Definition and Theorems (Chapter 8)

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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.

  1. The adjoint of T depends not only on T but on the inner product as well.
  2. 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

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}

Linear Algebra (2ed) Hoffman & Kunze 8.4

christangdt留下评论

如果在两个vector space之间有一个isomorphism，那么这个isomorphism是一个1-1对应的线性变换，而如果这两个vector space同时是inner product space，则isomorphism还要求内积不变，即preserve inner products。在两个有限维空间上，T是内积保留，isomorphism，标准正交积经T变换还是标准正交积，存在标准正交积经T变换还是标准正交积这四个部分是等价的(Theorem 10)，因此，有限维空间是isomorphism的充要条件是维数相同。对于任意空间，T保留内积的充要条件是模不变，即||T\alpha||=||\alpha||。
所谓的Unitary Operator是某一空间自己的isomorphism，I显然是一个unitary operator。Theorem 12是unitary等价的一个条件：adjoint存在且UU^{\ast}=U^{\ast}U=I。这一定理的重要性在于：其可以引出矩阵的unitary定义：如果矩阵A满足A^{\ast}A=I，那么A是unitary。Theorem 13是一个相对基础的结论：U是unitary当且仅当在某一正交基下的矩阵A是unitary的。
另一个相关的概念是orthogonal矩阵，如果矩阵A满足A^tA=I，那么A是orthogonal。real orthogonal矩阵一定是unitary的，而real unitary矩阵也是orthogonal的，但complex unitary矩阵则不是。
在某一inner product space上的所有unitary operators形成一个group，所有n阶unitary matrix也形成一个group，记为U(n)。在U(n)的矩阵中，Gram-Schmidt有比较有趣的结论，即Theorem 14：任意一个可逆矩阵B都可找到惟一的下三角矩阵M，其对角线元素都是正的，且MB是unitary的。unitary矩阵的一个等价条件是其行或者列构成orthonormal basis。如果把所有invertible的n阶矩阵形成的群记为GL(n)，那么定理14有如下推论：\forall B\in GL(n),\exists !N\in T^{+}(n),U\in U(n), s.t. B=N\cdot U。
最后简单提了unitary equivalent和orthogonally equivalent的含义。

Exercises

Exercise 1. Find a unitary matrix which is not orthogonal, and find an orthogonal matrix which is not unitary.
Solution: One unitary matrix which is not orthogonal may be

\displaystyle{\dfrac{1}{2}\begin{bmatrix}1+i&1-i\\-1-i&1-i\end{bmatrix}}

and one orthogonal matrix which is not unitary can be

\displaystyle{\dfrac{1}{2}\begin{bmatrix}1+i&1-i\\-1+i&1+i\end{bmatrix}}

Exercise 2. Let V be the space of complex n\times n matrices with inner product (A|B)=\text{tr}(AB^{\ast}). For each M\in V, let T_M be the linear operator defined by T_M(A)=MA. Show that T_M is unitary if and only if M is a unitary matrix.
Solution: We have

\displaystyle{(T_M(A)|T_M(B))=(MA|MB)=\text{tr}(MAB^{\ast}M^{\ast})=\text{tr}(B^{\ast}M^{\ast}MA)}

If M is a unitary matrix, then M^{\ast}M=I, thus (T_M(A)|T_M(B))=\text{tr}(B^{\ast}A)=\text{tr}(AB^{\ast})=(A|B), which means T_M is unitary. Converely, if T_M is unitary, then \text{tr}(B^{\ast}M^{\ast}MA)=\text{tr}(B^{\ast}A) for any matrix A,B\in V. It is easy to see that M^{\ast}M=I.

