Linear Algebra (2ed) Hoffman&Kunze

Linear Algebra (2ed) Hoffman & Kunze 后记

2020年8月2日christangdt2条评论

在我求学期间，高等代数和解析几何是相对分析学得比较好的一门课，但是这个比较好使得我在重学其他教材时产生了很大的困难。具体原因，当年用的北大王萼芳那本高等代数是标准的苏式教材，第一章就是行列式，且是用permutation和余子式计算行列式，我记得入学第一个月就懵圈许久，直到线性相关、线性无关时才反映过来。所以我对线性相关和向量空间没有什么困难，但是一直局限在解方程组的维度上，对矩阵的运算很熟，但完全无法理解矩阵、向量、空间等等的关系，后面的一些更抽象的定理，诸如spectral theorem等，则完全没有接触过。最近听说清华本科已经将线性代数教材改为英文版，我非常赞成。

后来在求学期间的末尾，试读了Curtis的Linear Algebra: An Introductory Approach。对线性代数的理解有所提升，因为忙于毕业，也没有读完。

2014年，我有机会接触了Sheldon Axler的Linear Algebra Done Right，这是在国内比较出名的一本小书，很薄，但我第一次读时颇有一种大开眼界之感，但当年读到特征值、特征多项式时就有些吃力，后面几章其实隶属于国内近世代数的内容（除了最后一章行列式，Axler一直比较自豪于把Determinant放到最后的处理，有兴趣的可以去查查他的官网，当然这种处理是有争议的，学线性代数想绕开行列式基本不可能）。2018年我又系统性的读了一次Linear Algebra Done Right，基本理清了这门学科的一些脉络，当时也投入了比较大的精力，读后的感觉是抽象化的建立了线性代数的学科内容，但是对一些比较难的概念还是不清楚为什么会设置，例如adjoint operator。

Hoffman的这一本书是MIT的教材，MIT的线性代数公开课久负盛名，这本书虽然成书于上世纪70年代，但是一直被奉为经典，北大近世代数选修课也用这本书的第6章之后的一些内容作为教材。这本书纸版我入手的很早（2011年），听说现在已经买不到一手书了，所以从去年开始抽空开始读，断断续续大致花了一年时间。总体的感觉，这本书的优缺点都很明显，优点是视点很高，对以后学泛函等很有帮助，内容中规中矩，该有的都有，不该有的也有（例如dual space，double dual，我在其他任何一本书里都没见过）。缺点是因为成书过早，符号体系很老，有点不适应，此外没有包括一些最新的进展（例如singular value decomposition，在Axler的书里提了），最后就是感觉从第8章开始有点混乱，一些概念例如positive operator是第9章定义，但第8章习题里已经有了，后面三章的笔误和错误习题也比较多。所以我舍弃了后面两章的内容。

这本书前八章质量仍是很高的，适应了符号体系后，有种大刀阔斧的感觉，但本书绝对不适用于第一次学线性代数的人，可能会完全云里雾里，本科高年级和研究生比较适合。学完这本书后的进阶教材是Roman的Advanced Linear Algebra，注重理论研究的则可以读Artin的Algebra。

Definition and Theorems (Chapter 8)

2020年8月2日christangdt留下评论

Definition. Let $F$ be the field of real numbers or the field of complex numbers, and $V$ a vector space over $F$ . An inner product on $V$ is a function which assigns to each ordered pair of vectors $\alpha,\beta\in V$ a scalar $(\alpha|\beta)$ in $F$ in such a way that for all $\alpha,\beta,\gamma\in V$ and all scalars $c$
( a ) $(\alpha+\beta|\gamma)=(\alpha|\gamma)+(\beta|\gamma)$ ;
( b ) $(c\alpha|\beta)=c(\alpha|\beta)$ ;
( c ) $(\beta|\alpha)=\overline{(\alpha|\beta)}$ , the bar denoting complex conjugation;
( d ) $(\alpha|\alpha)>0$ if $\alpha\neq 0$ .

The standard inner product on $F^n$ is defined on $\alpha=(x_1,\dots,x_n)$ and $\beta=(y_1,\dots,y_n)$ by $(\alpha|\beta)=\sum_jx_j\overline{y_j}$ .
The standard inner product on $F^{n\times 1}$ is defined for $X,Y\in F^{n\times 1}$ and an $n\times n$ invertible matrix $Q$ over $F$ as $(X|Y)=Y^{\ast}Q^{\ast}QX$ .

