IS Math-2019-01

题目来源：Problem 1 日期：2024-07-07 题目主题：Math-线性代数-酉矩阵与正交矩阵

具体题目

A complex square matrix $X$ is unitary if it holds that $X X^{*} = I$ , where $X^{*}$ is the conjugate transpose of $X$ (also known as the adjoint matrix of $X$ ) and $I$ is the appropriate identity matrix. Let $i$ be the imaginary unit. Answer the following questions.

For a positive integer $n$ , suppose that $A$ and $B$ are unitary matrices of size $n$ . Show that the matrix $AB$ is also unitary.
For a positive integer $n$ , suppose that $C$ and $D$ are real square matrices of size $n$ . Let $F$ be defined as $F = C + i D$ and $G$ be defined as

G = (C D - D C) .

Show that the matrix $G$ is orthogonal if and only if the matrix $F$ is unitary.

Find the eigenvalues of the following matrix.

\frac{1}{2} 1111 1 i - 1 - i 1 - 1 1 - 1 1 - i - 1 i

For a positive integer $n$ , suppose that the $(j, k)$ -th element $q_{jk}$ of a square matrix $Q$ of size $n$ is given by

q_{jk} = \frac{1}{n} exp (\frac{2 πi ( j - 1 ) ( k - 1 )}{n}) .

Show that the matrix $Q$ is unitary.

Show that a unitary matrix of size $2$ with determinant $1$ has a form of

H = (exp (i ψ) cos θ - exp (- i ψ) sin θ exp (i ψ) sin θ exp (- i ψ) cos θ),

where $θ$ and $ψ$ are real numbers.

Find the general form of the unitary matrices of size $2$ .

正确解答

Question 1

To show that the matrix $AB$ is also unitary if both $A$ and $B$ are unitary matrices, we need to verify that $(AB) (AB)^{*} = I$ .

Given:

A A^{*} = I and B B^{*} = I

Now consider $AB$ :

(AB) (AB)^{*} = AB (AB)^{*} = AB B^{*} A^{*} = AI A^{*} = A A^{*} = I

Thus, $AB$ is unitary.

Question 2

To show that $G$ is orthogonal if and only if $F$ is unitary, we need to verify that $G^{T} G = I$ if and only if $F F^{*} = I$ .

Consider $G$ :

G = (C D - D C)

Compute $G^{T}$ :

G^{T} = (C^{T} - D^{T} D^{T} C^{T})

Now compute $G^{T} G$ :

G^{T} G = (C^{T} - D^{T} D^{T} C^{T}) (C D - D C) = (C^{T} C + D^{T} D - D^{T} C + C^{T} D - C^{T} D + D^{T} C D^{T} D + C^{T} C)

Since $C$ and $D$ are real matrices, $C^{T} = C^{*}$ and $D^{T} = D^{*}$ . Thus, if $G$ is orthogonal, then:

C^{T} C + D^{T} D = I and - C^{T} D + D^{T} C = O

Now consider $F$ :

F = C + i D

Compute $F F^{*}$ :

F F^{*} = (C + i D) (C - i D) = C C^{*} + D D^{*} + i (C D^{*} - D C^{*})

For $F$ to be unitary:

C C^{*} + D D^{*} = I and C D^{*} - D C^{*} = O

Since $C^{*} = C^{T}$ and $D^{*} = D^{T}$ , the conditions match, and thus $G$ is orthogonal if and only if $F$ is unitary.

Question 3

To find the eigenvalues of the matrix

A = \frac{1}{2} 1111 1 i - 1 - i 1 - 1 1 - 1 1 - i - 1 i,

we can use the properties of its columns to simplify our calculations.