Exercise 3. Let V be the set of complex numbers, regarded as a real vector space.
( a ) Show that (\alpha|\beta)=\Re (\alpha\overline{\beta}) defines an inner product on V.
( b ) Exhibit an (inner product space) isomorphism of V onto R^2 with the standard inner product.
( c ) For each \gamma\in V, let M_{\gamma} be the linear operator on V defined by M_{\gamma}(\alpha)=\gamma\alpha. Show that (M_{\gamma})^{\ast}=M_{\overline{\gamma}}.
( d ) For which complex numbers \gamma is M_{\gamma} self-adjoint?
( e ) For which \gamma is M_{\gamma} unitary?
( f ) For which \gamma is M_{\gamma} positive?
( g ) What is \det (M_{\gamma})?
( h ) Find the matrix of M_{\gamma} in the basis \{1,i\}.
( i ) If T is a linear operator on V, find necessary and sufficient conditions for T to be an M_{\gamma}.
( j ) Find a unitary operator on V which is not an M_{\gamma}.
Solution:
We let \alpha=a+bi,\beta=c+di, then
( a ) We have (\alpha+\beta|\gamma)=\Re((\alpha+\beta)\overline{\gamma})=\Re(\alpha\overline{\gamma})+\Re(\beta\overline{\gamma})=(\alpha|\gamma)+(\beta|\gamma). If c\in R, then (c\alpha|\beta)=\Re(c\alpha\overline{\beta})=c\Re (\alpha\overline{\beta})=c(\alpha|\beta). Also, (\beta|\alpha)=\Re(\beta\overline{\alpha})=ac+bd, while \Re(\alpha\overline{\beta})=ac+bd, thus (\beta|\alpha)=(\alpha|\beta). Finally (\alpha|\alpha)=\Re(\alpha\overline{\alpha})=|\alpha|^2>0 if \alpha\neq 0.
( b ) Let T\alpha=T(a+bi)=(a,b), then (\alpha|\beta)=ac+bd, and (T\alpha|T\beta)=ac+bd.
( c ) We have (M_{\gamma}(\alpha)|\beta)=(\gamma\alpha|\beta)=\Re(\gamma\alpha\overline{\beta}), and (\alpha|M_{\overline{\gamma}}(\beta))=(\alpha|\overline{\gamma}\beta)=\Re(\alpha\overline{\overline{\gamma}\beta})=\Re(\gamma\alpha\overline{\beta}), thus (M_{\gamma}(\alpha)|\beta)=(\alpha|M_{\overline{\gamma}}(\beta)), from the uniqueness of (M_{\gamma})^{\ast} we finish the proof.
( d ) M_{\gamma} self-adjoint means (M_{\gamma})^{\ast}=M_{\gamma}, thus \gamma shall be real.
( e ) We need (M_{\gamma})^{\ast}M_{\gamma}=M_{\overline{\gamma}}M_{\gamma}=I, i.e., \overline{\gamma}\gamma=1, so \gamma shall have norm 1.
( f ) If M_{\gamma} is positive, then \gamma\in R, and (M_{\gamma}(\alpha)|\alpha)=\gamma||\alpha||^2>0 for all \alpha\neq 0, thus \gamma>0
( g ) V has a basis \{1,i\}, and M_{\gamma}(1)=\gamma=\Re(\gamma)+i\Im(\gamma), M_{\gamma}(i)=i\gamma=-\Im(\gamma)+i\Re(\gamma), so the matrix of M_{\gamma} in the basis \{1,i\} shall be

\displaystyle{\begin{bmatrix}\Re(\gamma)&-\Im(\gamma)\\{\Im(\gamma)}&\Re(\gamma)\end{bmatrix}}

thus \det (M_{\gamma})=(\Re(\gamma))^2+(\Im(\gamma))^2=||\gamma||^2.
( h ) See the process in (g)
( i ) The matrix of T in the basis \{1,i\} must satisfy the form \begin{bmatrix}a&-b\\{b}&a\end{bmatrix} where a,b\in R.
( j ) Let T(\alpha)=\overline{\alpha}, then (T\alpha|T\beta)=\Re(\overline{\alpha}\beta)=\Re(\alpha\overline{\beta})=(\alpha|\beta), thus T is unitary, but the matrix of T in the basis \{1,i\} is \begin{bmatrix}1&0\\{0}&-1\end{bmatrix}, which means T is not an M_{\gamma}.