The quadratic form determined by the inner product is the function that assigns to each vector $\alpha$ the scalar $||\alpha||^2$
The polarization identities are

$\displaystyle{(\alpha|\beta)=\frac{1}{4}\sum_{i=1}^4i^n||\alpha+i^n\beta||^2}$

in which in the real case

$\displaystyle{(\alpha|\beta)=\frac{1}{4}||\alpha+\beta||^2-\frac{1}{4}||\alpha-\beta||^2}$

in the complex case

$\displaystyle{(\alpha|\beta)=\frac{1}{4}||\alpha+\beta||^2-\frac{1}{4}||\alpha-\beta||^2+\frac{i}{4}||\alpha+i\beta||^2-\frac{i}{4}||\alpha-i\beta||^2}$

Suppose $V$ is finite-dimensional and $\mathfrak B=\{\alpha_1,\dots,\alpha_n\}$ is an ordered basis for $V$ , for any inner product $(\text{ }|\text{ })$ on $V$ , then the matrix $G$ with $G_{jk}=(\alpha_k|\alpha_j)$ is called the matrix of the inner product in the ordered basis $\mathfrak B$ . Since if $\alpha=\sum_kx_k\alpha_k$ and $\beta=\sum_jy_j\alpha_j$ , then

$\displaystyle{(\alpha|\beta)=(\sum_kx_k\alpha_k|\beta)=\sum_kx_k(\alpha_k|\beta)=\sum_kx_k\sum_j\overline{y_j}(\alpha_k|\alpha_j)=\sum_{j,k}\overline{y_j}G_{jk}x_k=Y^{\ast}GX}$

Definition. An inner product space is a real or complex vector space, together with a specified inner product on that space.
A finite-dimensional real inner product space is often called a Euclidean space. A complex inner product space is often referred to as a unitary space.

Theorem 1. If $V$ is an inner product space, then for any vectors $\alpha,\beta\in V$ and any scalar $c$
(i) $||c\alpha||=|c|\text{ }||\alpha||$ ;
(ii) $||\alpha||>0$ for $\alpha\neq 0$ ;
(iii) $|(\alpha|\beta)|\leq ||\alpha|| \text{ }||\beta||$ ;
(iv) $||\alpha+\beta||\leq ||\alpha||+||\beta||$ .
The inequality in (iii) is called the Cauchy-Schwarz inequality.

Definitions. Let $\alpha$ and $\beta$ be vectors in an inner product space $V$ . Then $\alpha$ is orthogonal to $\beta$ if $(\alpha|\beta)=0$ ; since this implies $\beta$ is orthogonal to $\alpha$ , we often simply say that $\alpha$ and $\beta$ are orthogonal. If $S$ is a set of vectors in $V$ , $S$ is called an orthogonal set provided all pairs of distinct vectors in $S$ are orthogonal. An orthonormal set is an orthogonal set $S$ with the additional property that $||\alpha||=1$ for every $\alpha\in S$ .

Theorem 2. An orthogonal set of non-zero vectors is linearly independent.
Corollary. If $\alpha$ and $\beta$ is a linear combination of an orthogonal sequence of non-zero vectors $\alpha_1,\dots,\alpha_m$ , then $\beta$ is the particular linear combination

$\displaystyle{\beta=\sum_{k=1}^m\frac{(\beta|\alpha_k)}{||\alpha_k||^2}\alpha_k.}$

Theorem 3. Let $V$ be an inner product space and let $\beta_1,\dots,\beta_n$ be any independent vectors in $V$ . Then one may construct orthogonal vectors $\alpha_1,\dots,\alpha_n$ in $V$ such that for each $k=1,2,\dots,n$ the set $\{\alpha_1,\dots,\alpha_k\}$ is a basis for the subspace spanned by $\beta_1,\dots,\beta_k$ .
Corollary. Every finite-dimensional inner product space has an orthonormal basis.

A best approximation to $\beta$ by vectors in $W$ is a vector $\alpha\in W$ such that $||\beta-\alpha||\leq ||\beta-\gamma||$ for every $\gamma\in W$ .