Observe Orthogonality and Symmetry: The columns (or rows) of matrix $A$ are orthogonal. This means that the dot product of any two different columns is zero. Also, observe that the complex columns are conjugates.
Calculate $A^{2}$ : To simplify finding the eigenvalues, we can compute $A^{2}$ :

A^{2} = \frac{1}{2} 1111 1 i - 1 - i 1 - 1 1 - 1 1 - i - 1 i^{2}

Perform Matrix Multiplication: Multiply $A$ by itself:

A \cdot A = \frac{1}{4} 1111 1 i - 1 - i 1 - 1 1 - 1 1 - i - 1 i 1111 1 i - 1 - i 1 - 1 1 - 1 1 - i - 1 i = \frac{1}{4} 4000000400400400 = 1000000100100100

Square the Result Again: To further simplify and find the eigenvalues, we compute $(A^{2})^{2}$ :

A^{4} = 1000000100100100^{2}

Perform the multiplication:

A^{4} = 10000001001001001000000100100100 = 1000010000100001

This indicates that $A^{4} = I$ , the identity matrix.

Determine the Eigenvalues: Since

$A^{4} = I$ , we know that the eigenvalues of $A^{2}$ are $\pm 1$ . Because the eigenvalues of $A^{2}$ are the squares of the eigenvalues of $A$ , we get:

λ^{4} = 1

Solving for $λ$ , we find:

λ = \pm 1, \pm i

Therefore, the eigenvalues of the matrix $A$ are $λ = 1, - 1, i, - i$ .

Question 4

To show that the matrix $Q$ is unitary:

q_{jk} = \frac{1}{n} exp (\frac{2 πi ( j - 1 ) ( k - 1 )}{n})

Let

Q = q_{11} q_{21} ⋮ q_{n 1} q_{12} q_{22} ⋮ q_{n 2} \dots \dots ⋱ \dots q_{1 n} q_{2 n} ⋮ q_{nn}

We need to show that $Q Q^{*} = I$ .

Compute $Q^{*}$ :

Q^{*} = \overline{q_{11}} \overline{q_{12}} ⋮ \overline{q_{1 n}} \overline{q_{21}} \overline{q_{22}} ⋮ \overline{q_{2 n}} \dots \dots ⋱ \dots \overline{q_{n 1}} \overline{q_{n 2}} ⋮ \overline{q_{nn}} = \frac{1}{n} exp (- \frac{2 πi ( 0 ) ( 0 )}{n}) exp (- \frac{2 πi ( 1 ) ( 0 )}{n}) ⋮ exp (- \frac{2 πi ( n - 1 ) ( 0 )}{n}) exp (- \frac{2 πi ( 0 ) ( 1 )}{n}) exp (- \frac{2 πi ( 1 ) ( 1 )}{n}) ⋮ exp (- \frac{2 πi ( n - 1 ) ( 1 )}{n}) \dots \dots ⋱ \dots exp (- \frac{2 πi ( 0 ) ( n - 1 )}{n}) exp (- \frac{2 πi ( 1 ) ( n - 1 )}{n}) ⋮ exp (- \frac{2 πi ( n - 1 ) ( n - 1 )}{n}) .

Now compute $Q Q^{*}$ :

(Q Q^{*})_{jk} = m = 1 \sum n q_{jm} \overline{q_{km}} = m = 1 \sum n \frac{1}{n} exp (\frac{2 πi ( j - 1 ) ( m - 1 )}{n}) exp (- \frac{2 πi ( k - 1 ) ( m - 1 )}{n}) .

Simplify the exponent:

= \frac{1}{n} m = 1 \sum n exp (\frac{2 πi ( j - k ) ( m - 1 )}{n}) .

For $j = k$ :

\frac{1}{n} m = 1 \sum n 1 = 1.

For $j \neq = k$ :

\frac{1}{n} m = 1 \sum n exp (\frac{2 πi ( j - k ) ( m - 1 )}{n}) = 0 (since it is a sum of roots of unity) .

Hence, $Q Q^{*} = I$ , and $Q$ is unitary.

