Let $V$ be an $n$ -dimensional inner product space.

1. Prove that a normal operator on a complex inner product space is self-adjoint if and only if all its eigenvalues are real.

Let $T$ be a normal operator on a complex inner product space. If $T$ is self-adjoint, as $\exists v\in V,\lambda\in\mathbb{C},Tv=\lambda v$ , we have $\lambda \|v\|^2=\langle \lambda v,v\rangle=\langle Tv,v\rangle=\langle v,Tv\rangle=\langle v,\lambda v\rangle=\overline\lambda\|v\|^2$ , which means $\lambda\in\R$ .
If all eigenvalues of $T$ are real. First of all, as $V$ is normal, it means there exists an orthonormal basis $e_1,\dots,e_n$ for $v$ such that $T$ has a diagonal matrix, $\mathrm{diag}\{\lambda_1,\dots,\lambda_n\},\lambda_i\in\R$ , with respect to it. So, the matrix of $T^*$ with respect to this basis equals to ${(\mathrm{diag}\{\lambda_1,\dots,\lambda_n\})}^*=\mathrm{diag}\{\lambda_1,\dots,\lambda_n\}$ , which means $T^*=T$ .
Combine the two conclusion, we have proved the statement.

2. Prove or give a counterexample: If $T \in \mathcal{L}(\mathbb{C}^3)$ is a diagonalizable operator, then $T$ is normal (with respect to the usual inner product).

Let $\alpha_i=(1,0,0)^T,(1,1,0)^T,(1,1,1)^T,e_i=(1,0,0)^T,(0,1,0)^T,(0,0,1)^T$ , which are bases for $V$ . And let the matrix of $T$ with respect to $\alpha_i$ equals to $\operatorname{diag}\{1,2,3\}$ , so absolutely $T$ is a diagonalizable operator. Notice that $e_i$ is an orthonormal basis for $\mathbb{C}^3$ and $\mathcal M|_{e_i}(T)=((1,0,0)^T,(1,2,0)^T,(1,1,3)^T),\mathcal{M}|_{e_i}(T^*)=((1,0,0)^T,(1,2,0)^T,(1,1,3)^T)^*=((1,1,1)^T,(0,2,1)^T,(0,0,3)^T)$ . So $\mathcal{M}|_{e_i}(TT^*-T^*T)\neq0$ , which means $T$ is not normal.

3. Suppose $F = \mathbb{C}$ and $T\in \mathcal{L}(V)$ . Prove that $T$ is normal if and only if every eigenvector of $T$ is also an eigenvector of $T^*$ .

Let $T$ is normal, notice that $T-\lambda I$ is normal, $\|(T-\lambda I)v\|=\|(T-\lambda I)^*v\|=\|(T^*-\overline\lambda I)v\|$ , so $Tv=\lambda v\Rightarrow T^*v=\overline\lambda v$ , which means every eigenvector of $T$ is also an eigenvector of $T^*$ .
Suppose that every eigenvector of $T$ is also an eigenvector of $T^*$ . According to Schur's theorem, there is an orthonormal basis $e_1,\dots,e_n$ of $V$ with respect to which T has an upper-triangular matrix $\mathcal M_e(T)$ . Notice that $Te_1=\mathcal M_e(T)_{1,1}e_1$ , which means $e_1$ is an eigenvector of $T$ , so $T^*e_1=\sum\overline{\mathcal{M}_e(T)_{1,i}}e_i=\lambda_1'e_1$ . Compare the two sides of equation, we have $\lambda_1'=\overline{\mathcal{M}_e(T)_{1,1}},{\mathcal{M}_e(T)_{1,i}}=0,\forall i\neq 1$ . Notice that with ${\mathcal{M}_e(T)_{1,i}}=0,\forall i\neq1$ , we have $Te_2=\mathcal M_e(T)_{2,2}e_2$ , which means $e_2$ is also an eigenvector of $T$ . By applying the method we deal with $e_1$ to $e_2$ , we can also get $\lambda_2'=\overline{\mathcal{M}_e(T)_{2,2}},{\mathcal{M}_e(T)_{2,i}}=0,\forall i\neq2$ . Continuing in this fashion, we see that all nondiagonal entries in the matrix $\mathcal M_e(T)$ equal $0$ . Thus $\mathcal M_e(T^*)=(\mathcal M_e(T))^*$ is also a diagonal matrix. So $\mathcal M_e(TT^*)=\mathcal M_e(T^*T)\Rightarrow T$ is normal.

4. Suppose $F = \mathbb{C}$ and $T \in \mathcal{L}(V)$ . Prove that $T$ is normal if and only if there exists a polynomial $p(x) \in \mathbb{C}[x]$ such that $T^* = p(T)$ .

