【转载】求解Hermitian特征向量的新方法

青青子衿

转自：3个搞物理的颠覆了数学常识，数学天才陶哲轩：我开始压根不相信
mp.weixin.qq.com/s?__biz=MzIzNjc1NzUzMw==&mid=2247533254& ... 94cd79d8a03e69a&
如下举三阶时的例子
\begin{gather*}
\boldsymbol{A}=
\begin{pmatrix}
a_{\overset{\,}11}&a_{\overset{\,}12}&a_{\overset{\,}13}\\
a_{\overset{\,}21}&a_{\overset{\,}22}&a_{\overset{\,}23}\\
a_{\overset{\,}31}&a_{\overset{\,}32}&a_{\overset{\,}33}\\
\end{pmatrix}\\
\\
\boldsymbol{A}_{\overset{\,}\{1\}}=
\begin{pmatrix}
a_{\overset{\,}22}&a_{\overset{\,}23}\\
a_{\overset{\,}32}&a_{\overset{\,}33}\\
\end{pmatrix}\qquad
\boldsymbol{A}_{\overset{\,}\{2\}}=
\begin{pmatrix}
a_{\overset{\,}11}&a_{\overset{\,}13}\\
a_{\overset{\,}31}&a_{\overset{\,}33}\\
\end{pmatrix}\qquad
\boldsymbol{A}_{\overset{\,}\{3\}}=
\begin{pmatrix}
a_{\overset{\,}11}&a_{\overset{\,}12}\\
a_{\overset{\,}21}&a_{\overset{\,}22}\\
\end{pmatrix}
\end{gather*}
\begin{gather*}
\boldsymbol{A}\,\boldsymbol{\xi}_{\overset{\,}1}=\lambda_{\overset{\,}1}\boldsymbol{\xi}_{\overset{\,}1}\\
\boldsymbol{A}\,\boldsymbol{\xi}_{\overset{\,}2}=\lambda_{\overset{\,}2}\boldsymbol{\xi}_{\overset{\,}2}\\
\boldsymbol{A}\,\boldsymbol{\xi}_{\overset{\,}3}=\lambda_{\overset{\,}3}\boldsymbol{\xi}_{\overset{\,}3}\\
\\
\begin{split}
\boldsymbol{A}_{\overset{\,}\{1\}}\boldsymbol{\eta}_{\overset{\,}11}=\mu_{\overset{\,}11}\boldsymbol{\eta}_{\overset{\,}11}\\
\boldsymbol{A}_{\overset{\,}\{1\}}\boldsymbol{\eta}_{\overset{\,}12}=\mu_{\overset{\,}12}\boldsymbol{\eta}_{\overset{\,}12}\\
\end{split}\qquad\quad
\begin{split}
\boldsymbol{A}_{\overset{\,}\{2\}}\boldsymbol{\eta}_{\overset{\,}21}=\mu_{\overset{\,}21}\boldsymbol{\eta}_{\overset{\,}21}\\
\boldsymbol{A}_{\overset{\,}\{2\}}\boldsymbol{\eta}_{\overset{\,}22}=\mu_{\overset{\,}22}\boldsymbol{\eta}_{\overset{\,}22}\\
\end{split}\qquad\quad
\begin{split}
\boldsymbol{A}_{\overset{\,}\{3\}}\boldsymbol{\eta}_{\overset{\,}31}=\mu_{\overset{\,}31}\boldsymbol{\eta}_{\overset{\,}31}\\
\boldsymbol{A}_{\overset{\,}\{3\}}\boldsymbol{\eta}_{\overset{\,}32}=\mu_{\overset{\,}32}\boldsymbol{\eta}_{\overset{\,}32}\\
\end{split}
\end{gather*}

