Suppose we have a hermitian matrix $M$ of form $$\left( \begin{matrix} \alpha& -\beta ^*\\ \beta& -\alpha ^*\\ \end{matrix} \right) $$ where $\alpha$ is a hermitian matrix and $\beta$ is an antisymmetric matrix with equal dimension, and I use the star $*$ for complex conjugate. Since $M$ is hermitian, then it can be diagonalized into $$TMT^{\dagger}=\left( \begin{matrix} d& 0\\ 0& -d\\ \end{matrix} \right)$$ where $d$ is diagonal, and I used the special form of $M$ to deduce that the eigenvalues are real and appear in matched pairs $\pm \lambda $.

But I want to say more about the unitary matrix $T$, i.e., can $T$ always be chosen to be of form $\left( \begin{matrix} \gamma& \mu\\ \mu ^*& \gamma ^*\\ \end{matrix} \right) $ where $\gamma$ and $\mu$ are two matrices with equal dimensions?


The question is found in this notes(from the bottom of page 6 to 7) for introduction of Jordan-Wigner transformation. The author states the above rules while I really can't see how to show that. I quoted the closely related part for your convenience:

One way of obtaining a rigorous proof is to find a $T$ satisfying $$ T M T^{\dagger}=\left[\begin{array}{cc} d & 0 \\ 0 & -d \end{array}\right] $$ and then to apply the cosine-sine (or CS) decomposition from linear algebra, which provides a beautiful way of representing block unitary matrices, and which, in this instance, allows us to obtain a $T$ of the desired form with just a little more work.

Thanks in advance!


2 Answers 2


Let $\dim(M)=2n$.
$\newcommand{\b}[1]{\begin{bmatrix}#1\end{bmatrix}}$ Let $S=\b{0 &I_n\\I_n &0}$. Then $S^{-1}=S^*=S$. $$S^{-1}MS=-M^*.$$

Recall that $\Bbb C^{2n\times1}$ is a Hermitian space with inner product defined by $\langle x, y\rangle=x^\dagger y$.
Let $\kappa:\Bbb C^{2n\times1}\to\Bbb C^{2n\times1}$, $\kappa(x)=Sx^*$, i.e., $\kappa\left(\b{a\\b}\right)=\b{b^*\\a^*}.$
$\kappa$ is an antiunitary operator and $\kappa^{-1}=\kappa$.

Let $\{0\}\not=M_\lambda=\{x\in\Bbb C^{2n\times1}\mid Mx=\lambda x\}$, the $\lambda$-eigenspace of $M$.
Since $M$ is Hermitian, $\lambda\in\Bbb R$ and $\Bbb C^{2n\times1}=\bigoplus M_{\lambda}$.
Suppose $x\in M_{\lambda}$. Since $$MSx^*=S(S^{-1}MS)x^*=S(-M^*)x^*=-S(Mx)^*=-S(\lambda x)^*=-\lambda Sx^*,$$ we have $\kappa(x)\in M_{-\lambda}$. Hence $\kappa$ induces an antiunitary isomorphism between $M_{\lambda}$ and $M_{-\lambda}$.

Let $\lambda_i$, where $1\le i\le k$ and $k\ge0$ be all positive eigenvalues of $M$.
Let $B_{\lambda_i}$ be an orthonormal basis of $M_{\lambda_i}$.
Let $B_{-\lambda_i}=\left\{\kappa(x)\mid x\in B_{\lambda_i}\right\}$.
We can see that $B_{-\lambda_i}$ is an orthonormal basis of $M_{-\lambda_i}$.

Note that $\kappa$ induces an antiunitary operator on $M_0$. Let us choose a "$\kappa$-paired" orthonormal basis of $M_0$.

Let $R_1=M_0$. Starting from $t=1$, repeat the following procedure as long as $R_t$ is not $\{0\}$. We will keep $R_t$ as an even-dimensional $\kappa$-invariant $\Bbb C$-linear subspace of $M_0$.

  • Suppose we can find $x\in R_t$ such that $x$ and $\kappa(x)$ are $\Bbb C$-linear independent.
    Let $\mathscr Z_t$ be the $\Bbb C$-linear space spanned by $x$ and $\kappa(x)$. Let $R_{t+1}$ be the $\Bbb C$-linear subspace of $R_t$ that is orthogonal to $\mathscr Z_t$. $\dim(R_{t+1})$ is even since it is $ \dim(R_{t})-2$.
    Verify that $\kappa$ induces an antiunitary operator on $\mathscr Z_t$ and $R_{t+1}$.
    Thanks to the linear independence of $x$ and $\kappa(x)$, we can find $0\not=y\in\mathscr Z_t$ such that $y$ and $\kappa(y)$ is orthogonal. Let $z_t=y/||y||$.
    $\{z_t, \kappa(z_t)\}$ is an orthonormal basis of $\mathscr Z_t$.
    Add $1$ to $t$.
  • Otherwise for all $x\in R_t$ and $x$ and $\kappa(x)$ are $\Bbb C$-linear dependent.
    We claim that if $0\not=x$ and $\kappa(x)$ are $\Bbb C$-linear dependent, then $x=\b{u\\e^{i\Theta}u}$ for some $u\in\Bbb C^{n\times1}$ and $\Theta\in \Bbb R$. The proof is straightforward.
    Since $R_t$ is even-dimensional, there are $x,y\in R_t$ that are $\Bbb C$-linear independent. Then either $x+y$ and $\kappa(x+y)=\kappa(x)+\kappa(y)$ are not $\Bbb C$-linear dependent, or $x-iy$ and $\kappa(x-iy)=\kappa(x)-i\kappa(y)$ are not $\Bbb C$-linear dependent.
    This contradicts the assumption of this case. Hence, this case does not exist.

