Advanced Algebra Assignment NO.4

1. Check the following functions are inner products:

(a) Let $C[-1,1]$ be the vector space of continuous complex-valued functions on the interval $[-1,1]$ . Define $\langle\cdot,\cdot\rangle:C[-1,1]\times C[-1,1]\rightarrow\mathbb{C}$ by $\langle f,g \rangle=\int_{-1}^1f(x)\overline{g(x)}\mathrm{d}x$ for all $f(x),g(x)\in C[-1,1]$ .

Defination: $\forall f,g\in C[-1,1],f(x)\overline{g(x)}:[-1,1]\rightarrow\mathbb{C}\Rightarrow\langle f,g\rangle\in\mathbb{C}$ .
Conjugate symmetry: $\forall f,g\in C[-1,1],\langle f,g\rangle=\int_{-1}^1f(x)\overline{g(x)}\mathrm{d}x=\overline{\int_{-1}^1g(x)\overline{f(x)}\mathrm{d}x}=\overline{\langle g,f\rangle}$ .
Linearity: $\langle f+kg,h\rangle=\int_{-1}^1(f+kg)(x)\overline{h(x)}\mathrm{d}x=\int_{-1}^1f(x)\overline{h(x)}+kg(x)\overline{h(x)}\mathrm{d}x=\int_{-1}^1f(x)\overline{h(x)}\mathrm{d}x+k\int_{-1}^1g(x)\overline{h(x)}\mathrm{d}x=\langle f,h\rangle+k\langle g,h \rangle,\forall f,g,h\in C[-1,1],k\in\mathbb{C}$ .
Positive-definiteness: $\forall f\in C[-1,1]/{0}\langle f,f\rangle=\int_{-1}^1f(x)\overline{f(x)}\mathrm{d}x\xRightarrow{f(x)\overline{(f(x)}\in\R^+}\langle f,f\rangle>0$ and $\langle 0,0\rangle=0$ .
$Q.E.D.$

(b) The function $\langle\cdot,\cdot\rangle:\mathbb{R}[x]\times\mathbb{R}[x]\rightarrow\R$ is defined by $\langle p,q\rangle=p(0)q(0)+\int_{-1}^{1}p'(x)q'(x)\mathrm{d}x$ for all $p(x),q(x)\in\R[x]$ .

Defination: $\forall p(x),q(x)\in\R[x],p(0)q(0)\in\R,p'(x)q'(x)\in\R[x]\Rightarrow\langle p,q\rangle\in\R$ .
Conjugate symmetry: $\forall p(x),q(x)\in\R[x],\langle p,q\rangle=p(0)q(0)+\int_{-1}^1p'(x)q'(x)\mathrm{d}x=q(0)p(0)+\int_{-1}^1q'(x)p'(x)\mathrm{d}x=\langle q,p\rangle$ .
Linearity: $\langle p+kq,r\rangle=(p+kq)(0)r(0)+\int_{-1}^1(p+kq)'(x)r'(x)\mathrm{d}x=p(0)r(0)+\int_{-1}^1p'(x)r'(x)\mathrm{d}x+k(q(0)r(0)+\int_{-1}^1q'(x)r'(x)\mathrm{d}x)=\langle p,r\rangle+k\langle q,r \rangle,\forall p(x),q(x),r(x)\in\R[x],k\in\R$ .
Positive-definiteness: $\forall p(x)\in(\R[x])^*,\langle p,p\rangle=(p(0))^2+\int_{-1}^1p^2(x)\mathrm{d}x>0$ and $\langle 0,0\rangle=0$ .
$Q.E.D.$

(c) The function $\langle\cdot,\cdot\rangle:\R[x]\times\R[x]\rightarrow\R$ is defined by $\langle p,q\rangle=\int_0^\infty p(x)q(x)e^{-x}\mathrm{d}x$ for all $p(x),q(x)\in\R[x]$ .