Exercise 4. Let V be R^2, with the standard inner product. If U is a unitary operator on V, show that the matrix of U in the standard ordered basis is either

\displaystyle{\begin{bmatrix}\cos\theta&-\sin\theta\\{\sin\theta}&\cos\theta\end{bmatrix}\text{ or }\begin{bmatrix}\cos\theta&\sin\theta\\{\sin\theta}&-\cos\theta\end{bmatrix}}

for some real \theta,0\leq \theta<2\pi. Let U_{\theta} be the linear operator corresponding to the first matrix, i.e., U_{\theta} is rotation through the angle \theta. Now convince yourself that every unitary operator on V is either a rotation, or reflection about the \epsilon_1-axis followed by a rotation.
( a ) What is U_{\theta}U_{\phi}?
( b ) Show that U_{\theta}^{\ast}=U_{-\theta}.
( c ) Let \phi be a fixed real number, and let \mathfrak B=\{\alpha_1,\alpha_2\} be the orthonormal basis obtained by rotating \{\epsilon_1,\epsilon_2\} through the angle \phi, i.e., \alpha_j=U_{\phi}\epsilon_j. If \theta is another real number, what is the matrix of U_{\theta} in the ordered basis \mathfrak B?
Solution:
( a ) It is rotation throught the angle \theta+\phi if \theta+\phi<2\pi, and \theta+\phi-2\pi if \theta+\phi\geq2\pi.
( b ) The matrix of U^{\ast} in the standard basis is the conjugate transpose of the original matrix, which is

\displaystyle{\begin{bmatrix}\cos\theta&\sin\theta\\{-\sin\theta}&\cos\theta\end{bmatrix}=\begin{bmatrix}\cos(-\theta)&-\sin(-\theta)\\{\sin(-\theta)}&\cos(-\theta)\end{bmatrix}}

which is precisely the matrix of U_{-\theta} in the standard basis.
( c ) We have U_{\theta}\alpha_1=U_{\theta}U_{\phi}\epsilon_1=U_{\phi}U_{\theta}\epsilon_1=U_{\phi}(\cos\theta\epsilon_1+\sin\theta\epsilon_2)=\cos\theta\alpha_1+\sin\theta\alpha_2, and similarly U_{\theta}\alpha_2=-\sin\theta\alpha_1+\cos\theta\alpha_2, thus the matrix of U_{\theta} in the ordered basis \mathfrak B is the same as the matrix of U_{\theta} in the standard ordered basis.

Exercise 5. Let V be R^3, with the standard inner product. Let W be the plane spanned by \alpha=(1,1,1) and \beta=(1,1,-2). Let U be the linear operator defined, geometrically as follows: U is rotation through the angle \theta, about the straight line through the origin which is orthogonal to W. There are actually two such rotations–choose one. Find the matrix of U in the standard ordered basis.
Solution: We have (\alpha|\beta)=1+1-2=0, thus they are orthogonal, the vector \gamma=(1,-1,0) is orthogonal to both \alpha and \beta, if we let \alpha_1=\alpha/||\alpha||,\alpha_2=\beta/||\beta|| and \alpha_3=\gamma/||\gamma||, then the matrix of U in the basis \{\alpha_1,\alpha_2,\alpha_3\} is

\displaystyle{A=\begin{bmatrix}\cos\theta&\sin\theta&0\\{-\sin\theta}&\cos\theta&0\\0&0&1\end{bmatrix}}

the matrix B we required can be obtained by

B=P^{-1}AP,\quad P=\begin{bmatrix}\dfrac{1}{\sqrt{3}}&\dfrac{1}{\sqrt{6}}&\dfrac{1}{\sqrt{2}}\\{\dfrac{1}{\sqrt{3}}}&\dfrac{1}{\sqrt{6}}&-\dfrac{1}{\sqrt{2}}\\{\dfrac{1}{\sqrt{3}}}&-\dfrac{2}{\sqrt{6}}&0\end{bmatrix}

