正规矩阵的不同特征值对应的特征向量正交

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For Hermitian matrices

page 3
Proof of b). Suppose we have two distinct eigenvalues $λ≠μ$. Then
\begin{equation}Ax = λx, Ay = μy, \label3\end{equation}
where $x, y$ are eigenvectors. Multiply the first equation on $y$, use $(Ax, y) = (x, Ay)$ and the fact that $λ$ is real which was just established.
\[λ(x, y) = (λx, y) = (Ax, y) = (x, Ay) = (x, μy) = μ(x, y).\]
As $λ≠ μ$, we conclude that $(x, y) = 0$, which proves b).

For unitary matrices

page 4
Proof of b). Begin with \eqref{3}, and multiply the two equations in \eqref{3}:
\[(x, y) = (Ax, Ay) = \bar λμ(x, y).\]
As $|λ| = 1$, we conclude that $\bar λ = λ^{−1}$, so $\bar λμ=μ/λ ≠ 1$ by our assumption that $μ ≠ λ$. So $(x, y) = 0$, which proves b).

一般地, 对于正规矩阵, 如何证明「不同特征值对应的特征向量正交」

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正规性是必要的. 见MSE.
Let $A$ be a $N×N$ matrix which has $k<N$ distinct eigenvalues. Are eigenspaces corresponding to different eigenvalues orthogonal in general ?
No.
Counterexample: $\begin{pmatrix}1&0&0\\0&1&1\\0&0&2\end{pmatrix}$ has eigenspaces $\{(t,u,0)^T\}$ with eigenvalue $1$ and $\{(0,t,t)^T\}$ with eigenvalue $2$, and they are not orthogonal.

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hbghlyj 发表于 2023-1-6 18:33
正规性是必要的. 见MSE.
Let $A$ be a $N×N$ matrix which has $k

$\pmatrix{1&1\\0&2}$ 可对角化, 但特征向量 $\pmatrix{1\\0},\pmatrix{1\\1}$ 不是正交的.
抄自What's an example of a matrix that's diagonalizable but not normal?

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引理:
$A$正规, 则$A v = \lambda v \implies A^* v = \bar{\lambda} v$

证明

$T$正规$⇔\|Tx\|^{2}=\|T^{\star}x\|^{2},\;\;∀ x \in X.$
因为$T-\lambda I$正规, 且$(\lambda I)^\star=\overline{\lambda}I$, 所以$\|(T-\lambda I)x\|=\|(T^{\star}-\overline{\lambda}I)x\|,\;∀x\in X$
$$\ker(T-\lambda I)=\ker(T^{\star}-\overline{\lambda}I)$$证毕.
又见Adjoint matrix eigenvalues and eigenvectors
$A v = \lambda v \implies A^* v = \bar{\lambda} v$ if $A$ is normal

正规矩阵的不同特征值对应的特征向量正交

证明

设$\lambda\neq \mu$, 由引理, $Av=\lambda v\implies A^*v=\overline \lambda v$,\begin{cases}\langle v,Aw\rangle=\langle v,\mu w\rangle=\overline\mu\langle v,w\rangle\\
\langle v,Aw\rangle=\langle A^*v,w\rangle=\langle\overline\lambda v,w\rangle=\overline\lambda\langle v,w\rangle \end{cases}由$\overline\mu\neq\overline\lambda$得$\langle v,w\rangle =0$

搬运MSE

hbghlyj

逆命题成立吗
不同特征值对应的特征向量正交, 则为正规矩阵

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Gilbert Strang - Introduction to Linear Algebra (2016)
Chapter 6. Eigenvalues and Eigenvectors page 348

$N$ has $n$ orthonormal eigenvectors$$N=Q \Lambda \bar{Q}^{\mathrm{T}}$$if and only if $N$ is normal. (a) Start from $N=Q \Lambda \bar{Q}^{\mathrm{T}}$ with $\bar{Q}^{\mathrm{T}} Q=I$. Show that $\bar{N}^{\mathrm{T}} N=N \bar{N}^{\mathrm{T}}: N$ is normal. (b) Now start from $\bar{N}^{\mathrm{T}} N=N \bar{N}^{\mathrm{T}}$. Schur found $A=Q T \bar{Q}^{\mathrm{T}}$ for every matrix $A$, with a triangular $T$. For normal matrices $A=N$ we must show (in 3 steps) that this triangular matrix $T$ will actually be diagonal. Then $T=\Lambda$. Step 1. Put $N=Q T \bar{Q}^{\mathrm{T}}$ into $\bar{N}^{\mathrm{T}} N=N \bar{N}^{\mathrm{T}}$ to find $\bar{T}^{\mathrm{T}} T=T \bar{T}^{\mathrm{T}}$. Step 2. Suppose $T =\begin{bmatrix}a&b\\c&d\end{bmatrix}$ has $\bar T^{\rm T} T = T\bar T^{\rm T}$. Prove that $b = 0$. Step 3. Extend Step 2 to size $n$. Any normal triangular $T$ must be diagonal.

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Proposition — A normal  triangular matrix is diagonal.

Proof

Let \(A\) be any normal upper triangular matrix. Since $ {\displaystyle (A^{*}A)_{ii}=(AA^{*})_{ii},} $ using subscript notation, one can write the equivalent expression using instead the \(i\)th unit vector ($ {\hat {\mathbf {e} }}_{i} $) to select the \(i\)th row and \(i\)th column: $ {\displaystyle {\hat {\mathbf {e} }}_{i}^{\intercal }\left(A^{*}A\right){\hat {\mathbf {e} }}_{i}={\hat {\mathbf {e} }}_{i}^{\intercal }\left(AA^{*}\right){\hat {\mathbf {e} }}_{i}.} $ The expression $ {\displaystyle \left(A{\hat {\mathbf {e} }}_{i}\right)^{*}\left(A{\hat {\mathbf {e} }}_{i}\right)=\left(A^{*}{\hat {\mathbf {e} }}_{i}\right)^{*}\left(A^{*}{\hat {\mathbf {e} }}_{i}\right)} $ is equivalent, and so is $ {\displaystyle \left\|A{\hat {\mathbf {e} }}_{i}\right\|^{2}=\left\|A^{*}{\hat {\mathbf {e} }}_{i}\right\|^{2},} $

which shows that the \(i\)th row must have the same norm as the \(i\)th column.

Consider $i = 1$. The first entry of row 1 and column 1 are the same, and the rest of column 1 is zero (because of triangularity). This implies the first row must be zero for entries 2 through \(n\). Continuing this argument for row–column pairs 2 through \(n\) shows \(A\) is diagonal.