Homework 8

(30 p.) For each of the following $\mathbb{F}$-vector spaces $V$ and $\Phi: V \times V \rightarrow \mathbb{F}$, show that $\Phi$ is not an inner product on $V$.
(a) $\mathbb{F}=\mathbb{C}, V=\mathbb{C}^2, \Phi(x, y)=x_1 \bar{y}_2+x_2 \bar{y}_1+k x_2 \bar{y}_2$ with $k \in \mathbb{R}$ fixed.
(b) $\mathbb{F}=\mathbb{R}, V=\mathbb{R}^3, \Phi(u, v)=\|u \times v\|$, where $u \times v$ the usual cross product of vectors in space.
(c) $\mathbb{F}=\mathbb{R}, V=\mathbb{R}_3[x], \Phi(p, q)=p(0)q(0)+p(2)q(2)$.

(a) (1) consider $\Phi(x+z, y)=(x_1+z_1)\bar{y}_2+(x_2+z_2)\bar{y}_1+k(x_2+z_2)\bar{y}_2$
$=x_1\bar{y}_2+x_2\bar{y}_1+kx_2\bar{y}_2+z_1\bar{y}_2+z_2\bar{y}_1+kz_2\bar{y}_2$ $=\Phi(x, y)+\Phi(z, y)$

$\Phi(\lambda x, y) = \lambda x_1 \bar{y}_2 + \lambda x_2 \bar{y}_1 + \lambda k x_2 \bar{y}_2$ $= \lambda(x_1 \bar{y}_2 + x_2 \bar{y}_1 + k x_2 \bar{y}_2)$ $= \lambda \Phi(x, y)$

(2) $\Phi(x,y)=\overline{x_1}y_{2}+x_{2}\overline{y_1}+kx_{2}\overline{y_2}=\overline {\overline{x_1}y_2+\overline{x_2}y_1+\overline{kx_2}y_2}=\overline{\overline{x_1}y_2+\overline{x_2}y_1+k\overline{x_2}y_2} $ since $k\in \R$

$=\overline{y_1\overline{x_2}+y_2\overline{x_1}+ky_2\overline{x_2}}=\overline{\Phi(y,x)}$

(3) For $\forall x \neq 0$, $\Phi(x,x)=x_{1}\bar{x}_{2}+x_{2}\bar{x}_{1}+kx_{2}\bar{x}_{2}=x_{1}\bar{x}_{2}+x _{2}\bar{x}_{1}+k||x_{2}||^{2}$ is not always $>0$.
Counter example: $x = (1,1)$ and $k = -1000$. $\Phi(x,x) = 1+1 - 1000 = -998 < 0$.
Thus this is not inner product.

(b) Consider $\Phi(u+w, v) = ||(u+w) \times v|| = ||u \times v + w \times v|| \le ||u \times v|| + ||w \times v|| = \Phi(u,v) + \Phi(w,v)$
Counter example: $u \times v = (1,1,2)$, $w \times v = (0,2,0)$
$||u \times v + w \times v|| = ||(1,3,2)|| = \sqrt{14}$, $||u\times v||+||w\times v||=\sqrt{6}+2$
Thus $||u \times v + w \times v|| \ne ||u \times v|| + ||w \times v||$.
Thus this is not inner product.

(c) (1) Consider $\Phi(p+t, q) = (p+t)(0)q(0) + (p+t)(2)q(2)$
$= [p(0)+t(0)]q(0) + [p(2)+t(2)]q(2)$
$= p(0)q(0) + t(0)q(0) + p(2)q(2) + t(2)q(2)$
$= \Phi(p,q) + \Phi(t,q)$

$\Phi(\lambda p, q) = \lambda p(0)q(0) + \lambda p(2)q(2) = \lambda [p(0)q(0) + p(2)q(2)] = \lambda \Phi(p,q)$

(2) $\Phi(p,q) = p(0)q(0) + p(2)q(2) = q(0)p(0) + q(2)p(2)$. Since $F=\mathbb{R}, V=\mathbb{R}_3[x]$
$= \overline{q(0)p(0) + q(2)p(2)} = \overline{\Phi(q,p)}$