Exercise 6. Let V be a finite-dimensional inner product space, and let W be a subspace of V. Then V=W\oplus W^{\perp}, that is, each \alpha\in V is uniquely expressible in the form \alpha=\beta+\gamma, with \beta\in W and \gamma\in W^{\perp}. Define a linear operator U by U\alpha=\beta-\gamma.
( a ) Prove that U is both self-adjoint and unitary.
( b ) If V is R^3 with the standard inner product and W is the subspace spanned by (1,0,1), find the matrix of U in the standard ordered basis.
Solution:
( a ) To prove U is self adjoint, let \alpha=\beta+\gamma and \alpha'=\beta'+\gamma', then (U\alpha|\alpha')=(\beta-\gamma|\beta'+\gamma')=(\beta|\beta')-(\gamma|\gamma'), and (\alpha|U\alpha')=(\beta+\gamma|\beta'-\gamma')=(\beta|\beta')-(\gamma|\gamma'), thus (U\alpha|\alpha')=(\alpha|U\alpha'), which means U^{\ast}=U.
To prove U is unitary, let U\alpha=0, then \beta=\gamma, which means \beta=\gamma=0, or U is isomorphism, since (U\alpha|U\alpha')=(\alpha|\alpha')=(\beta|\beta')+(\gamma|\gamma'), U preserves inner product.
( b ) W is spanned by \beta=\epsilon_1+\epsilon_3, and W^{\perp} is spanned by \gamma_1=\epsilon_2,\gamma_2=\epsilon_1-\epsilon_3, so

\displaystyle{U\epsilon_1=U(\frac{1}{2}\beta+\frac{1}{2}\gamma_2)=\frac{1}{2}\beta-\frac{1}{2}\gamma_2=\epsilon_3 \\ U\epsilon_2=U\gamma_1=-\gamma_1=-\epsilon_2 \\ U\epsilon_3=U(\frac{1}{2}\beta-\frac{1}{2}\gamma_2)=\frac{1}{2}\beta+\frac{1}{2}\gamma_2=\epsilon_1}

the matrix of U in the standard ordered basis is \begin{bmatrix}0&0&1\\0&1&0\\1&0&0\end{bmatrix}.

Exercise 7. Let V be a complex inner product space and T a self-adjoint linear operator on V. Show that
( a ) ||\alpha+iT\alpha||=||\alpha-iT\alpha|| for every \alpha\in V.
( b ) \alpha+iT\alpha=\beta+iT\beta if and only if \alpha=\beta.
( c ) I+iT is non-singular.
( d ) I-iT is non-singular.
( e ) Now suppose V is finite-dimensional, and prove that U=(I-iT)(I+iT)^{-1} is a unitary operator; U is called the Cayley transform of T. In a certain sense, U=f(T), where f(x)=(1-ix)/(1+ix).
Solution:
( a ) We have (T\alpha|\alpha)=(\alpha|T\alpha) and

\displaystyle{\begin{aligned}||\alpha+iT\alpha||^2&=(\alpha+iT\alpha|\alpha+iT\alpha)=(\alpha|\alpha)+i(T\alpha|\alpha)+(\alpha|iT\alpha)+(iT\alpha|iT\alpha)\\&=(\alpha|\alpha)+(T\alpha|T\alpha)/||\alpha-iT\alpha||^2\\&=(\alpha-iT\alpha|\alpha-iT\alpha)=(\alpha|\alpha)-i(T\alpha|\alpha)-(\alpha|iT\alpha)+(iT\alpha|iT\alpha)\\&=(\alpha|\alpha)+(T\alpha|T\alpha)\end{aligned}}