If there exists $p(x)=\sum_{i=0}^ka_ix^i,T^*=p(T)$ , we have $T^*T-TT^*=(\sum_{i=0}^ka_iT^i)T-T(\sum_{i=0}^ka_iT^i)=0$ , which means $T$ is normal.
Let $T$ be normal. Suppose that the eigenvalues of $T$ are $\lambda_1,\dots,\lambda_k$ . We know $T$ is also diagonalizable, which means $V=\mathcal N(T-\lambda_1)\oplus\cdots\oplus\mathcal N(T-\lambda_k)$ . So for all $v\in V$ , we have $v=\sum_{m=1}^kv_m,v_m\in\mathcal N(T-\lambda_m)$ . Let $P_i(x):=\prod_{t\neq i}(\frac{x-\lambda_t}{\lambda_i-\lambda_t})$ , consider that $T-\lambda_t$ is exchangeable, which implies the order is not important. So, we have $P_i(\lambda_j)=\delta_{ij},P_i(T)v_m=P_i(\lambda_m)v_m=\delta_{im}v_m,P_i(T)v=\sum_{m=1}^kP_i(T)v_m=v_i$ . Additionally, $Tv_i=\lambda_i v_i\Leftrightarrow T^*v_i=\overline{\lambda_i} v_i$ , which means $T^*v_i=\overline{\lambda_i}v_i=\overline{\lambda_i}P_i(T)v$ . Above all, we have $T^*v=\sum_{i=1}^kT^*v_i=\sum_{i=1}^k\overline{\lambda_i}P_i(T)v$ , which means $T^*=\sum_{i=1}^k\overline{\lambda_i}P_i(T)$ .

5. Suppose $F = \mathbb{C}$ and $T \in\mathcal L(V)$ . Let $\lambda_1,\dots,\lambda_n$ be eigenvalues of $T$ . Prove that $T$ is normal if and only if the singular values of $T$ are $|\lambda_1|,|\lambda_2|,\dots,|\lambda_n|$ .

$T$ is normal $\Rightarrow$ $\exists$ orthonormal basis $e_1,\dots,e_n\in V,\mathcal M(T,(e_1,\dots,e_n))=\operatorname{diag}\{\lambda_1,\lambda_2,\dots,\lambda_n\}\Rightarrow\mathcal M(T^*,(e_1,\dots,e_n))=\operatorname{diag}\{\overline{\lambda_1},\dots,\overline{\lambda_n}\}\Rightarrow\mathcal M(T^*T,(e_1,\dots,e_n))=\operatorname{diag}\{|\lambda_1|^2,\dots,|\lambda_n|^2\}=(\operatorname{diag}\{|\lambda_1|,\dots,|\lambda_n|\})^2\Rightarrow$ the singular values of $T$ are $|\lambda_1|,|\lambda_2|,\dots,|\lambda_n|\Rightarrow$ $|\lambda_1|^2,\dots,|\lambda_n|^2$ are eigenvalues of $T^*T$ $\Rightarrow\exists$ orthonormal basis $f_1,\dots,f_n\in V,\mathcal M(T^*T,(f_1,\dots,f_n))=\operatorname{diag}\{|\lambda_1|^2,\dots,|\lambda_n|^2\}\Rightarrow\mathcal M(TT^*,(f_1,\dots,f_n))=\operatorname{diag}\{\overline{|\lambda_1|^2},\dots,\overline{|\lambda_n|^2}\}=\mathcal M(T^*T,(f_1,\dots,f_n))\Rightarrow T^*T=TT^*\Rightarrow T$ is normal.

6. Find the singular values of the differentiation operator $D \in \mathcal L(\R_2[x])$ defined by $D(p(x)) = p'(x)$ , where the inner product on $\R_2[x]$ is defined by $\langle p(x),q(x)\rangle =\int^1_{−1}p(x)q(x)\mathrm dx$ .

Notice that $\frac{\sqrt2}{2},\frac{\sqrt6}{2}x,\frac{3\sqrt{10}}{4}(x^2-\frac{1}{3})$ is an orthonormal basis for $(\R_2[x],\langle\cdot,\cdot\rangle)$ . So $\mathcal M(D)=((0,0,0)^T ,(\sqrt3,0,0)^T,(0,\sqrt{15},0)^T),\mathcal M(T^*)=((0,\sqrt3,0)^T,(0,0,\sqrt{15})^T,(0,0,0)^T),\mathcal M(T^*T)=\operatorname{diag}\{0,3,15\}=(\operatorname{diag}\{0,\sqrt3,\sqrt{15}\})^2$ , which means the singular values of $D$ are $\sqrt{15},\sqrt3,0$ .