(3) $\Phi(P,P) = P(0)^2 + P(2)^2$. This can be $0$.
Counter example: $P(x)=x^2-2x$. $\implies P(0)=0$, $P(2)=0$, $P(x) \ne 0$
$\implies \Phi(P,P) = 0+0=0$
Thus this is not inner product.
(30 p.) Let $V = C([a, b])$ be the $\mathbb{R}$-vector space of continuous real-valued functions on $[a, b]$, and let $w \in V$ be a fixed function such that $w(x) > 0$ for all $x \in [a, b]$.
(a) Show that the formula $\langle f, g \rangle = \int_a^b f(x)g(x)w(x) \, dx$ defines an inner product on $V$.
(b) Let $S$ be the subspace of $C([0, 1])$ generated by the set $\mathcal{B} = \{1, x, e^x\}$. Show that $\mathcal{B}$ is a basis for $S$ and compute the matrix $[\langle -, - \rangle]_{\mathcal{B}}$ with $w(x) = 1$ for all $x \in [0, 1]$.

a) (1) $\langle f+h, g \rangle = \int_a^b (f+h)(x) g(x) w(x) dx = \int_a^b [f(x)+h(x)] g(x) w(x) dx$
$= \int_a^b f(x) g(x) w(x) dx + \int_a^b h(x) g(x) w(x) dx = \langle f, g \rangle + \langle h, g \rangle$
$\langle \lambda f, g \rangle = \int_a^b \lambda f(x) g(x) w(x) dx = \lambda \int_a^b f(x) g(x) w(x) dx = \lambda \langle f, g \rangle$
(2) $\langle f, g \rangle = \int_a^b f(x) g(x) w(x) dx = \int_a^b g(x) f(x) w(x) dx$
Since $F = \mathbb{R}$ and functions are real-valued, then $\langle f, g \rangle = \overline{\int_a^b g(x) f(x) w(x) dx} = \overline{\langle g, f \rangle}$

(3) $\forall f\ne0,\langle f,f\rangle=\int_{a}^{b}f(x)^{2}w(x)dx$.
let $t \in [a,b]$ s.t. $f(t) \ne 0$. Then $f$ is continuous at $t$.
① $f(t) > 0$. Then $\begin{cases} f(t)>0 \\ w(t)>0 \end{cases} \forall x \in (t-\delta, t+\delta)$.
Then $\lang f,f\rang = \int_{a}^{b}f(x)^{2} w(x) dx = \int_{t-\delta}^{t+\delta}f(x)^{2} w(x) d x + \int_{[a,b]\setminus (t-\delta, t+\delta)}f(x)^{2} w(x) dx$.
Then the first term $>0$. the second term $\ge 0$. Then $\langle f,f\rangle>0$.
② $f(t)<0$. Then $\begin{cases} f(t)<0 \\ w(t)>0 \end{cases} \forall x \in (t-\delta, t+\delta)$.
Then $\lang f,f\rang = \int_{a}^{b}f(x)^{2} w(x) dx = \int_{t-\delta}^{t+\delta}f(x)^{2} w(x) d x + \int_{[a,b]\setminus (t-\delta, t+\delta)}f(x)^{2} w(x) dx$.
Then the first term $>0$. the second term $\ge 0$. Then $\lang f,f\rang >0$.
Thus $\lang f,f\rang >0, \forall f \ne 0$. Thus this is a inner product.

(b) To prove $B$ is a basis. Since $S$ is generated by $B$, then just need to prove $B$ is linearly independent.
Consider $0 = a_1 + a_2x + a_3e^x$, when $x=0$: $a_1 + a_3 = 0$, $x=1$: $a_1 + a_2 + ea_3 = 0$
$x=\frac{1}{2}$: $a_1 + \frac{1}{2}a_2 + \sqrt{e}a_3 = 0$, then $a_3 = a_2 = a_1 = 0$. Thus $B$ is basis.

