane.
Proof : Clearly \(0 \in S^{\perp }\) and hence \(S^{\perp } \neq \phi \).
Now, let \(x, y \in S^{\perp }\) and \(\alpha , \beta \in F\). Then
\begin{align*} \langle x, u\rangle &=\langle y, u\rangle =0 \text { for all } u \in S . \\ \therefore \quad \langle \alpha x+\beta y, u\rangle & =\alpha \langle x, u\rangle +\beta \langle y, u\rangle =0 \text { for all } u \in S . \\ \therefore \quad \alpha x+\beta y & \in S^{\perp } \end{align*} Hence \(S^{\perp } \) is a subspace of \(V \). □
Proof : We shall prove that
(i) \(W \cap W^{\perp }=\{0\},\) and
(ii) \(W+W^{\perp }=V .\)
(i) Let \(v \in W \cap W^{\perp }\). Then \(v \in W\) and \(v \in W^{\perp }\).
Now, \(v \in W^{\perp } \Rightarrow v\) is orthogonal to every vector in \(W\).
In particular, \(v\) is orthogonal to itself.
\[ \therefore \langle v, v\rangle =0 \text { and hence } v=0. \]
Hence \(W \cap W^{\perp }=\{0\}\).
(ii) Let \(\left \{v_{1}, v_{2}, \cdots , v_{r}\right \}\) be an orthonormal basis for \(W\). Let \(v \in V\). Consider
\(\seteqnumber{0}{3.}{0}\)\begin{align*} v_{0} & =v-\left \langle v, v_{1}\right \rangle v_{1}-\left \langle v, v_{2}\right \rangle v_{2} \ldots ,-\left \langle v, v_{r}\right \rangle v_{r} \\ \therefore \quad \left \langle v_{0}, v_{i}\right \rangle & =\left \langle v, v_{i}\right \rangle -\left \langle v, v_{i}\right \rangle \left \langle v_{i}, v_{i}\right \rangle \text { (since }\left \langle v_{i}, v_{j}\right \rangle =0 \text { if } i \neq j ) \\ & =\left \langle v, v_{i}\right \rangle -\left \langle v, v_{i}\right \rangle \left (\text { since }\left \langle v_{i}, v_{i}\right \rangle =1\right ) \\ &=0 \end{align*} \(v_{0}\) is orthogonal to each of \(v_{1}, v_{2}, \ldots , v_{r}\) and hence is orthogonal to every vector in \(W\). Hence \(v_{0} \in \mathbf {W}^{\perp }\) and
\(\seteqnumber{0}{3.}{0}\)\begin{align*} v &=\left [\left \langle v, v_{1}\right \rangle v_{1}+\left \langle v, v_{2}\right \rangle v_{2}+\cdots +\left \langle v, v_{r}\right \rangle v_{r}\right ]+v_{0} \in W+W^{\perp } \\ V=& W \oplus W^{\perp } \end{align*}
This completes the proof. □
Proof : \(\operatorname {dim} V=\operatorname {dim}\left (W \oplus W^{\perp }\right )=\operatorname {dim} W+\operatorname {dim} W^{\perp }\). □
Proof : Let \(w \in W\). Then for any \(u \in W^{\perp },\langle w, u\rangle =0\). Hence \(w \in \left ( \left (W^{\perp }\right )^{\perp }\right )\). Thus
\(\seteqnumber{0}{3.}{0}\)
\begin{align}
\label {thm67eq1} W \subseteq \left (W^{\perp }\right )^{\perp }
\end{align}
Now by Theorem 3.3.4, \(V=W \oplus W^{\perp }\).
Also \(V=W^{\perp } \oplus \left (W^{\perp }\right )^{\perp }\). Hence
\begin{align} \label {thm67eq2} \operatorname {dim} W=\operatorname {dim}\left (W^{\perp }\right )^{\perp } \end{align} From (3.1) and (3.2) we get \(W=\left (W^{\perp }\right )^{\perp }\). □
Solved Problems
Solution : Let \(u \in S_{2}^{\perp }\).
Then \(\langle u, v\rangle =0\) for all \(v \in S_{2}\).
But \(S_{1} \subseteq S_{2}\). Hence \(\langle u, v\rangle =0\) for all \(v \in S_{1}\).
Hence \(u \in S_{1}^{\perp }\). Thus \(S_{2}^{\perp } \subseteq S_{1}^{\perp }\).
Solution :
(i) We know that \(W_{1} \subset W_{1}+W_{2}\).
\[\therefore \left (W_{1}+W_{2}\right )^{\perp } \subseteq W_{1}^{\perp } \]
Similarly,
\[\left (W_{1}+W_{2}\right )^{\perp } \subseteq W_{2}^{\perp }\]
Hence
\(\seteqnumber{0}{3.}{2}\)
\begin{align}
\label {sec63prob2eq1} \left (W_{1}+W_{2}\right )^{\perp } \subseteq W_{1}^{\perp } \cap W_{2}^{\perp }
\end{align}
Now, let \(w \in W_{1}^{\perp } \cap W_{2}^{\perp }\).
Then \(w \in W_{1}^{\perp }\) and \(w \in W_{2}^{\perp }\).
\(\langle w, u\rangle =0\) for all \(u \in W_{1}\) and \(W_{2}\).
Now, let \(v \in W_{1}+W_{2}\).
Then \(v=v_{1}+v_{2}\) where \(v_{1} \in W_{1}\) and \(v_{2} \in W_{2}\).
\begin{align*} \langle w, v\rangle &=\left \langle w, v_{1}+v_{2}\right \rangle \\ &=\left \langle w, v_{1}\right \rangle +\left \langle w, v_{2}\right \rangle \\ &=0+0\left (\text { since } v_{1} \in W_{1} \text { and } v_{2} \in W_{2}\right ) \\ &=0 \end{align*} Hence \(w \in \left (W_{1}+W_{2}\right )^{\perp }\).
\(\seteqnumber{0}{3.}{3}\)\begin{align} \label {sec63prob2eq2} \therefore \quad W_{1}^{\perp } \cap W_{2}^{\perp } \subseteq \left (W_{1}+W_{2}\right )^{\perp } \end{align} From (3.3) and (3.4) we get
\[ \left (W_{1}+W_{2}\right )^{\perp }=W_{1}^{\perp } \cap W_{2}^{\perp } . \]
(ii) Proof is similar to that of (i).
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