[矩阵论]正规矩阵可酉相似对角化 |
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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: 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 |
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