Theorem 4. Let $W$ be a subspace of an inner product space $V$ and let $\beta\in V$ .
(i) $\alpha\in W$ is a best approximation to $\beta$ by vectors in $W$ if and only if $\beta-\alpha$ is orthogonal to every vector in $W$ .
(ii) If a best approximation to $\beta$ by vectors in $W$ exists, it is unique.
(iii) If $W$ is finite-dimensional and $\{\alpha_1,\dots,\alpha_n\}$ is any orthogonal basis for $W$ , then the vector

$\displaystyle{\alpha=\sum_k\frac{(\beta|\alpha_k)}{||\alpha_k||^2}\alpha_k}$

is the (unique) best approximation to $\beta$ by vectors in $W$ .

Definition. Let $V$ be an inner product space and $S$ any set of vectors in $V$ . The orthogonal complement of $S$ is the set $S^{\perp}$ of all vectors in $V$ which are orthogonal to every vector in $S$ .

Definition. Whenever the vector $\alpha$ in Theorem 4 exists it is called the orthogonal projection of $\beta$ on $W$ . If every vector in $V$ has an orthogonal projection on $W$ , the mapping that assigns to each vector in $V$ its orthogonal projection on $W$ is called the orthogonal projection of $V$ on $W$

Corollary. Let $V$ be an inner product space, $W$ a finite-dimensional subspace, and $E$ the orthogonal projection of $V$ on $W$ . Then the mapping $\beta\to\beta-E\beta$ is the orthogonal projection of $V$ on $W^{\perp}$ .

Theorem 5. Let $W$ be a finite-dimensional subspace of an inner product space $V$ and let $E$ be the orthogonal projection of $V$ on $W$ . Then $E$ is an idempotent linear transformation of $V$ onto $W,W^{\perp}$ is the null space of $E$ , and $V=W\oplus W^{\perp}$ .
Corollary. Under the conditions of the theorem, $I-E$ is the orthogonal projection of $V$ on $W^{\perp}$ . It is an idempotent linear transformation of $V$ onto $W^{\perp}$ with null space $W$ .
Corollary. (Bessel’s inequality) Let $\{\alpha_1,\dots,\alpha_n\}$ be an orthogonal set of non-zero vectors in an inner product space $V$ . If $\beta$ is any vector in $V$ , then

$\displaystyle{\sum_k\frac{|(\beta|\alpha_k)|^2}{||\alpha_k||^2}\leq ||\beta||^2}$

and equality holds if and only if

$\displaystyle{\beta=\sum_k\frac{(\beta|\alpha_k)}{||\alpha_k||^2}\alpha_k.}$

Theorem 6. Let $V$ be a finite-dimensional inner product space, and $f$ a linear functional on $V$ . Then there exists a unique vector $\beta\in V$ such that $f(\alpha)=(\alpha|\beta)$ for all $\alpha\in V$ .

Theorem 7. For any linear operator $T$ on a finite-dimensional inner product space $V$ , there exists a unique linear operator $T^{\ast}$ on $V$ such that $(T\alpha|\beta)=(\alpha|T^{\ast}\beta)$ for all $\alpha,\beta\in V$ .

Theorem 8. Let $V$ be a finite-dimensional inner product space and let $\mathscr B=\{\alpha_1,\dots,\alpha_n\}$ be an (ordered) orthonormal basis for $V$ . Let $T$ be a linear operator on $V$ and let $A$ be the matrix of $T$ in the ordered basis $\mathscr B$ . Then $A_{kj}=(T\alpha_j|\alpha_k)$ .
Corollary. Let $V$ be a finite-dimesional inner product space, and let $T$ be a linear operator on $V$ . In any orthonormal basis for $V$ , the matrix of $T^{\ast}$ is the conjugate transpose of the matrix of $T$ .

Definition. Let $T$ be a linear operator on an inner product space $V$ . Then we say that $T$ has an adjoint on $V$ if there exists a linear operator $T^{\ast}$ on $V$ such that $(T\alpha|\beta)=(\alpha|T^{\ast}\beta)$ for all $\alpha,\beta\in V$ .

The adjoint of $T$ depends not only on $T$ but on the inner product as well.
in an arbitrary ordered basis $\mathscr B$ , the relation between $[T]_{\mathscr B}$ and $[T^{\ast}]_{\mathscr B}$ is more complicated than that given in the corollary above.