7. Suppose $D : \R_8[x] \rightarrow \R_8[x]$ is the differentiation operator defined by $D(p(x)) = p'(x)$ . Prove that there does not exist an inner product on $R_8[x]$ that makes $D$ a normal operator.

Notice that $D(1)=0,D(x)=1$ , which means $0\neq 1\in\mathcal N(D)\cap\mathcal R(D)$ . So $\R_8[x]\neq \mathcal N(D)\oplus\mathcal R(D)$ . As we all know $T\in\mathcal L(V)$ is normal $\Rightarrow V=\mathcal N(T)\oplus\mathcal R(T)$ . So $D$ can't be normal.

8. Suppose $T \in\mathcal L(V)$ . Then there exists a unitary operator $S \in\mathcal L(V)$ such that $T =S\sqrt{T^*T}$ .

(Prove by SVD): Consider $T=U\Sigma V^*$ , where $U,V$ are both with orthonormal columns, and $\Sigma$ is diagonal. Define $S=UV^*$ , we have $S^*S=SS^*=\operatorname{id}$ , which means $S$ is a unitary operator, and $T=UV^*V\Sigma V^*=SV\Sigma V^*$ . Notice that $T^*T=V\Sigma^2V^*=(V\Sigma V^*)^2$ . So $\sqrt{T^*T}=V\Sigma V^*$ , which means $T=S\sqrt{T^*T}$ .

Following is another approach to prove the statement.

(Prove by Linear Algebra Done Right): Let $s_1,\dots,s_m$ be the positive singular values of $T$ , and let $e_1,\dots,e_m$ and $f_1,\dots,f_m$ be orthonormal lists in $V$ such that $Tv=s_1\langle v,e_1\rangle f_1 + \cdots+s_m\langle v,e_m\rangle f_m$ for every $v\in V$ . Extend $e_1,\dots,e_m$ and $f_1,\dots, f_m$ to orthonormal bases $e_1,\dots,e_n$ and $f_1,\dots, f_n$ of $V$ .
Define $S\in\mathcal L(V)$ by $Sv = \langle v,e_1\rangle f_1 + \cdots+\langle v,e_n\rangle f_n$ for each $v \in V$ . Then $\|Sv\|^2 = \|\langle v,e_1\rangle f_1+ \cdots +\langle v,e_n\rangle f_n\|^2 =|\langle v,e_1\rangle|^2 + \cdots + |\langle v,e_n\rangle|^2=\|v\|^2$ . Thus $S$ is a unitary operator.
Applying $T^*$ to both sides of $Tv=s_1\langle v,e_1\rangle f_1 + \cdots+s_m\langle v,e_m\rangle f_m$ and then using the formula for $T^*$ given by $T^*v=s_1\langle v, f_1\rangle e_1 + \cdots+s_m\langle v,f_m\rangle e_m$ shows that $T^*Tv=s_1^2\langle v,e_1\rangle e_1+\cdots+s_m^2\langle v,e_m\rangle e_m$ for every $v \in V$ . Thus if $v\in V$ , then $\sqrt{T^*T}v = s_1\langle v,e_1\rangle e_1 + \cdots+s_m\langle v,e_m\rangle e_m$ because the operator that sends $v$ to the right side of the equation above is a positive operator whose square equals $T^*T$ . Now $S\sqrt{T^*T}v =S(s_1\langle v,e_1\rangle e_1 + \cdots +s_m\langle v,e_m\rangle e_m) =s_1\langle v,e_1\rangle f_1 + \cdots +s_m\langle v,e_m\rangle f_m =Tv$ .

人は繋がっているのよ

Advanced Algebra Assignment NO.5

Let $V$ be an $n$ -dimensional inner product space.

1. Prove that a normal operator on a complex inner product space is self-adjoint if and only if all its eigenvalues are real.

2. Prove or give a counterexample: If $T \in \mathcal{L}(\mathbb{C}^3)$ is a diagonalizable operator, then $T$ is normal (with respect to the usual inner product).

3. Suppose $F = \mathbb{C}$ and $T\in \mathcal{L}(V)$ . Prove that $T$ is normal if and only if every eigenvector of $T$ is also an eigenvector of $T^*$ .

4. Suppose $F = \mathbb{C}$ and $T \in \mathcal{L}(V)$ . Prove that $T$ is normal if and only if there exists a polynomial $p(x) \in \mathbb{C}[x]$ such that $T^* = p(T)$ .