\begin{align*}
\boldsymbol{\xi}_{\overset{\,}1}=\bigg({\xi}_{\overset{\,}11},\,{\xi}_{\overset{\,}12},\,{\xi}_{\overset{\,}13}\bigg)
\quad\quad
\begin{split}
\left|{\xi}_{\overset{\,}11}\right|^2=\dfrac{\left(\lambda_{\overset{\,}1}-\mu_{\overset{\,}11}\right)\left(\lambda_{\overset{\,}1}-\mu_{\overset{\,}12}\right)}{\left(\lambda_{\overset{\,}1}-\lambda_{\overset{\,}2}\right)\left(\lambda_{\overset{\,}1}-\lambda_{\overset{\,}3}\right)}\\
\\
\left|{\xi}_{\overset{\,}12}\right|^2=\dfrac{\left(\lambda_{\overset{\,}1}-\mu_{\overset{\,}21}\right)\left(\lambda_{\overset{\,}1}-\mu_{\overset{\,}22}\right)}{\left(\lambda_{\overset{\,}1}-\lambda_{\overset{\,}2}\right)\left(\lambda_{\overset{\,}1}-\lambda_{\overset{\,}3}\right)}\\
\\
\left|{\xi}_{\overset{\,}13}\right|^2=\dfrac{\left(\lambda_{\overset{\,}1}-\mu_{\overset{\,}31}\right)\left(\lambda_{\overset{\,}1}-\mu_{\overset{\,}32}\right)}{\left(\lambda_{\overset{\,}1}-\lambda_{\overset{\,}2}\right)\left(\lambda_{\overset{\,}1}-\lambda_{\overset{\,}3}\right)}
\end{split}
\end{align*}

\begin{align*}
\boldsymbol{\xi}_{\overset{\,}2}=\bigg({\xi}_{\overset{\,}21},\,{\xi}_{\overset{\,}22},\,{\xi}_{\overset{\,}23}\bigg)
\quad\quad
\begin{split}
\left|{\xi}_{\overset{\,}21}\right|^2=\dfrac{\left(\lambda_{\overset{\,}2}-\mu_{\overset{\,}11}\right)\left(\lambda_{\overset{\,}2}-\mu_{\overset{\,}12}\right)}{\left(\lambda_{\overset{\,}2}-\lambda_{\overset{\,}1}\right)\left(\lambda_{\overset{\,}2}-\lambda_{\overset{\,}3}\right)}\\
\\
\left|{\xi}_{\overset{\,}22}\right|^2=\dfrac{\left(\lambda_{\overset{\,}2}-\mu_{\overset{\,}21}\right)\left(\lambda_{\overset{\,}2}-\mu_{\overset{\,}22}\right)}{\left(\lambda_{\overset{\,}2}-\lambda_{\overset{\,}1}\right)\left(\lambda_{\overset{\,}2}-\lambda_{\overset{\,}3}\right)}\\
\\
\left|{\xi}_{\overset{\,}23}\right|^2=\dfrac{\left(\lambda_{\overset{\,}2}-\mu_{\overset{\,}31}\right)\left(\lambda_{\overset{\,}2}-\mu_{\overset{\,}32}\right)}{\left(\lambda_{\overset{\,}2}-\lambda_{\overset{\,}1}\right)\left(\lambda_{\overset{\,}2}-\lambda_{\overset{\,}3}\right)}
\end{split}
\end{align*}

\begin{align*}
\boldsymbol{\xi}_{\overset{\,}3}=\bigg({\xi}_{\overset{\,}31},\,{\xi}_{\overset{\,}32},\,{\xi}_{\overset{\,}33}\bigg)
\quad\quad
\begin{split}
\left|{\xi}_{\overset{\,}31}\right|^2=\dfrac{\left(\lambda_{\overset{\,}3}-\mu_{\overset{\,}11}\right)\left(\lambda_{\overset{\,}3}-\mu_{\overset{\,}12}\right)}{\left(\lambda_{\overset{\,}3}-\lambda_{\overset{\,}1}\right)\left(\lambda_{\overset{\,}3}-\lambda_{\overset{\,}2}\right)}\\
\\
\left|{\xi}_{\overset{\,}32}\right|^2=\dfrac{\left(\lambda_{\overset{\,}3}-\mu_{\overset{\,}21}\right)\left(\lambda_{\overset{\,}3}-\mu_{\overset{\,}22}\right)}{\left(\lambda_{\overset{\,}3}-\lambda_{\overset{\,}1}\right)\left(\lambda_{\overset{\,}3}-\lambda_{\overset{\,}2}\right)}\\
\\
\left|{\xi}_{\overset{\,}33}\right|^2=\dfrac{\left(\lambda_{\overset{\,}3}-\mu_{\overset{\,}31}\right)\left(\lambda_{\overset{\,}3}-\mu_{\overset{\,}32}\right)}{\left(\lambda_{\overset{\,}3}-\lambda_{\overset{\,}1}\right)\left(\lambda_{\overset{\,}3}-\lambda_{\overset{\,}2}\right)}
\end{split}
\end{align*}