Since the dimension of $R_t$ decreases by $2$ at each iteration, the procedure above will run $\dim(M_0)/2$ times.
We obtain that $M_0$, as an orthogonal sum of $\mathscr Z_t$, $t=1,\cdots, \dim(M_0)/2$, has an orthonormal basis $z_1, \cdots, z_{\dim(M_0)/2}, \kappa(z_1), \cdots, \kappa(z_{\dim(M_0)/2})$.

Let $\cup B_{\lambda_i}=\{w_1, \cdots, w_{n-\dim(M_0)/2}\}$. Let $T$ be the following unitary matrix $$\b{w_1, \cdots, w_{n-\dim(M_0)/2}, z_1, \cdots, z_{\dim(M_0)/2}, \kappa(w_1), \cdots, \kappa(w_{n-\dim(M_0)/2}), \kappa(z_1), \cdots, \kappa(z_{\dim(M_0)/2})}$$ Then $T$ is of the required form $\b{\gamma &\mu\\\mu^* &\gamma^*}$ and $$MT=\b{\Lambda & 0 & 0 & 0\\0 & 0 & 0 & 0\\0&0 &-\Lambda& 0\\0 & 0 & 0 &0}T=\b{d &0 \\ 0 & -d}$$ where $\Lambda$ is a diagonal real matrix with $\lambda_i$'s on the diagonal, and $d$ is the $n$-dimensional diagonal real matrix $\b{\Lambda & 0\\ 0 & 0 }$.

Since $T^\dagger M(T^\dagger)^\dagger=T^\dagger MT=\b{d &0 \\ 0 & -d}$, and $T^\dagger$ is an unitary matrix of the form $\b{\gamma &\mu\\\mu^* &\gamma^*}$ as well, we have done.

  • $\begingroup$ Thanks for the answer, but I think I need sometime to digest the answer and added some references to my original post. And $\beta$ is antisymmetric so a number might not be a choice, I think.. $\endgroup$
    – Sherlock
    Sep 1, 2022 at 5:41
  • 1
    $\begingroup$ @Sherlock You are right. Since $\beta$ is antisymmetric, a one-dimensional $\beta$ must be $0$. My answer was completely wrong. Please check my updated answer. It takes quite some time for me to figure out how we can handle the $0$-eigenspace of $M$. Hopefully, my answer is good enough for you to accept. $\endgroup$
    – Apass.Jack
    Sep 12, 2022 at 3:02

First all blocks are considered matrices of equal size. Set $M=\begin{pmatrix}A&B\\-B&-A\end{pmatrix}$ a $2n\times 2n$ complexe matrix with $A^*=A$ and $B^*=-B$. Let $S=\frac{1}{\sqrt{2}}\begin{pmatrix}I&-I\\I&I\end{pmatrix}$, $X=A+B$ so $X^*=A-B$. Here take the polar decomposition of $X$ as $X=U|X|$ with $H^*|X|H=D$ where $D$ is a diagonal matrix and $H$ unitary. Now we apply the following $K:=(LSRS)M(LSRS)^*$ where $R=\begin{pmatrix}U^*&0\\0&I\end{pmatrix}$ and $L=\begin{pmatrix}H^*&0\\0&H^*\end{pmatrix}$ are unitaries. We check that $K=\begin{pmatrix}-D&0\\0&D\end{pmatrix}$ and that $LSRS=\begin{pmatrix}E&F\\-F&-E\end{pmatrix}$. This is different than the form presented for $T$, in dimension $2$ ($E$ and $F$ are scalars) this is almost invalid for $b\neq 0$, $M=\begin{pmatrix}a&ib\\-ib&-a\end{pmatrix}$, the two eigenvectors are up to a factor: $\begin{pmatrix}\frac{-i(\sqrt{a^2+b^2}-a)}{b}\\1\end{pmatrix}$ and $\begin{pmatrix}\frac{i(\sqrt{a^2+b^2}+a)}{b}\\1\end{pmatrix}$...

  • $\begingroup$ Sorry, I want to ask do you use $*$ for complex conjugate? I use $*$ as complex conjugate instead of conjugate transpose. $\endgroup$
    – Sherlock
    Aug 16, 2022 at 12:50
  • $\begingroup$ Hey it is for $A^*=\bar{A}^T$ (conjugate transpose as adjoint). $\endgroup$
    – Toni Mhax
    Aug 16, 2022 at 13:06
  • 1
    $\begingroup$ I'm sorry that I use $A^*$ to mean $\bar A$. I just get used to use $A^\dagger$ stands for $\bar A^T$... Sorry about the ambiguity. $\endgroup$
    – Sherlock
    Aug 16, 2022 at 13:13

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