( b ) One direction is obvious, for the other direction, we have \alpha-\beta+iT(\alpha-\beta)=0 and so ||\alpha-\beta+iT(\alpha-\beta)||=0=||\alpha-\beta-iT(\alpha-\beta)||, which means \alpha-\beta-iT(\alpha-\beta)=0, thus \alpha-\beta=0.
( c ) If (I+iT)\alpha=0, then \alpha+iT\alpha=0+iT0, thus \alpha=0.
( d ) follows from ( c ) and ( a ).
( e ) U is invertible by ( c ) and ( d ). We have (I-iT)^{\ast}=I^{\ast}+iT^{\ast}=I+iT and (I-iT)(I+iT)=I+T, also (I+iT)(I-iT)=I+T, so

\displaystyle{\begin{aligned}(U\alpha|U\beta)&=((I-iT)(I+iT)^{-1}\alpha|(I-iT)(I+iT)^{-1}\beta)\\&=((I+iT)^{-1}\alpha|(I+T)(I+iT)^{-1}\beta)\\&=((I+iT)^{-1}\alpha|(I-iT)\beta)=(\alpha|\beta)\end{aligned}}

Exercise 8. If \theta is a real number, prove that the following matrices are unitarily equivalent.

\displaystyle{\begin{bmatrix}{\cos\theta}&{-\sin\theta}\\{\sin\theta}&{\cos\theta}\end{bmatrix},\quad \begin{bmatrix}e^{i\theta}&0\\0&e^{-i\theta}\end{bmatrix}}

Solution: Let

\displaystyle{P=\frac{1}{\sqrt{2}}\begin{bmatrix}1&1\-i&i\end{bmatrix}}

We can verify that P is unitary and

\displaystyle{\begin{bmatrix}{\cos\theta}&{-\sin\theta}\\{\sin\theta}&{\cos\theta}\end{bmatrix}P=P \begin{bmatrix}e^{i\theta}&0\\0&e^{-i\theta}\end{bmatrix}}

Exercise 9. Let V be a finite-dimensional inner product space and T a positive linear operator on V. Let p_T be the inner product on V defined by p_T(\alpha,\beta)=(T\alpha|\beta). Let U be a linear operator on V and U^{\ast} its adjoint with respect to (|). Prove that U is unitary with respect to the inner product p_T if and only if T=U^{\ast}TU.
Solution: We have p_T(U\alpha,U\beta)=(TU\alpha|U\beta)=(U^{\ast}TU\alpha|\beta) and p_T(\alpha,\beta)=(T\alpha|\beta), thus U is unitary with respect to p_T if and only if (U^{\ast}TU\alpha|\beta)=(T\alpha|\beta) for all \alpha,\beta\in V, this happens if and only if T=U^{\ast}TU.

Exercise 10. Let V be a finite-dimensional inner product space. For each \alpha,\beta\in V, let T_{\alpha,\beta} be the linear operator on V defined by T_{\alpha,\beta}(\gamma)=(\gamma|\beta)\alpha. Show that
( a ) T^{\ast}_{\alpha,\beta}=T_{\beta,\alpha}.
( b ) \text{trace }(T_{\alpha,\beta})=(\alpha|\beta).
( c ) T_{\alpha,\beta}T_{\gamma,\delta}=T_{\alpha,(\beta|\gamma)\delta}.
( d ) Under what conditions is T_{\alpha,\beta} self-adjoint?
Solution:
( a ) We have (T_{\alpha,\beta}(a)|b)=((a|\beta)\alpha|b)=(a|\beta)(\alpha|b), and also (a|T_{\beta,\alpha}(b))=(a|(b|\alpha)\beta)=(\alpha|b)(a|\beta), thus (T_{\alpha,\beta}(a)|b)=(a|T_{\beta,\alpha}(b)), which proves the result.
( b ) Let \{\epsilon_1,\dots,\epsilon_n\} be a basis for V, and \alpha=\sum_1^nc_i\epsilon_i. By definition we have T_{\alpha,\beta}(\epsilon_i)=(\epsilon_i|\beta)\alpha=\sum_{j=1}^nc_j(\epsilon_i|\beta)\alpha_j, which means \text{trace }(T_{\alpha,\beta})=\sum_{j=1}^nc_j(\epsilon_j|\beta)=(\alpha|\beta).
( c ) For any a\in V, we have T_{\alpha,\beta}T_{\gamma,\delta}(a)=T_{\alpha,\beta}(a|\delta)\gamma=(a|\delta)(\gamma|\beta)\alpha, and T_{\alpha,(\beta|\gamma)\delta}(a)=(a|(\beta|\gamma)\delta)\alpha=(\gamma|\beta)(a|\delta)\alpha, thus they are equal.
( d ) We shall have T_{\alpha,\beta}=T_{\beta,\alpha} if T is self-adjoint, which means (a|\beta)\alpha=(a|\alpha)\beta, this requires \alpha=k\beta with k\in R, or \beta=0.