Defination: $\forall p(x),q(x)\in\R[x],p(x)q(x)e^{-x}\in\R\Rightarrow\langle p,q\rangle\in\R$ .
Conjugate symmetry: $\forall p(x),q(x)\in\R[x],\langle p,q\rangle=\int_0^\infty p(x)q(x)e^{-x}\mathrm{d}x=\int_0^\infty q(x)p(x)e^{-x}\mathrm{d}x=\langle q,p\rangle$ .
Linearity: $\langle p+kq,r\rangle=\int_0^\infty (p+kq)(x)r(x)e^{-x}\mathrm{d}x=\int_0^\infty p(x)r(x)e^{-x}\mathrm{d}x+k\int_0^\infty q(x)r(x)e^{-x}\mathrm{d}x=\langle p,r\rangle+k\langle q,r \rangle,\forall p(x),q(x),r(x)\in\R[x],k\in\R$ .
Positive-definiteness: $\forall p(x)\in(\R[x])^*,\langle p,p\rangle=\int_0^\infty p^2(x)e^{-x}\mathrm{d}x\xRightarrow{p^2(x)>0,e^{-x}>0}\langle p,p\rangle>0$ and $\langle 0,0\rangle=0$ .
$Q.E.D.$

2. Let $(V,\langle\cdot,\cdot\rangle)$ be an inner product space. Show that the function $\|\cdot\|:V\rightarrow\R$ defined by $\|v\|=\sqrt{\langle v,v\rangle}$ , for any $v\in V$ , is a norm.

Defination: $\forall v\in V,\langle v,v\rangle\ge0\Rightarrow\|v\|=\sqrt{\langle v,v\rangle}\in[0,\infty)$ .
Positive-definiteness: $\|v\|=0\Leftrightarrow\sqrt{\langle v,v\rangle}=0\Leftrightarrow\langle v,v\rangle=0\Leftrightarrow v=0$ .
Absolute homogeneity: $\forall v\in V,\lambda\in \mathbb{K},\|\lambda v\|=\sqrt{\langle \lambda v,\lambda v\rangle}=\sqrt{\lambda^2\langle v,v\rangle}=|\lambda|\|v\|$ .
Subadditivity: Notice that $\forall u,v\in V,\lambda\in\mathbb{K}$ , we have $0\leq\langle u+\lambda v,u+\lambda v\rangle=\langle u,u\rangle+\overline\lambda\langle u,v\rangle+\lambda\overline{\langle u,v\rangle}+\lambda\overline\lambda\langle v,v\rangle$ . Without loss of generality, let $v\neq0,\lambda=-\frac{\langle u,v\rangle}{\langle v,v\rangle}$ . (When $v=0$ , $|\langle u,0\rangle|^2=0=\langle u,u\rangle\langle0,0\rangle$ . As $\langle u,v\rangle\in\mathbb{K},\langle v,v\rangle\in\mathbb{K}^*$ , we can say $\lambda\in\mathbb{K}$ .) So $\langle u,u\rangle-2\frac{|\langle u,v\rangle|^2}{\langle v,v\rangle}+\frac{|\langle u,v\rangle|^2}{\langle v,v\rangle}\ge0\Rightarrow|\langle u,v\rangle|^2\leq\langle u,u\rangle\langle v,v\rangle$ . In this way, $\forall u,v\in V,\|u+v\|=\sqrt{\langle u+v,u+v\rangle}=\sqrt{\langle u,u\rangle+\langle v,v\rangle+2\mathrm{Re}\langle u,v\rangle}\leq\sqrt{\|u\|^2+\|v\|^2+2|\langle u,v\rangle|}\leq\sqrt{\|u\|^2+\|v^2\|+2\|u\|\|v\|}=\|u\|+\|v\|$ .
$Q.E.D.$

3. Find a polynomial $q(x)\in\R_2[x]$ such that $\int_0^1p(x)\cos(\pi x)\mathrm{d}x=\int_0^1p(x)q(x)\mathrm{d}x$ for every $p(x)\in\R_2[x]$ .