Question 5

Let $H$ be a unitary matrix of size $2$ with determinant $1$ . Let $H$ has the form:

H = (a c b d)

We know that $H H^{*} = I$ , so:

(a c b d) (\overline{a} \overline{b} \overline{c} \overline{d}) = (1001)

Also, $H^{*} H = I$ , so:

(\overline{a} \overline{b} \overline{c} \overline{d}) (a c b d) = (1001)

Solving these equations, we get:

a \overline{a} + b \overline{b} c \overline{a} + d \overline{b} a \overline{c} + b \overline{d} c \overline{c} + d \overline{d} \overline{a} a + \overline{c} c \overline{b} a + \overline{d} c \overline{a} b + \overline{c} d \overline{b} b + \overline{d} d = 1 = 0 = 0 = 1 = 1 = 0 = 0 = 1

Therefore,

b \overline{b} a \overline{a} = c \overline{c} = d \overline{d}

So we can let $c$ , $d$ as:

c d = b exp (i θ) = a exp (i ϕ)

Substitute $c$ and $d$ back into the equations, we get:

ab exp (i θ) + ab exp (- i ϕ) ab + ab exp (i (θ - ϕ)) = 0 = 0

Therefore, $θ - ϕ = (2 k + 1) π$ for some integer $k$ .

Then, the general form of $H$ is:

H = (a - b exp (i ϕ) b a exp (i ϕ))

From $a \overline{a} + b \overline{b} = 1$ , we have:

a b = exp (i ψ) cos θ = exp (i ψ) sin θ

Therefore, the general form of $H$ is:

H = (exp (i ψ) cos θ - exp (i (ϕ - ψ)) sin θ exp (i ψ) sin θ exp (i (ϕ - ψ)) cos θ)

with a determinant of $d e t (H) = e x p (i (ϕ))$ .

Since $H$ has a determinant of $1$ , we have that $ϕ = 0$ . Therefore, the general form of $H$ is:

H = (exp (i ψ) cos θ - exp (- i ψ) sin θ exp (i ψ) sin θ exp (- i ψ) cos θ)

Question 6

To find the general form of unitary matrices of size $2$ , we can use the results from Question 5. The general form of a unitary matrix of size $2$ is:

H = (exp (i ψ) cos θ - exp (i (ϕ - ψ)) sin θ exp (i ψ) sin θ exp (i (ϕ - ψ)) cos θ)

知识点

酉矩阵正交矩阵特征值和特征向量 Cayley-Hamilton定理特征多项式

特征多项式 Characteristic Polynomial

Cayley-Hamilton 定理 Cayley-Hamilton Theorem

复数的根和单位根 Roots and Roots of Unity

复共轭 Hermitian Conjugate

复矩阵的特征值和特征向量 Eigenvalues and Eigenvectors of Complex Matrices

Complex Matrices

难点解题思路

验证矩阵是酉矩阵的步骤通常涉及计算其共轭转置并验证乘积是否为单位矩阵。
对于正交矩阵，需要验证其转置乘以自身是否为单位矩阵。
特征值计算可能涉及观察矩阵的结构或使用代数方法简化计算。

解题技巧和信息

处理酉矩阵和正交矩阵时，通常需要利用矩阵的共轭转置和单位矩阵的性质。
对于复数矩阵的特征值问题，识别矩阵的特殊结构有助于简化计算。
通过参数化方法，可以找到特定类型的矩阵的一般形式，例如 2×2 酉矩阵。

重点词汇

Unitary Matrix (酉矩阵): A matrix $X$ such that $X X^{*} = I$ .
Orthogonal Matrix (正交矩阵): A real matrix $G$ such that $G^{T} G = I$ .
Eigenvalue (特征值): A scalar $λ$ such that $Av = λ v$ for some vector $v$ .
Conjugate Transpose (共轭转置): The transpose of a matrix with its elements replaced by their complex conjugates.

参考资料

Linear Algebra and Its Applications by Gilbert Strang, Chap. 7.
Introduction to Linear Algebra by Gilbert Strang, Chap. 5.

Zephyr's Notes on ISCS & CBMS, UTokyo

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