Theorem 9. Let $V$ be a finite-dimensional inner product space. If $T$ and $U$ are linear operators on $V$ and $c$ is a scalar,
(i) $(T+U)^{\ast}=T^{\ast}+U^{\ast}$ ;
(ii) $(cT)^{\ast}=\overline{c}T^{\ast}$ ;
(iii) $(TU)^{\ast}=U^{\ast}T^{\ast}$ ;
(iv) $(T^{\ast})^{\ast}=T$ .

A linear operator $T$ such that $T=T^{\ast}$ is called self-adjoint (or Hermitian).

Definition. Let $V$ and $W$ be inner product spaces over the smae field, and let $T$ be a linear transformation from $V$ into $W$ . We say that $T$ preserves inner products if $(T\alpha|T\beta)=(\alpha|\beta)$ for all $\alpha,\beta\in V$ . An isomorphism of $V$ onto $W$ is a vector space isomorphism $T$ of $V$ onto $W$ which also preserves inner products.
When such a $T$ exists, we shall say $V$ and $W$ are isomorphic.

Theorem 10. Let $V$ and $W$ be finite-dimensional inner product spaces over the same field, having the same dimension. If $T$ is a linear transformation from $V$ into $W$ , the following are quivalent.
(i) $T$ preserves inner products.
(ii) $T$ is an (inner product space) isomorphism.
(iii) $T$ carries every orthonormal basis for $V$ onto an orthonormal basis for $W$ .
(iv) $T$ carries some orthonormal basis for $V$ onto an orthonormal basis for $W$ .
Corollary. Let $V$ and $W$ be finite-dimensional inner product space over the same field. Then $V$ and $W$ are isomorphic if and only if they have the same dimension.

Theorem 11. Let $V$ and $W$ be inner product spaces over the same field, and let $T$ be a linear transformation from $V$ into $W$ . Then $T$ preserves inner products if and only if $||T\alpha||=||\alpha||$ for every $\alpha\in V$ .

Definition. A unitary operator on an inner product space is an isomorphism of the space onto itself.

Theorem 12. Let $U$ be a linear operator on an inner product space $V$ . Then $U$ is unitary if and only if the adjoint $U^{\ast}$ of $U$ exists and $UU^{\ast}=U^{\ast}U=I$ .

Definition. A complex $n\times n$ matrix $A$ is called unitary, if $A^{\ast}A=I$ .

Theorem 13. Let $V$ be a finite-dimensional inner product space and let $U$ be a linear operator on $V$ . Then $U$ is unitary if and only if the matrix of $U$ in some (or every) ordered orthonormal basis is a unitary matrix.

Definition. A real or complex $n\times n$ matrix $A$ is said to be orthogonal, if $A^tA=I$ .

Theorem 14. For every invertible complex $n\times n$ matrix $B$ there exists a unique lower-triangular matrix $M$ with positive entries on the main diagonal such that $MB$ is unitary.
Let $GL(n)$ denote the set of all invertible complex $n\times n$ matrices. Then $GL(n)$ is a group under matrix multiplication. This group is called the general linear group.
Corollary. For each $B$ in $GL(n)$ there exist unique matrices $N$ and $U$ such that $N$ is in $T^{+}(n)$ , the set of all complex $n\times n$ lower-triangular matrices with positive entries on the main diagonal, and $U$ is in $U(n)$ , and $B=N\cdot U$ .

Definition. Let $A$ and $B$ be complex $n\times n$ matrices. We say that $B$ is unitarily equivalent to $A$ if there is an $n\times n$ unitary matrix $P$ such that $B=P^{-1}AP$ . We say that $B$ is orthogonally equivalent to $A$ if there is an $n\times n$ orthogonal matrix $P$ such that $B=P^{-1}AP.$

Definition. Let $V$ be a finite-dimensional inner product space and $T$ a linear operator on $V$ . We say that $T$ is normal if it commutes with its adjoint, i.e., $TT^{\ast}=T^{\ast}T$ .

Theorem 15. Let $V$ be an inner product space and $T$ a self-adjoint linear operator on $V$ . Then each chareacteristic value of $T$ is real, and characteristic vectors of $T$ associated with distinct characteristic values are orthogonal.

Theorem 16. On a finite-dimensional inner product space of positive dimension, every self-adjoint operator has a (non-zero) characteristic vector.

Theorem 17. Let $V$ be a finite-dimensional inner product space, and let $T$ be any linear operator on $V$ . Suppose $W$ is a subspace of $V$ which is invariant under $T$ . Then the orthogonal complement of $W$ is invariant under $T^{\ast}$ .