Consider $\langle p,q\rangle=\int_0^1p(x)q(x)\mathrm{d}x\in\R$ . Notice that $\alpha=\{1,x-\frac{1}{2},x^2-x+\frac{1}{6}\}$ is an orthogonal basis for $\R_2[x]$ . So the $q(x)$ satisfied that $\langle p,\cos(\pi x)\rangle=\langle p,q\rangle,\forall p\in\R_2[x]$ , which means $\langle \alpha_i,\cos(\pi x)\rangle=\langle\alpha_i,q\rangle,\forall \alpha_i\in\alpha$ . So $q(x)=\sum_{i=1}^3\frac{\langle \alpha_i,q\rangle}{\langle\alpha_i,\alpha_i\rangle}\alpha_i=\sum_{i=1}^3\frac{\langle \alpha_i,\cos(\pi x)\rangle}{\langle\alpha_i,\alpha_i\rangle}\alpha_i=0\cdot1-\frac{24}{\pi^2}(x-\frac{1}{2})+0\cdot(x^2-x+\frac{1}{6})=-\frac{24}{\pi^2}x+\frac{12}{\pi^2}$ .

4. Suppose $v_1,\dots,v_m$ is a linearly independent list in $V$ . Explain why the orthonormal list produced by the formulas of the Gram-Schmidt procedure is the only orthonormal list $e_1,\dots,e_m$ in $V$ such that $\langle v_k,e_k\rangle > 0$ and $\mathrm{Span}(v_1,\dots,v_k) = \mathrm{Span}(e_1,\dots,e_k)$ for each $k = 1,\dots,m$ .

Existence: $f_k:=v_k-\sum_{i=1}^{k-1}\frac{\langle v_k,f_i\rangle}{\langle f_i,f_i\rangle}f_i$ , we have $e_k=\frac{f_k}{\|f_k\|}$ . According to the defination of $e_k$ , we can find $(e_1,e_2,\dots,e_k)=(v_1,v_2,\dots,v_k)M_{k\times k}$ , and $M_{k\times k}$ is an upper triangular matrix. So $M_{k\times k}$ is invertible, which proves that $\mathrm{Span}(v_1,\dots,v_k) = \mathrm{Span}(e_1,\dots,e_k)$ for each $k = 1,\dots,m$ . So $\langle v_k,e_k\rangle=\frac{1}{\|f_k\|}\langle v_k,f_k\rangle=\frac{1}{\|f_k\|}\langle f_k+\sum_{i=1}^{k-1}\frac{\langle v_k,f_i\rangle}{\langle f_i,f_i\rangle}f_i,f_k\rangle=\frac{1}{\|f_k\|}\langle f_k,f_k\rangle>0$
Uniqueness: Consider that an orthonormal list $e'1,\dots,e'_m\neq e_1,\dots,e_m$ such that $\langle v_k,e'_k\rangle > 0$ and $\mathrm{Span}(v_1,\dots,v_k) = \mathrm{Span}(e'_1,\dots,e'_k)$ for each $k = 1,\dots,m$ . When $k=1$ , we get $\mathrm{Span}(v_1)=\mathrm{Span}(e_1)=\mathrm{Span}(e'_1)\Rightarrow e'_1=\lambda e_1\xRightarrow{\|e'_1\|=1}|\lambda|=1\xRightarrow{\R\ni\langle v_1,e'_1\rangle=\overline\lambda\langle v_1,e_1\rangle>0\Rightarrow\R\ni\lambda>0}\lambda=1\Rightarrow e'_1=e_1$ . Suppose that $\forall i\in \{1,\dots,t\},t<m,e'_i=e_i$ , when $k=t+1$ , we get $\mathrm{Span}(v_1,\dots,v{t+1})=\mathrm{Span}(e_1,\dots,e_{t+1})=\mathrm{Span}(e'_1,\dots,e'_{t+1})=\mathrm{Span}(e_1,\dots,e_t,e'_{t+1})\Rightarrow e'_{t+1}=\lambda e_{t+1}+\sum_{i=1}^t\langle e'_{t+1},e_i\rangle e_i=\lambda e_{t+1}+\sum_{i=1}^t\langle e'_{t+1},e'i\rangle e_i=\lambda e_{t+1}\xRightarrow{\|e'_{t+1}\|=|\lambda|\|e_{t+1}\|=1}|\lambda|=1\xRightarrow{\langle v{t+1},e'_{t+1}\rangle=\overline\lambda\langle v{t+1},e_{t+1}\rangle>0\Rightarrow\lambda>0}\lambda=1\Rightarrow e'_{t+1}=e_{t+1}$ . So, according to the Mathematical Induction, we can say $\forall i,e'_i=e_i$ , which is contrary to our assumption. So we have proved the uniqueness.