Exercise 11. Let V be an n-dimensional inner product space over the field F, and let L(V,V) be the space of linear operators on V. Show that there is a unique inner product on L(V,V) with the property that ||T_{\alpha,\beta}||^2=||\alpha||^2||\beta||^2 for all \alpha,\beta\in V. Find an isomorphism between L(V,V) with this inner product and the space of n\times n matrices over F, with the inner product (A|B)=\text{tr}(AB^{\ast}).
Solution: The inner product (T|U)=\text{trace}(TU^{\ast}) satisfies the condition since

\displaystyle{||T_{\alpha,\beta}||^2=(T_{\alpha,\beta}|T_{\alpha,\beta})=\text{trace}(T_{\alpha,\beta}T_{\beta,\alpha})=\text{trace}(T_{\alpha,(\beta|\beta)\alpha})=||\alpha||^2||\beta||^2}

the uniqueness follows from the polarization identities. If we link T\in L(V,V) with the matrix of T in the standard basis of V, then suppose the matrix of T and U are A and B, then (M(T)|M(U))=(A|B)=\text{tr}(AB^{\ast}).

Exercise 12. Let V be a finite-dimensional inner product space. In Exercise 6, we showed how to construct some linear operators on V which are both self-adjoint and unitary. Now we prove that there are no others, i.e., that every self-adjoint unitary operator arises from some subspace W as we described in Exercise 6.
Solution: Let U be a self-adjoint unitary operator, then U^{\ast}=U and UU^{\ast}=I, which means U^2=I or (U+I)(U-I)=0, it follows that the minimal polynomial for U divides (x-1)(x+1), thus U is diagonalizable, the possible characteristic value being 1,-1. Choose a basis for V consisting of characteristic vectors of U, namely \mathfrak B, let W be the subspace spanned by characteristic vectors in \mathfrak B associated with 1, and W' be the subspace spanned by characteristic vectors in \mathfrak B associated with -1, we only have to prove W'=W^{\perp}, choose a\in W and b\in W', then Ua=a and Ub=-b, since U is unitary we have

\displaystyle{(a|b)=(Ua|Ub)=(a|-b)=-(a|b)\implies(a|b)=0}

Exercise 13. Let V and W be finite-dimensional inner product spaces having the same dimension. Let U be an isomorphism of V onto W. Show that
( a ) The mapping T\rightarrow UTU^{-1} is an isomorphism of the vector space L(V,V) onto the vector space L(W,W).
( b ) \text{trace}(UTU^{-1})=\text{trace}(T) for each T\in L(V,V).
( c ) UT_{\alpha,\beta}U^{-1}=T_{U\alpha,U\beta} (T_{\alpha,\beta} described in Exercise 10).
( d ) (UTU^{-1})^{\ast}=UT^{\ast}U^{-1}.
( e ) If we equip L(V,V) with inner product (T_1|T_2)=\text{trace}(T_1T_2^{\ast}), and similarly for L(W,W), then T\rightarrow UTU^{-1} is an inner product space isomorphism.
Solution:
( a ) If UTU^{-1} is the zero operator on W, then since U is an isomorphism, we have TU^{-1}w=0 for all w\in W, as U^{-1} has range V, we see T is the zero operator, thus the mapping is injective. Let T'\in L(W,W) be any operator, we choose a basis for W to be \{w_1,\dots,w_n\}, then let v_i=U^{-1}w_i, we see \{v_1,\dots,v_n\} is a basis for V. Define T as Tv_i=U^{-1}T'w_i for i=1,\dots,n, then UTU^{-1}w_i=T'w_i, thus T'=UTU^{-1} and the mapping is surjective.
( b ) Let \mathfrak B=\{v_1,\dots,v_n\} be a basis for V, then \text{trace}(T)=\text{trace}([T]_{\mathfrak B}). If w_i=Uv_i for i=1,\dots,n, then \mathfrak B'=\{w_1,\dots,w_n\} is a basis for W, notice that