Theorem 18. Let $V$ be a finite-dimensional inner product space, and let $T$ be a self-adjoint linear operator on $V$ . Then there is an orthonormal basis for $V$ , each vector of which is a characteristic vector for $T$ .
Corollary. Let $A$ be an $n\times n$ Hermitan (self-adjoint) matrix. Then there is a unitary matrix $P$ such that $P^{-1}AP$ is diagonal ( $A$ is unitarily equivalent to a diagonal matrix). If $A$ is a real symmetric matrix, there is a real orthogonal matrix $P$ such that $P^{-1}AP$ is diagonal.

Theorem 19. Let $V$ be a finite-dimensional inner product space and $T$ a normal operator on $V$ . Suppose $\alpha$ is a vector in $V$ . Then $\alpha$ is a characteristic vector for $T$ with characteristic value $c$ if and only if $\alpha$ is a characteristic vector for $T^{\ast}$ with characteristic value $\overline{c}$ .

Definition. A complex $n\times n$ matrix $A$ is called normal if $AA^{\ast}=A^{\ast}A$ .

Theorem 20. Let $V$ be a finite-dimensional inner product space, $T$ a linear operator on $V$ , and $\mathfrak B$ an orthonormal basis for $V$ . Suppose that the matrix $A$ of $T$ in the basis $\mathfrak B$ is upper triangular. Then $T$ is normal if and only if $A$ is a diagonal matrix.

Theorem 21. Let $V$ be a finite-dimensional complex inner product space, $T$ a linear operator on $V$ . Then ther eis an orthonormal basi for $V$ in which the matrix of $T$ is upper triangular.
Corollary. For every complex $n\times n$ matrix $A$ there is a unitary matrix $U$ such that $U^{-1}AU$ is upper-triangular.

Theorem 22. Let $V$ be a finite-dimensional complex inner product space and $T$ a normal operator on $V$ . Then $V$ has an orthonormal basis consisting of characteristic vectors for $T$ .
Corollary. For every normal matrix $A$ there is a unitary matrix $P$ such that $P^{-1}AP$ is a diagonal matrix.

Linear Algebra (2ed) Hoffman & Kunze 8.5

2020年8月2日christangdt留下评论

这一节的核心任务是解决如下问题：If $T$ is a linear operator on a finite-dimensional inner product space $V$ , under what conditions does $V$ have an orthonormal basis consisting of characteristic vectors for $T$ ?即标准正交基下的对角化问题。答案就是 $T$ 是normal的，在real情形下， $T$ 要求是self-adjoint的。
需要说明的是，这一节（或者这一章）的习题并不理想，一方面很多题目需要在finite-dimensional的假设下才可作出，另一方面部分题目涉及到了positive operator的定义，这一定义在第9章才给出。

Exercises

Exercise 1. For each of the following real symmetric matrices $A$ , find a real orthogonal matrix $P$ such that $P^tAP$ is diagonal.

$\displaystyle{\begin{bmatrix}1&1\\1&1\end{bmatrix},\quad\begin{bmatrix}1&2\\2&1\end{bmatrix},\quad\begin{bmatrix}\cos\theta&\sin\theta\\{\sin\theta}&-\cos\theta\end{bmatrix}}$

Solution: For the first matrix, we have

$\displaystyle{\begin{bmatrix}1&1\\1&1\end{bmatrix}\begin{bmatrix}1\\1\end{bmatrix}=2\begin{bmatrix}1\\1\end{bmatrix},\quad\begin{bmatrix}1&1\\1&1\end{bmatrix}\begin{bmatrix}1\\-1\end{bmatrix}=0\begin{bmatrix}1\\-1\end{bmatrix}}$

Also we have $(1,1)^T$ and $(1,-1)^T$ be orthogonal, thus the matrix $P=\dfrac{1}{\sqrt{2}}\begin{bmatrix}1&1\\1&-1\end{bmatrix}$ .
For the second matrix, the characteristic value are $-1,3$ , and we have

$\displaystyle{\begin{bmatrix}1&2\\2&1\end{bmatrix}\begin{bmatrix}1\\1\end{bmatrix}=3\begin{bmatrix}1\\1\end{bmatrix},\quad\begin{bmatrix}1&2\\2&1\end{bmatrix}\begin{bmatrix}1\\-1\end{bmatrix}=-\begin{bmatrix}1\\-1\end{bmatrix}}$

thus the required matrix $P$ is the same as the first matrix.
For the third matrix, the characteristic value are $1,-1$ , and we have