5. Suppose $\langle\cdot,\cdot\rangle_1$ and $\langle\cdot,\cdot\rangle_2$ are inner products on $V$ such that $\langle u,v\rangle_1 = 0$ if and only if $\langle u,v\rangle_2 = 0$ . Prove that there is a positive number $c$ such that $\langle u,v\rangle_1 = c\langle u,v\rangle_2$ for every $u,v \in V$ .

First of all, we defaultly accept AC, so according to Zorn's lemma, we can prove for all vector spaces, which is not zero space, there exists an orthonormal basis. Let $\alpha$ be an orthonormal basis for $V,|\cdot|_1$ ,which $|u|_i:=\sqrt{\langle u,u\rangle_i}$ . In this way, $\forall i\neq j,\langle \alpha_i,\alpha_j\rangle_1=0\Rightarrow\langle \alpha_i,\alpha_j\rangle_2=0;\langle\alpha_i,\alpha_i\rangle_1=1\Rightarrow\langle\alpha_i,\alpha_i\rangle_2=c_i\neq0$ . Notice that $\langle\alpha_i+\alpha_j,\alpha_i-\alpha_j\rangle_1=\langle\alpha_i,\alpha_i\rangle_1-\langle\alpha_j,\alpha_j\rangle_1=0\Rightarrow\langle\alpha_i+\alpha_j,\alpha_i-\alpha_j\rangle_2=\langle\alpha_i,\alpha_i\rangle_2-\langle\alpha_j,\alpha_j\rangle_2=0\Rightarrow c_i=c_j$ . So $\forall i,j,\langle\alpha_i,\alpha_j\rangle_1=c\langle\alpha_i,\alpha_j\rangle_2$ , which means there is a positive number $c$ such that $\langle u,v\rangle_1 = c\langle u,v\rangle_2$ for every $u,v \in V$ .

6. Define an operator $S:F^2\rightarrow F^2$ by $S(w,z)=(-z,w)$ .

(a) Find a formula for $S^*$ .

$\langle S^*(a,b),(c,d)\rangle=\langle(a,b),S(c,d)\rangle=\langle(a,b),(-d,c)\rangle=-a\overline d+b\overline c=\langle(b,-a),(c,d)\rangle\Rightarrow S^*(a,b)=(b,-a)$ .

(b) Show that $S$ is normal but not self-adjoint.

Notice that $S^*S(w,z)=S^*(-z,w)=(w,z)=S(z,-w)=SS^*(w,z)$ , so $SS^*=S^*S=\mathrm{id}$ , which means $S$ is normal but not self-adjoint.

(c) Find all eigenvalues of $S$ .

$S(w,z)=\lambda(w,z)=(-z,w)\Rightarrow\lambda w=-z,\lambda z=w\Rightarrow -\lambda^2w=w\Rightarrow\lambda=i,-i,(F=\mathbb{C})$ or $\lambda$ doesn't exist, ( $F=\R$ ).

7. Suppose $T \in \mathcal{L}(V)$ and $a_0 +a_1z +a_2z^2 +\cdots+a_{m−1}z^{m−1} +z^m$ is the minimal polynomial of $T$ . Find the minimal polynomial of $T^*$ .

Let $f(z):=a_0 +a_1z +a_2z^2 +\cdots+a_{m−1}z^{m−1} +z^m,\overline f(z):=\overline{a_0}+\overline{a_1}z+\overline{a_2}z^2+\cdots+\overline{a_{m-1}}z^{m-1}+z^m$ . Notice that $\langle a_k T^ku,v\rangle=a_k\langle T^{k-1}u,T^*v\rangle=\cdots=\langle u,\overline{a_k}(T^*)^kv\rangle$ . So, $(a_kT^k)^*=\overline{a_k}(T^*)^k,(f(T))^*=\overline f(T^*)=0$ . On the other hand, $\forall g\in\mathbb{K}[z],\deg g<m$ , we know $g(z)$ is not the zero polynomial of $T$ , which means $(g(T))^*=\overline g(T^*)\neq0$ . Combining the two conclusions, we can say the minimal polynomial of $T^*$ is $\overline f(z)$ .