\displaystyle{UTU^{-1}w_j=UTv_j=U\left(\sum_{i=1}^n{A_{ij}v_i}\right)=\sum_{i=1}^n{A_{ij}w_i}}

We can say that [UTU^{-1}]_{\mathfrak B'}=[T]_{\mathfrak B}, and the conclusion is obvious.
( c ) Since U is an isomorphism, we have (\alpha|\beta)=(U\alpha|U\beta), while the first inner product is on V and the second on W. Again we choose a basis for W to be \{w_1,\dots,w_n\} and let v_i=U^{-1}w_i, then

\displaystyle{UT_{\alpha,\beta}U^{-1}w_j=UT_{\alpha,\beta}(v_j)=U(v_j|\beta)\alpha=(v_j|\beta)U\alpha \\ T_{U\alpha,U\beta}(w_j)=(w_j|U\beta)U\alpha=(v_j|\beta)U\alpha}

( d ) Use the fact that (\alpha|\beta)=(U\alpha|U\beta), we see that for any a,b\in W

\displaystyle{(UTU^{-1}a|b)=(UTU^{-1}a|UU^{-1}b)=(TU^{-1}a|U^{-1}b)=(U^{-1}a|T^{\ast}U^{-1}b)=(a|UT^{\ast}U^{-1}b)}

by the uniqueness of (UTU^{-1})^{\ast} we see that (UTU^{-1})^{\ast}=UT^{\ast}U^{-1}.
( e ) From (a), T\rightarrow UTU^{-1} is already a vector space isomorphism, If T_1,T_2\in L(V,V), then

\displaystyle{\begin{aligned}(UT_1U^{-1}|UT_2U^{-1})&=\text{trace}(UT_1U^{-1}(UT_2U^{-1})^{\ast})=\text{trace}(UT_1U^{-1}UT_2^{\ast}U^{-1})\\&=\text{trace}(UT_1T_2^{\ast}U^{-1})=\text{trace}(T_1T_2^{\ast})=(T_1|T_2)\end{aligned}}

Exercise 14. If V is an inner product space, a rigid motion is any function T from V into V (not necessarily linear) such that ||T\alpha-T\beta||=||\alpha-\beta|| for all \alpha,\beta\in V. One example of a rigid motion is a linear unitary operator. Another example is translation by a fixed vector \gamma: T_{\gamma}(\alpha)=\alpha+\gamma
( a ) Let V be R^2 with the standard inner product. Suppose T is a rigid motion of V and that T(0)=0. Prove that T is linear and a unitary operator.
( b ) Use the result of part (a) to prove that every rigid motion of R^2 is composed of a translation, followed by a unitary operator.
( c ) Now show that a rigid motion of R^2 is either a translation followed by a rotation, or a translation followed by a reflection followed by a rotation.
Solution:
( a ) We have ||T\alpha-T(0)||=||\alpha-0||=||\alpha|| for all \alpha\in V, thus (T\alpha|T\beta)=(\alpha|\beta) due to the polarization identity. Thus we have

\displaystyle{(T(c\alpha+\beta)|T\gamma)=(c\alpha+\beta|\gamma)=c(\alpha|\gamma)+(\beta|\gamma)=c(T\alpha|T\gamma)+(T\beta|T\gamma)=(cT\alpha+T\beta|T\gamma)}