$\begin{bmatrix}\cos\theta&\sin\theta\\{\sin\theta}&-\cos\theta\end{bmatrix}\begin{bmatrix}{\sin\theta}\\{1-\cos\theta}\end{bmatrix}=\begin{bmatrix}{\sin\theta}\\{1-\cos\theta}\end{bmatrix}\\ \begin{bmatrix}\cos\theta&\sin\theta\\{\sin\theta}&-\cos\theta\end{bmatrix}\begin{bmatrix}{-\sin\theta}\\{1+\cos\theta}\end{bmatrix}=-\begin{bmatrix}{-\sin\theta}\\{1+\cos\theta}\end{bmatrix}$

the two vectors are orthogonal, thus the required matrix $P$ is

$\displaystyle{\begin{bmatrix}\dfrac{\sin\theta}{\sqrt{2-2\cos\theta}}&\dfrac{-\sin\theta}{\sqrt{2+2\cos\theta}}\\{\dfrac{1-\cos\theta}{\sqrt{2-2\cos\theta}}}&\dfrac{1+\cos\theta}{\sqrt{2+2\cos\theta}}\end{bmatrix}}$

Exercise 2. Is a complex symmetric matrix self-adjoint? Is it normal?
Solution: Let $A=\begin{bmatrix}1&i\\i&1\end{bmatrix}$ , then $A^{\ast}=\begin{bmatrix}1&-i\-i&1\end{bmatrix}$ , thus $A$ is not self-adjoint.
Let $B=\begin{bmatrix}i&1\\1&-i\end{bmatrix}$ , then $B^{\ast}=\begin{bmatrix}-i&1\\1&i\end{bmatrix}$ , we have $BB^{\ast}=\begin{bmatrix}2&2i\\-2i&2\end{bmatrix}$ and $B^{\ast}B=\begin{bmatrix}2&-2i\\2i&2\end{bmatrix}$ , thus $B$ is not normal.

Exercise 3. For

$\displaystyle{A=\begin{bmatrix}1&2&3\\2&3&4\\3&4&5\end{bmatrix}}$

there is a real orthogonal matrix $P$ such that $P^tAP=D$ is diagonal. Find such a diagonal matrix $D$ .
Solution: We have

$\displaystyle{\begin{aligned}\det(xI-A)&=\begin{vmatrix}x-1&-2&-3\\-2&x-3&-4\\-3&-4&x-5\end{vmatrix}\\&=(x-1)\begin{vmatrix}x-3&-4\\-4&x-5\end{vmatrix}-2\begin{vmatrix}2&3\\-4&x-5\end{vmatrix}+3\begin{vmatrix}2&3\\x-3&-4\end{vmatrix}\\&=(x-1)(x^2-8x-1)-2(2x-10+12)+3(-8-3x+9)\\&=x^3-9x^2+7x+1-(4x+4)+3(1-3x)\\&=x^3-9x^2-6x=x(x^2-9x-6)\end{aligned}}$

the solution of the equation $\det(xI-A)=0$ is $0,\frac{9+\sqrt{105}}{2},\frac{9-\sqrt{105}}{2}$ , thus the matrix $D$ is

$\displaystyle{D=\begin{bmatrix}\dfrac{9+\sqrt{105}}{2}\\&\dfrac{9-\sqrt{105}}{2}\\&&0\end{bmatrix}}$

Exercise 4. Let $V$ be $C^2$ , with the standard inner product. Let $T$ be the linear operator on $V$ which is represented in the standard ordered basis by the matrix $A=\begin{bmatrix}1&i\\i&1\end{bmatrix}$ . Show that $T$ is normal, and find an orthonormal basis for $V$ , consisting of characteristic vectors for $T$ .
Solution: We have $A^{\ast}=\begin{bmatrix}1&-i\-i&1\end{bmatrix}$ , thus $AA^{\ast}=\begin{bmatrix}1&i\\i&1\end{bmatrix}\begin{bmatrix}1&-i\\-i&1\end{bmatrix}=\begin{bmatrix}2&0\\0&2\end{bmatrix}$ , and $A^{\ast}A=\begin{bmatrix}1&-i\\-i&1\end{bmatrix}\begin{bmatrix}1&i\\i&1\end{bmatrix}=\begin{bmatrix}2&0\\0&2\end{bmatrix}$ , thus $A$ is normal, and so is $T$ . We have $\det(xI-A)=(x-1)^2+1$ , which means the characteristic value of $A$ is $1-i,1+i$ , and we can see that