8. Prove or give a counterexample: If $T \in \mathcal{L}(V)$ and there is an orthonormal basis $e_1,\dots, e_n$ of $V$ such that $\|Te_k\| = \|T^∗e_k\|$ for each $k = 1,\dots,n$ , then $T$ is normal.

Counterexample: Consider $V=R^2,T:e_1\mapsto e_1+e_2,e_2\mapsto -e_1-e_2$ , we have $\langle T^*(a,b),(c,d)\rangle=\langle(a,b),T(c,d)\rangle=\langle(a,b),(c-d,c-d)\rangle=\langle(a+b,-a-b),(c,d)\rangle$ . So, $T^*:e_k\mapsto e_1-e_2,k=1,2$ . Notice that $\|Te_k\|=\sqrt2=\|T^*e_k\|,k=1,2;TT^*:e_k\mapsto2e_1+2e_2,k=1,2;T^*T:e_1\mapsto2e_1-2e_2,e_2\mapsto-2e_1+2e_2$ . Therefore, $TT^*\neq T^*T$ , which means $T$ is not normal.

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Advanced Algebra Assignment NO.4

1. Check the following functions are inner products:

(a) Let $C[-1,1]$ be the vector space of continuous complex-valued functions on the interval $[-1,1]$ . Define $\langle\cdot,\cdot\rangle:C[-1,1]\times C[-1,1]\rightarrow\mathbb{C}$ by $\langle f,g \rangle=\int_{-1}^1f(x)\overline{g(x)}\mathrm{d}x$ for all $f(x),g(x)\in C[-1,1]$ .

(b) The function $\langle\cdot,\cdot\rangle:\mathbb{R}[x]\times\mathbb{R}[x]\rightarrow\R$ is defined by $\langle p,q\rangle=p(0)q(0)+\int_{-1}^{1}p'(x)q'(x)\mathrm{d}x$ for all $p(x),q(x)\in\R[x]$ .

(c) The function $\langle\cdot,\cdot\rangle:\R[x]\times\R[x]\rightarrow\R$ is defined by $\langle p,q\rangle=\int_0^\infty p(x)q(x)e^{-x}\mathrm{d}x$ for all $p(x),q(x)\in\R[x]$ .

2. Let $(V,\langle\cdot,\cdot\rangle)$ be an inner product space. Show that the function $\|\cdot\|:V\rightarrow\R$ defined by $\|v\|=\sqrt{\langle v,v\rangle}$ , for any $v\in V$ , is a norm.

3. Find a polynomial $q(x)\in\R_2[x]$ such that $\int_0^1p(x)\cos(\pi x)\mathrm{d}x=\int_0^1p(x)q(x)\mathrm{d}x$ for every $p(x)\in\R_2[x]$ .

6. Define an operator $S:F^2\rightarrow F^2$ by $S(w,z)=(-z,w)$ .

(a) Find a formula for $S^*$ .

(b) Show that $S$ is normal but not self-adjoint.

(c) Find all eigenvalues of $S$ .

7. Suppose $T \in \mathcal{L}(V)$ and $a_0 +a_1z +a_2z^2 +\cdots+a_{m−1}z^{m−1} +z^m$ is the minimal polynomial of $T$ . Find the minimal polynomial of $T^*$ .

8. Prove or give a counterexample: If $T \in \mathcal{L}(V)$ and there is an orthonormal basis $e_1,\dots, e_n$ of $V$ such that $\|Te_k\| = \|T^∗e_k\|$ for each $k = 1,\dots,n$ , then $T$ is normal.

Advanced Algebra Assignment NO.3

Advanced Algebra Assignment NO.5

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1. Check the following functions are inner products:

(a) Let C[-1,1] be the vector space of continuous complex-valued functions on the interval [-1,1]. Define \langle\cdot,\cdot\rangle:C[-1,1]\times C[-1,1]\rightarrow\mathbb{C} by \langle f,g \rangle=\int_{-1}^1f(x)\overline{g(x)}\mathrm{d}x for all f(x),g(x)\in C[-1,1].

(b) The function \langle\cdot,\cdot\rangle:\mathbb{R}[x]\times\mathbb{R}[x]\rightarrow\R is defined by \langle p,q\rangle=p(0)q(0)+\int_{-1}^{1}p'(x)q'(x)\mathrm{d}x for all p(x),q(x)\in\R[x].

