满足:
A
H
A
=
A
A
H
A^H A = AA^H
AHA=AAH
的矩阵,被称为正规矩阵
证明
A
A
A可以酉相似对角化的充要条件是,
A
A
A是正规矩阵
A
H
A
=
A
A
H
A^H A = AA^H
AHA=AAH
这里插一句:
一般矩阵可以对角化是:
P
−
1
A
P
=
Λ
P^{-1}AP = \Lambda
P−1AP=Λ
Λ
\Lambda
Λ 是对角阵,而对角化只要求P是个可逆矩阵即可
这里的酉对角化,是一个更强的条件,要求 P P P是一个酉矩阵
再插一句:
酉矩阵
U
U
U,可以看做正交矩阵的推广,即:
U
H
U
=
U
U
H
=
I
U^HU = UU^H = I
UHU=UUH=I
要证明,上述条件是充要的
先证 A H A = A A H A^H A = AA^H AHA=AAH ⇐ \Leftarrow ⇐ A 可以酉相似对角化 A\text{可以酉相似对角化} A可以酉相似对角化
则,存在酉矩阵
U
U
U,
s
.
t
.
s.t.
s.t.:
U
−
1
A
U
=
U
H
A
U
=
[
λ
1
λ
2
.
.
.
.
λ
n
]
U^{-1} A U = U^H A U = \begin{bmatrix} \lambda_1 & & & \\ & \lambda_2 \\ & & .... \\ &&& \lambda_n \end{bmatrix}
U−1AU=UHAU=⎣⎢⎢⎡λ1λ2....λn⎦⎥⎥⎤
给上式取转置共轭,则有:
U
H
A
H
U
=
[
λ
ˉ
1
λ
ˉ
2
.
.
.
.
λ
ˉ
n
]
U^H A^H U = \begin{bmatrix} \bar{\lambda}_1 & & & \\ & \bar{\lambda}_2 \\ & & .... \\ &&& \bar{\lambda}_n \end{bmatrix}
UHAHU=⎣⎢⎢⎡λˉ1λˉ2....λˉn⎦⎥⎥⎤
然后两式子相乘
(
U
H
A
U
)
(
U
H
A
H
U
)
=
U
H
A
A
H
U
=
[
∣
λ
1
∣
2
∣
λ
2
∣
2
.
.
.
.
∣
λ
n
∣
2
]
(U^H A U) (U^H A^H U) = U^H A A^H U = \begin{bmatrix} | \lambda_1|^2 & & & \\ & | \lambda_2|^2 \\ & & .... \\ &&& |\lambda_n|^2 \end{bmatrix}
(UHAU)(UHAHU)=UHAAHU=⎣⎢⎢⎡∣λ1∣2∣λ2∣2....∣λn∣2⎦⎥⎥⎤
而 ( U H A H U ) ( U H A U ) = U H A H A U (U^H A^H U) (U^H A U)=U^H A^H A U (UHAHU)(UHAU)=UHAHAU 的值也是上述对角阵
故而:
U
H
A
H
A
U
=
U
H
A
A
H
U
U^H A^H A U = U^H A A^H U
UHAHAU=UHAAHU
有:
A
H
A
=
A
A
H
A^H A = A A^H
AHA=AAH
再证 A H A = A A H A^H A = AA^H AHA=AAH ⇒ \Rightarrow ⇒ A 可以酉相似对角化 A\text{可以酉相似对角化} A可以酉相似对角化
由附录的Schur分解定理,显然存在酉矩阵 U U U, s . t . s.t. s.t.
U
H
A
U
=
[
t
11
t
12
.
.
.
t
1
n
t
22
.
.
.
t
2
n
.
.
.
.
.
.
t
n
n
]
U^H A U = \begin{bmatrix} t_{11} & t_{12} & ... & t_{1n} \\ & t_{22} & ... & t_{2n} \\ & & ... & ... \\ && & t_{nn} \\ \end{bmatrix}
UHAU=⎣⎢⎢⎡t11t12t22.........t1nt2n...tnn⎦⎥⎥⎤
对该式子进行共轭转置,有:
U
H
A
H
U
=
[
t
ˉ
11
t
ˉ
12
t
ˉ
22
.
.
.
.
.
.
.
.
.
t
ˉ
1
n
t
ˉ
2
n
.
.
.
t
ˉ
1
n
]
U^H A^H U = \begin{bmatrix} \bar{t}_{11} & & & \\ \bar{t}_{12} & \bar{t}_{22} & & \\ ... & ... & ... & \\ \bar{t}_{1n} & \bar{t}_{2n} & ... & \bar{t}_{1n} \\ \end{bmatrix}
UHAHU=⎣⎢⎢⎡tˉ11tˉ12...tˉ1ntˉ22...tˉ2n......tˉ1n⎦⎥⎥⎤
根据之前的证明
(
U
H
A
H
U
)
(
U
H
A
U
)
=
U
H
A
H
A
U
=
U
H
A
A
H
U
=
(
U
H
A
U
)
(
U
H
A
H
U
)
(U^H A^H U) (U^H A U)=U^H A^H A U = U^H A A^H U = (U^H A U)(U^H A^H U)
(UHAHU)(UHAU)=UHAHAU=UHAAHU=(UHAU)(UHAHU)
有:
[
t
11
t
12
.
.
.
t
1
n
t
22
.
.
.
t
2
n
.
.
.
.
.
.
t
n
n
]
[
t
ˉ
11
t
ˉ
12
t
ˉ
22
.
.
.
.
.
.
.
.
.
t
ˉ
1
n
t
ˉ
2
n
.
.
.
t
ˉ
1
n
]
=
[
t
ˉ
11
t
ˉ
12
t
ˉ
22
.