5. Suppose $F = \mathbb{C}$ and $T \in\mathcal L(V)$ . Let $\lambda_1,\dots,\lambda_n$ be eigenvalues of $T$ . Prove that $T$ is normal if and only if the singular values of $T$ are $|\lambda_1|,|\lambda_2|,\dots,|\lambda_n|$ .

6. Find the singular values of the differentiation operator $D \in \mathcal L(\R_2[x])$ defined by $D(p(x)) = p'(x)$ , where the inner product on $\R_2[x]$ is defined by $\langle p(x),q(x)\rangle =\int^1_{−1}p(x)q(x)\mathrm dx$ .

7. Suppose $D : \R_8[x] \rightarrow \R_8[x]$ is the differentiation operator defined by $D(p(x)) = p'(x)$ . Prove that there does not exist an inner product on $R_8[x]$ that makes $D$ a normal operator.

8. Suppose $T \in\mathcal L(V)$ . Then there exists a unitary operator $S \in\mathcal L(V)$ such that $T =S\sqrt{T^*T}$ .

Following is another approach to prove the statement.

Advanced Algebra Assignment NO.4

Advanced Algebra Assignment NO.6

Comments NOTHING

取消回复

人は繋がっているのよ

Let V be an n-dimensional inner product space.

1. Prove that a normal operator on a complex inner product space is self-adjoint if and only if all its eigenvalues are real.

2. Prove or give a counterexample: If T \in \mathcal{L}(\mathbb{C}^3) is a diagonalizable operator, then T is normal (with respect to the usual inner product).

3. Suppose F = \mathbb{C} and T\in \mathcal{L}(V). Prove that T is normal if and only if every eigenvector of T is also an eigenvector of T^∗.

4. Suppose F = \mathbb{C} and T \in \mathcal{L}(V). Prove that T is normal if and only if there exists a polynomial p(x) \in \mathbb{C}[x] such that T^* = p(T).

5. Suppose F = \mathbb{C} and T \in\mathcal L(V). Let \lambda_1,\dots,\lambda_n be eigenvalues of T. Prove that T is normal if and only if the singular values of T are |\lambda_1|,|\lambda_2|,\dots,|\lambda_n|.

6. Find the singular values of the differentiation operator D \in \mathcal L(\R_2[x]) defined by D(p(x)) = p'(x), where the inner product on \R_2[x] is defined by \langle p(x),q(x)\rangle =\int^1_{−1}p(x)q(x)\mathrm dx.

7. Suppose D : \R_8[x] \rightarrow \R_8[x] is the differentiation operator defined by D(p(x)) = p'(x). Prove that there does not exist an inner product on R_8[x] that makes D a normal operator.

8. Suppose T \in\mathcal L(V). Then there exists a unitary operator S \in\mathcal L(V) such that T =S\sqrt{T^*T}.

Following is another approach to prove the statement.

Advanced Algebra Assignment NO.4

Advanced Algebra Assignment NO.6

Comments NOTHING

取消回复

Let $V$ be an $n$ -dimensional inner product space.

2. Prove or give a counterexample: If $T \in \mathcal{L}(\mathbb{C}^3)$ is a diagonalizable operator, then $T$ is normal (with respect to the usual inner product).

3. Suppose $F = \mathbb{C}$ and $T\in \mathcal{L}(V)$ . Prove that $T$ is normal if and only if every eigenvector of $T$ is also an eigenvector of $T^*$ .

4. Suppose $F = \mathbb{C}$ and $T \in \mathcal{L}(V)$ . Prove that $T$ is normal if and only if there exists a polynomial $p(x) \in \mathbb{C}[x]$ such that $T^* = p(T)$ .

5. Suppose $F = \mathbb{C}$ and $T \in\mathcal L(V)$ . Let $\lambda_1,\dots,\lambda_n$ be eigenvalues of $T$ . Prove that $T$ is normal if and only if the singular values of $T$ are $|\lambda_1|,|\lambda_2|,\dots,|\lambda_n|$ .

6. Find the singular values of the differentiation operator $D \in \mathcal L(\R_2[x])$ defined by $D(p(x)) = p'(x)$ , where the inner product on $\R_2[x]$ is defined by $\langle p(x),q(x)\rangle =\int^1_{−1}p(x)q(x)\mathrm dx$ .

7. Suppose $D : \R_8[x] \rightarrow \R_8[x]$ is the differentiation operator defined by $D(p(x)) = p'(x)$ . Prove that there does not exist an inner product on $R_8[x]$ that makes $D$ a normal operator.

8. Suppose $T \in\mathcal L(V)$ . Then there exists a unitary operator $S \in\mathcal L(V)$ such that $T =S\sqrt{T^*T}$ .