This is true for any choice of \alpha,\beta,\gamma\in V, thus we can conclude T(c\alpha+\beta)=cT\alpha+T\beta, or T is linear. Now ||T\alpha||=||\alpha|| shows T is injective and preserves norms, thus T is unitary.
( b ) If T is any rigid motion of R^2, we let \gamma=-T(0), then T_{\gamma}(\alpha)=\alpha-T(0), thus T_{\gamma}T(\alpha)=T\alpha-T(0) and T_{\gamma}T(0)=0, moreover T_{\gamma}T is still a rigid motion, thus unitary. Let U=T_{\gamma}T, then T_{-\gamma}U=T_{-\gamma}T_{\gamma}T=T, which proves the result.
( c ) The conclusion follows from (b) and Exercise 4.

Exercise 15. A unitary operator on R^4 (with the standard inner product) is simply a linear operator which preserves the quadratic form

\displaystyle{||(x,y,z,t)||^2=x^2+y^2+z^2+t^2}

that is, a linear operator U such that ||U\alpha||^2=||\alpha||^2 for all \alpha\in R^4. In a certain part of the theory of relativity, it is of interest to find the linear operators T which preserve the form

\displaystyle{||(x,y,z,t)||^2_L=t^2-x^2-y^2-z^2.}

Now ||\text{ }||^2_L does not come from an inner product, but from something called the ‘Lorentz metric’ (which we shall not go into). For that reason, a linear operator T on R^4 such that ||T\alpha||^2_L=||\alpha||^2_L for every \alpha\in R^4 is called a Lorentz transformation.
( a ) Show that the function U defined by

\displaystyle{U(x,y,z,t)=\begin{bmatrix}t+x&y+iz\\y-iz&t-x\end{bmatrix}}

is an isomorphism of R^4 onto the real vector space H of all self-adjoint 2\times 2 complex matrices.
( b ) Show that ||\alpha||^2_L=\det (U\alpha).
( c ) Suppose T is a (real) linear operator on the space H of 2\times 2 self-adjoint matrices. Show that L=U^{-1}TU is a linear operator on R^4.
( d ) Let M be any 2\times 2 complex matrix. Show that T_M(A)=M^{\ast}AM defines a linear operator T_M on H. (Be sure you check that T_M maps H into H.)
( e ) If M is a 2\times 2 matrix such that |\det M|=1, show that L_M=U^{-1}T_MU is a Lorentz transformation on R^4.
( f ) Find a Lorentz transformation which is not an L_M.
Solution:
( a ) First if U(x,y,z,t)=0, then we have t+x=0,t-x=0, which gives t=x=0, similarly y+iz=0,y-iz=0 gives y=z=0, this shows U is injective. Also U is surjectiv since for any M\in H we have

\displaystyle{M=\begin{bmatrix}a&b\\{\overline{b}}&c\end{bmatrix}, \quad a,c\in R\implies U\left(\frac{a-c}{2},\Re(b),\Im(b),\frac{a+c}{2}\right)=M}

( b ) Let \alpha=(x,y,z,t), then ||\alpha||^2_L=t^2-x^2-y^2-z^2, and \det(U\alpha)=t^2-x^2-(y^2+z^2).
( c ) Sicne U is linear, we can easily see U^{-1}TU is linear.
( d ) First check T_M maps H into H, let A\in H, then A^{\ast}=A, and so (T_M(A))^{\ast}=M^{\ast}A^{\ast}M=T_M(A), and we obviously have T_M(cA+B)=M^{\ast}(cA+B)M=cT_M(A)+T_M(B).
( e ) We have to show ||L_M(\alpha)||_L^2=||\alpha||^2_L, let \alpha=(x,y,z,t), then ||\alpha||^2_L=\det(U\alpha), and

\displaystyle{||L_M(\alpha)||_L^2=||U^{-1}T_MU\alpha||_L^2=\det(UU^{-1}T_MU\alpha)=\det(M^{\ast}(U\alpha)M)=\det(U\alpha)}