$\displaystyle{\begin{bmatrix}1&i\\i&1\end{bmatrix}\begin{bmatrix}1\\-1\end{bmatrix}=(1-i)\begin{bmatrix}1\\-1\end{bmatrix},\begin{bmatrix}1&i\\i&1\end{bmatrix}\begin{bmatrix}1\\1\end{bmatrix}=(1+i)\begin{bmatrix}1\\1\end{bmatrix}}$

thus an orthonormal basis for $V$ consisting of characteristic vectors for $T$ can be $(1/\sqrt{2},-1/\sqrt{2})$ and $(1/\sqrt{2},1/\sqrt{2})$ .

Exercise 5. Give an example of a $2\times 2$ matrix $A$ such that $A^2$ is normal but $A$ is not.
Solution: Let $A=\begin{bmatrix}1&i\\i&-1\end{bmatrix}$ , then $A^2=0$ and thus is normal, but $A^{\ast}=\begin{bmatrix}1&-i\-i&-1\end{bmatrix}$ , thus we have

$\displaystyle{AA^{\ast}=\begin{bmatrix}1&i\\i&-1\end{bmatrix}\begin{bmatrix}1&-i\\\-i&-1\end{bmatrix}=\begin{bmatrix}2&-2i\\2i&2\end{bmatrix},A^{\ast}A=\begin{bmatrix}1&-i\\-i&-1\end{bmatrix}\begin{bmatrix}1&i\\i&-1\end{bmatrix}=\begin{bmatrix}2&2i\\-2i&2\end{bmatrix}}$

thus $A$ is not normal.

Exercise 6. Let $T$ be a normal operator on a finite-dimensional complex inner product space. Prove that $T$ is self-adjoint, positive, or unitary according as every characteristic value of $T$ is real, positive, or of absolute value $1$ .
Solution: Let $\mathfrak B$ be an orthonormal basis such that $[T]_{\mathfrak B}=A$ is a diagonal matrix, then if $T$ is self-adjoint, positive, or unitary, the same is $A$ . Write $A$ as $\text{diag}(a_1,\dots,a_n)$ , we see that the every characteristic value of $T$ must be one of $a_i$ .
If $A$ is self-adjoint, then $a_i=\overline{a_i}$ , thus $a_i\in R$
If $A$ is positive, then $a_i>0$ for all $i$ .
If $A$ is unitary, then $A^{\ast}A=I$ , thus $\overline{a_i}a_i=|a_i|^2=1$ , which gives $|a_i|=1$ .

Exercise 7. Let $T$ be a linear operator on the finite-dimensional inner product space $V$ , and suppose $T$ is both positive and unitary. Prove $T=I$ .
Solution: Find an orthonormal basis $\mathfrak B$ such that $[T]_{\mathfrak B}=A$ is upper triangular, since $T$ is positive, $A$ is positive and thus $A^{\ast}=A$ , which means $A$ is diagonal, and the entires $A_{ii}>0$ for $i=1,\dots,n$ . If $A$ is unitary, then $A_{ii}^2=1$ , which means $A_{ii}=1$ .

Exercise 8. Prove $T$ is normal if and only if $T=T_1+iT_2$ , where $T_1$ and $T_2$ are self-adjoint operators which commute.
Solution: If $T_1$ and $T_2$ are self-adjoint operators which commute, we have $(T_1+iT_2)^{\ast}=T_1^{\ast}-iT_2^{\ast}=T_1-iT_2$ , thus

$\displaystyle{(T_1+iT_2)(T_1-iT_2)=T_1^2+T_2^2=(T_1-iT_2)(T_1+iT_2)}$

Conversely, if $T$ is normal, let $\mathfrak B=\{\beta_1,\dots,\beta_n\}$ be an orthonormal basis such that $T\beta_i=a_i\beta_i$ , define $T_1\beta_i=\Re(a_i)\beta_i$ and $T_2\beta_i=\Im(a_i)\beta_i$ , then $T=T_1+iT_2$ , the matrix of $T_1$ and $T_2$ are real diagonal matrix, thus they are self-adjoint and commute.