(c) The function \langle\cdot,\cdot\rangle:\R[x]\times\R[x]\rightarrow\R is defined by \langle p,q\rangle=\int_0^\infty p(x)q(x)e^{-x}\mathrm{d}x for all p(x),q(x)\in\R[x].

2. Let (V,\langle\cdot,\cdot\rangle) be an inner product space. Show that the function \|\cdot\|:V\rightarrow\R defined by \|v\|=\sqrt{\langle v,v\rangle}, for any v\in V, is a norm.

3. Find a polynomial q(x)\in\R_2[x] such that \int_0^1p(x)\cos(\pi x)\mathrm{d}x=\int_0^1p(x)q(x)\mathrm{d}x for every p(x)\in\R_2[x].

5. Suppose \langle\cdot,\cdot\rangle_1 and \langle\cdot,\cdot\rangle_2 are inner products on V such that \langle u,v\rangle_1 = 0 if and only if \langle u,v\rangle_2 = 0. Prove that there is a positive number c such that \langle u,v\rangle_1 = c\langle u,v\rangle_2 for every u,v \in V.

6. Define an operator S:F^2\rightarrow F^2 by S(w,z)=(-z,w).

(a) Find a formula for S^*.

(b) Show that S is normal but not self-adjoint.

(c) Find all eigenvalues of S.

7. Suppose T \in \mathcal{L}(V) and a_0 +a_1z +a_2z^2 +\cdots+a_{m−1}z^{m−1} +z^m is the minimal polynomial of T. Find the minimal polynomial of T^*.

8. Prove or give a counterexample: If T \in \mathcal{L}(V) and there is an orthonormal basis e_1,\dots, e_n of V such that \|Te_k\| = \|T^∗e_k\| for each k = 1,\dots,n, then T is normal.

Advanced Algebra Assignment NO.3

Advanced Algebra Assignment NO.5

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(a) Let $C[-1,1]$ be the vector space of continuous complex-valued functions on the interval $[-1,1]$ . Define $\langle\cdot,\cdot\rangle:C[-1,1]\times C[-1,1]\rightarrow\mathbb{C}$ by $\langle f,g \rangle=\int_{-1}^1f(x)\overline{g(x)}\mathrm{d}x$ for all $f(x),g(x)\in C[-1,1]$ .

(b) The function $\langle\cdot,\cdot\rangle:\mathbb{R}[x]\times\mathbb{R}[x]\rightarrow\R$ is defined by $\langle p,q\rangle=p(0)q(0)+\int_{-1}^{1}p'(x)q'(x)\mathrm{d}x$ for all $p(x),q(x)\in\R[x]$ .

(c) The function $\langle\cdot,\cdot\rangle:\R[x]\times\R[x]\rightarrow\R$ is defined by $\langle p,q\rangle=\int_0^\infty p(x)q(x)e^{-x}\mathrm{d}x$ for all $p(x),q(x)\in\R[x]$ .

2. Let $(V,\langle\cdot,\cdot\rangle)$ be an inner product space. Show that the function $\|\cdot\|:V\rightarrow\R$ defined by $\|v\|=\sqrt{\langle v,v\rangle}$ , for any $v\in V$ , is a norm.

3. Find a polynomial $q(x)\in\R_2[x]$ such that $\int_0^1p(x)\cos(\pi x)\mathrm{d}x=\int_0^1p(x)q(x)\mathrm{d}x$ for every $p(x)\in\R_2[x]$ .

6. Define an operator $S:F^2\rightarrow F^2$ by $S(w,z)=(-z,w)$ .

(a) Find a formula for $S^*$ .

(b) Show that $S$ is normal but not self-adjoint.

(c) Find all eigenvalues of $S$ .

7. Suppose $T \in \mathcal{L}(V)$ and $a_0 +a_1z +a_2z^2 +\cdots+a_{m−1}z^{m−1} +z^m$ is the minimal polynomial of $T$ . Find the minimal polynomial of $T^*$ .

8. Prove or give a counterexample: If $T \in \mathcal{L}(V)$ and there is an orthonormal basis $e_1,\dots, e_n$ of $V$ such that $\|Te_k\| = \|T^∗e_k\|$ for each $k = 1,\dots,n$ , then $T$ is normal.