hbghlyj

arxiv.org/abs/1908.03795
$n$阶 Hermitian matrix $A$ 的特征值为实数$\lambda_1(A),\dots,\lambda_n(A)$, 特征向量为 $(v_{i,1},\dots,v_{i,n}),i=1,\dots,n$.
$A$ 的minor(余子式) $M_j$ 的特征值为$λ_k(M_j)$。
\begin{equation}\label{eq:wts}
|v_{i,j}|^2\prod_{k=1;k\neq
i}^{n}\left(\lambda_i(A)-\lambda_k(A)\right)=\prod_{k=1}^{n-1}\left(\lambda_i(A)-\lambda_k(M_j)\right)\end{equation}
The adjugate proof
Recall that if $A$ is an $n \times n$ matrix, the adjugate matrix $\mathrm{adj}(A)$ is given by the formula
\begin{equation}\label{adj}
\mathrm{adj}( A ) \coloneqq \left( (-1)^{i+j} \mathrm{det}(M_{ji}) \right)_{1 \leq i, j \leq n}
\end{equation}
where $M_{ji}$ is the $n-1 \times n-1$ matrix formed by deleting the $j^{\mathrm{th}}$ row and $i^{\mathrm{th}}$ column from $A$.  From Cramer's rule we have the identity
$$ \mathrm{adj}( A ) A = A \mathrm{adj}( A ) = \mathrm{det}(A) I_n.$$
If $A$ is a diagonal matrix with (complex) entries $\lambda_1(A),\dots,\lambda_n(A)$, then $\mathrm{adj}(A)$ is also a diagonal matrix with $ii$ entry $\prod_{k=1; k \neq i}^n \lambda_k(A)$.  More generally, if $A$ is a normal matrix with diagonalization
\begin{equation}\label{adiag}
A = \sum_{i=1}^n \lambda_i(A) v_i v_i^*
\end{equation}
where $v_1,\dots,v_n$ are an orthonormal basis of eigenvectors of $A$ and $v_i^*$ is the conjugate transpose of $v_i$, then $\mathrm{adj}(A)$ has the same basis of eigenvectors with diagonalization
\begin{equation}\label{adja}
\mathrm{adj}(A) = \sum_{i=1}^n \left(\prod_{k=1; k \neq i}^n \lambda_k(A)\right) v_i v_i^*.
\end{equation}
If one replaces $A$ by $\lambda I_n - A$ for any complex number $\lambda$, we therefore have
$$ \mathrm{adj}(\lambda I_n - A) = \sum_{i=1}^n \left(\prod_{k=1; k \neq i}^n (\lambda - \lambda_k(A)) \right) v_i v_i^*.$$
If one specializes to the case $\lambda = \lambda_i(A)$ for some $i=1,\dots,n$, then all but one of the summands on the right-hand side vanish, and the adjugate matrix becomes a scalar multiple of the rank one projection $v_i v_i^*$:
\begin{equation}\label{adji}
\mathrm{adj}(\lambda_i(A) I_n - A) = \left(\prod_{k=1; k \neq i}^n (\lambda_i(A) - \lambda_k(A)) \right) v_i v_i^*.
\end{equation}
Extracting out the $jj$ component of this identity using \eqref{adj}, we conclude that
\begin{equation}\label{adji-2}
\mathrm{det}(\lambda_i(A) I_{n-1} - M_j) = \left(\prod_{k=1; k \neq i}^n (\lambda_i(A) - \lambda_k(A)) \right) |v_{i,j}|^2
\end{equation}
which is equivalent to \eqref{eq:wts}.  In fact this shows that the eigenvector-eigenvalue identity holds for normal matrices $A$ as well as Hermitian matrices (despite the fact that the minor $M_j$ need not be Hermitian or normal in this case).  Of course in this case the eigenvalues are not necessarily real and thus cannot be arranged in increasing order, but the order of the eigenvalues plays no role in the identity \eqref{eq:wts}.