Exercise 9. Prove that a real symmetric matrix has a real symmetric cube root; i.e., if $A$ is real symmetric, there is a real symmetric $B$ such that $B^3=A$ .
Solution: Real symmetric means $A$ is Hermitan, thus there is a real orthogonal matrix $P$ such that $D=P^{-1}AP$ is diagonal, let $D'=\text{diag}(\sqrt[3]{D_{11}},\dots,\sqrt[3]{D_{nn}})$ and $B=PD'P^{-1}$ , then $B^3=PDP^{-1}=P(P^{-1}AP)P^{-1}=A$ .

Exercise 10. Prove that every positive matrix is the square of a positive matrix.
Solution: If $A$ is positive, we can find unitary matrix $P$ such that $D=P^{-1}AP$ is diagonal, and $D_{ii}>0$ , We let $B=PD^2P^{-1}$ , then $B=A^2$ , and since $D^2$ is positive, so is $B$ .

Exercise 11. Prove that a normal and nilpotent operator is the zero operator.
Solution: If $T$ is normal, then there is an orthonormal basis $\mathfrak B$ such that $A=[T]_{\mathfrak B}$ is diagonal, if $T$ is nilpotent, so is $A$ , which means $A=0$ .

Exercise 12. If $T$ is a normal operator, prove that characteristic vectors for $T$ which are associated with distinct characteristic values are orthogonal.
Solution: Let $T\alpha=m\alpha,T\beta=n\beta$ , then $T^{\ast}\alpha=\overline{m}\alpha,T^{\ast}\beta=\overline{n}\beta$ , if $m\neq n$ , then we have

$\displaystyle{((TT^{\ast}-T^{\ast}T)\alpha|\beta)=(T^{\ast}\alpha|T^{\ast}\beta)-(T\alpha|T\beta)=(\overline{m}-\overline{n})(\alpha|T^{\ast}\beta)=(n-m)(T\alpha|\beta)}$

Since $TT^{\ast}=T^{\ast}T$ , we see that $(\alpha|T^{\ast}\beta)=(T\alpha|\beta)$ , which means $n(\alpha|\beta)=m(\alpha|\beta)=0$ , as $n\neq m$ , we see $(\alpha|\beta)=0$ .

Exercise 13. Let $T$ be a normal operator on a finite-dimensional complex inner product space. Prove that there is a polynomial $f$ , with complex coefficients, such that $T^{\ast}=f(T)$ .
Solution: Let $\mathfrak B$ be an orthonormal basis such that

$\displaystyle{[T]_{\mathfrak B}=\begin{bmatrix}d_1\\&{\ddots}\\&&d_n\end{bmatrix}}$

then we see

$\displaystyle{[T^{\ast}]_{\mathfrak B}=\begin{bmatrix}\overline{d_1}\\&{\ddots}\\&&\overline{d_n}\end{bmatrix}}$

the polynomial $f$ we need must satisfy $f(d_i)=\overline{d_i}$ for all $i$ , we can suppose the first $k(k\leq n)$ elements of $d_1,\dots,d_n$ are distinct, and let

$\displaystyle{\begin{bmatrix}f_0\\{\vdots}\\f_{k-1}\end{bmatrix}=\begin{bmatrix}1&d_1&\cdots&d_1^{k-1}\\&\\1&d_k&{\cdots}&d_k^{k-1}\end{bmatrix}^{-1}\begin{bmatrix}\overline{d_1}\\{\vdots}\\\overline{d_k}\end{bmatrix}}$

since $d_1,\dots,d_k$ are distinct, the matrix above is indeed invertible, if we let $f=\sum_{i=0}^{k-1}f_ix^i$ , then $f(d_j)=\overline{d_j}$ for $j=1,\dots,k$ .

Exercise 14. If two normal operators commute, prove that their product is normal.
Solution: Let $T$ and $U$ be two normal operators which commute, then by Exercise 13, we have $T^{\ast}=f(T)$ and $U^{\ast}=g(U)$ for some $f,g\in C[x]$ , thus we have $(TU)^{\ast}=U^{\ast}T^{\ast}=g(U)f(T)$ , since $T$ and $U$ commute, we have $T$ commute with $g(U)$ and $U$ commute with $f(T)$ , so $TU$ is normal since

$\displaystyle{TU(TU)^{\ast}=TUg(U)f(T)=g(U)f(T)TU=(TU)^{\ast}TU}$

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