.
.
.
.
.
.
.
.
t
ˉ
1
n
t
ˉ
2
n
.
.
.
t
ˉ
1
n
]
[
t
11
t
12
.
.
.
t
1
n
t
22
.
.
.
t
2
n
.
.
.
.
.
.
t
n
n
]
\begin{bmatrix} t_{11} & t_{12} & ... & t_{1n} \\ & t_{22} & ... & t_{2n} \\ & & ... & ... \\ && & t_{nn} \\ \end{bmatrix}\begin{bmatrix} \bar{t}_{11} & & & \\ \bar{t}_{12} & \bar{t}_{22} & & \\ ... & ... & ... & \\ \bar{t}_{1n} & \bar{t}_{2n} & ... & \bar{t}_{1n} \\ \end{bmatrix} = \begin{bmatrix} \bar{t}_{11} & & & \\ \bar{t}_{12} & \bar{t}_{22} & & \\ ... & ... & ... & \\ \bar{t}_{1n} & \bar{t}_{2n} & ... & \bar{t}_{1n} \\ \end{bmatrix} \begin{bmatrix} t_{11} & t_{12} & ... & t_{1n} \\ & t_{22} & ... & t_{2n} \\ & & ... & ... \\ && & t_{nn} \\ \end{bmatrix}
⎣⎢⎢⎡t11t12t22.........t1nt2n...tnn⎦⎥⎥⎤⎣⎢⎢⎡tˉ11tˉ12...tˉ1ntˉ22...tˉ2n......tˉ1n⎦⎥⎥⎤=⎣⎢⎢⎡tˉ11tˉ12...tˉ1ntˉ22...tˉ2n......tˉ1n⎦⎥⎥⎤⎣⎢⎢⎡t11t12t22.........t1nt2n...tnn⎦⎥⎥⎤
仅仅看等式左右结果的对角线元素:
{
∣
t
11
∣
2
+
∣
t
12
∣
2
+
.
.
.
+
∣
t
1
n
∣
2
=
∣
t
11
∣
2
∣
t
22
∣
2
+
∣
t
23
∣
2
+
.
.
.
+
∣
t
2
n
∣
2
=
∣
t
22
∣
2
.
.
.
.
.
.
∣
t
n
n
∣
2
=
∣
t
1
n
∣
2
+
∣
t
2
n
∣
2
+
.
.
.
+
∣
t
n
n
∣
2
\left\{\begin{matrix} |t_{11}|^2 + |t_{12}|^2 + ... + |t_{1n}|^2 & = & |t_{11}|^2 \\ |t_{22}|^2 + |t_{23}|^2 + ... + |t_{2n}|^2 & = & |t_{22}|^2 \\ ... && ...\\ |t_{nn}|^2 & = & |t_{1n}|^2 + |t_{2n}|^2 + ... + |t_{nn}|^2& \end{matrix}\right.
⎩⎪⎪⎨⎪⎪⎧∣t11∣2+∣t12∣2+...+∣t1n∣2∣t22∣2+∣t23∣2+...+∣t2n∣2...∣tnn∣2===∣t11∣2∣t22∣2...∣t1n∣2+∣t2n∣2+...+∣tnn∣2
得出右上角元素(对角线除外)全为0,故:
U
H
A
U
=
[
t
11
0
.
.
.
0
t
22
.
.
.
0
.
.
.
.
.
.
t
n
n
]
U^H A U = \begin{bmatrix} t_{11} & 0 & ... & 0 \\ & t_{22} & ... & 0 \\ & & ... & ... \\ && & t_{nn} \\ \end{bmatrix}
UHAU=⎣⎢⎢⎡t110t22.........00...tnn⎦⎥⎥⎤
证毕
附录:Schur分解定理
设 ∀ A ∈ C n × n \forall A \in C^{n \times n} ∀A∈Cn×n,存在酉矩阵 U ∈ C n × n U\in C^{n\times n} U∈Cn×n,使得
(划重点,这里是任意矩阵!!)
U − 1 A U = U H A U = T = [ λ 1 ∗ . . . ∗ λ 2 . . . ∗ . . . ∗ λ n ] U^{-1}AU = U^H A U = T = \begin{bmatrix} \lambda_1 & * & ... & * \\ & \lambda_2 & ... & * \\ & & ... & * \\ &&& \lambda_n \end{bmatrix} U−1AU=UHAU=T=⎣⎢⎢⎡λ1∗λ2.........∗∗∗λn⎦⎥⎥⎤
其中, λ 1 \lambda_1 λ1, λ 2 \lambda_2 λ2,…, λ n \lambda_n λn是 A A A的特征值,即 ∀ A \forall A ∀A都可以酉相似于一个上三角矩阵 T T T
证明自己搜吧哈哈哈哈
给出酉相似的标准定义:
设
A
,
B
∈
C
n
×
n
A,B\in C^{n\times n}
A,B∈Cn×n,若存在酉矩阵
U
U
U使得
U
−
1
A
U
=
U
H
A
U
=
B
U^{-1} A U = U^H A U = B
U−1AU=UHAU=B
则称
A
A
A与
B
B
B酉相似
有参考自:文章来源:https://www.toymoban.com/news/detail-421860.html
可对角化的矩阵一定是正规矩阵吗? - junjun的回答 - 知乎 https://www.zhihu.com/question/361598765/answer/1606553052文章来源地址https://www.toymoban.com/news/detail-421860.html
到了这里,关于[矩阵论]正规矩阵可酉相似对角化的文章就介绍完了。如果您还想了解更多内容,请在右上角搜索TOY模板网以前的文章或继续浏览下面的相关文章,希望大家以后多多支持TOY模板网!