$w(x)=1$. $[\langle-,- \rangle]_B = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ \vdots & \vdots & \vdots \\ a_{31} & a_{32} & a_{33} \end{pmatrix}$ where $a_{ij}= \langle v_{j}, v_{i} \rangle$, $1 \leq i, j \leq 3$.
$a_{11}= \langle v_{1}, v_{1}\rangle = \langle 1, 1 \rangle = \int_{0}^{1}w(x) dx = \int_{0}^{1}1 dx = 1$
$a_{12}=\langle v_{2},v_{1}\rangle=\langle x,1\rangle=\int_{0}^{1}xdx=\frac{1}{2} x^{2}|_{0}^{1}=\frac{1}{2}$
$a_{13}=\langle v_{3},v_{1}\rangle=\langle e^{x},1\rangle=\int_{0}^{1}e^{x}dx=e^{x} |_{0}^{1}=e-1$
$a_{21}=\langle v_{1},v_{2}\rangle=\langle1,x\rangle=\int_{0}^{1}xdx=\frac{1}{2}$
$a_{22}= \langle v_{2}, v_{2} \rangle = \langle x, x \rangle = \int_{0}^{1} x^{2} dx = \frac{1}{3}x^{3}|_{0}^{1} = \frac{1}{3}$
$a_{23}=\langle v_{3},v_{2}\rangle=\langle e^{x},x\rangle=\int_{0}^{1}xe^{x}dx=xe ^{x}|_{0}^{1}-\int_{0}^{1}e^{x}dx=e-e+1=1$
$a_{31}= \langle v_{1}, v_{3}\rangle = e-1$
$a_{32}= \langle v_{2}, v_{3}\rangle = 1$
$a_{33}= \langle v_{3}, v_{3}\rangle = \int_{0}^{1}e^{2x}dx = \frac{1}{2}e^{2x}|_{0} ^{1}= \frac{1}{2}e^{2}-\frac{1}{2}$
Thus $[\langle-,- \rangle]_{B}= \begin{pmatrix} 1 & \frac{1}{2} & e -1 \\ \frac{1}{2} & \frac{1}{3} & 1 \\ e -1 & 1 & \frac{1}{2}e^{2}-\frac{1}{2} \end{pmatrix}$
(40 p.) Let $V$ be a vector space of dimension $n$, and let $B = \{v_1, \dots, v_n\}$ be a basis for $V$.
(a) Show that there exists a unique inner product on $V$ for which $B$ is orthonormal.
(b) Find the inner product in the following cases:
i. $V = \mathbb{R}^2$ and $B = \{(-2, 1), (1, 1)\}$,
ii. $V = \mathbb{C}^3$ and $B = (1, i, 1), (0, 1, i), (0, 0, 1)$.

(a) Uniqueness:
Since $\Phi(v,w)=\overline{[w]_{B}^{t}}[\Phi]_{B}[v]_{B}$. Then inner product is unique iff $[\Phi]_B$ is unique. And we know when $B$ is orthonormal, $[\Phi]_B = I_{n \times n}$.
Thus, it exists, inner product is unique.
Existence.
Since $[\Phi]_B = I_{n \times n}$, then $[\Phi]_E = C(E, B)^T [\Phi]_B \overline{C(E, B)} = (C(B, E)^{-1})^T \overline{C(B, E)^{-1}}$.
Since $B$ is basis and $E$ is basis, then $C(B, E)$ exists, then $[\Phi]_E$ exists. Then inner product exists

(b)
①Since $[\Phi]_{E}=(C(B,E)^{-1})^{t}\cdot\overline{C(B,E)^{-1}}$ and $C(B,E) = \begin{pmatrix} -2 & 1 \\ 1 & 1 \end{pmatrix}$
Then $[\Phi]_{E}$
$= \begin{pmatrix} -\frac{1}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{2}{3} \end{pmatrix}^{t} \begin{pmatrix} -\frac{1}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{2}{3} \end{pmatrix}$ $= \begin{pmatrix} \frac{2}{9} & \frac{1}{9} \\ \frac{1}{9} & \frac{5}{9} \end{pmatrix}$
Then $\Phi((x_{1},x_{2}),(y_{1},y_{2}))=\frac{2}{9}x_1y_1+\frac{1}{9}x_1y_2+\frac{1}{9}x_2y_1+\frac{5}{9}x_2y_2$
② Since $[\Phi]_{E}=(C(B,E)^{-1})^{t}\cdot\overline{C(B,E)^{-1}}$ and $C(B,E)= \begin{pmatrix} 1 & 0 & 0 \\ i & 1 & 0 \\ 1 & i & 1 \end{pmatrix}$
Then $[\Phi]_{E}= \begin{pmatrix} 1 & 0 & 0 \\ -i & 1 & 0 \\ -2 & -i & 1 \end{pmatrix}^{t}\overline{\begin{pmatrix}1 & 0 & 0\\ -i & 1 & 0\\ -2 & -i & 1\end{pmatrix}}$ $= \begin{pmatrix} 6 & -3i & -2 \\ 3i & 2 & -i \\ -2 & i & 1 \end{pmatrix}$
Then $\Phi\left((x_{1},x_{2},x_{3}\right),\left(y_{1},y_{2},y_{3}))=6x_{1}\bar{y_1}-3i x_{1}\bar{y_2}-2x_{1}\bar{y_3}+3ix_{2}\bar{y_1}+2x_{2}\bar{y_2}-ix_{2}\bar{y_3}-2 x_{3}\bar{y_1}+ix_{3}\bar{y_2}+x_{3}\bar{y_3}\right.$

by formula $\Phi(v,w)=\overline{[w]_{B}^{t}}[\Phi]_{B}[v]_{B}$

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