I have a Hermitian matrix of size $n$, $H_{n\times n}$ with only off diagonal entries, actually all the entries are real in $H$ (very special case). Now my job is to find the anti-commuting matrix say $O_{n\times n}$ to $H_{n\times n}$,
$O H O^{-1} = -H $, and with $O^{2} = 1 $ which is unitary matrix

There are two cases, (i) when $n$ is odd, (ii) when $n$ is even.
(i) If $H$ has entries where only at the odd sum of indices($i+j$ $\epsilon$ odd) exists then we can find $O$,
e.g. $$H_{1} =\begin{bmatrix}0_{1,1} & a_{1,2} & 0_{1,3} \\a_{2,1} & 0_{2,2} & b_{2,3} \\ 0_{3,1} & b_{3,2} & 0_{3,3} \end{bmatrix}$$
$O = \{\{\sigma_{z},0\},\{0,1\}\} $, where $\sigma_{z}$ is Pauli Matrix
If we have element also at the even site $(1,3)$ and $(3,1)$, say $c$, then $$H_{2} =\begin{bmatrix}0_{1,1} & a_{1,2} & c_{1,3} \\a_{2,1} & 0_{2,2} & b_{2,3} \\ c_{3,1} & b_{3,2} & 0_{3,3} \end{bmatrix}$$ Then we can't find $O$.
Then the story is same for case (ii) even $n$.

Is there a way to show this?
(My background is in Physics, I'm not very good with maths. I strongly apologize for that). Both a sketch solution or a good starting point would help me.

  • $\begingroup$ @Desura if $O$ is unitary, then $OO^{\dagger} =I$ but here, $OO =I$ $\endgroup$ – L.K. Jun 3 '18 at 22:14

I don't know if this will help you but here is a sufficient and necessary condition for the existence of the matrix $O$. The proof also provides a way of computing explicitely $O$.

Let $H \in \Bbb C^{n \times n}$ be a diagonalizable matrix (which is the case if $H$ is hermitian). Then there exists an invertible $O \in \Bbb C^{n \times n}$ such that $O^2 = I$ and $HO = -OH$ if and only if for all eigenvalues $\lambda$ of $H$, $- \lambda$ is also an eigenvalue of $H$ and the eigenspaces $E_{\lambda}$ and $E_{-\lambda}$ have the same dimension (which is equivalent to saying that $\lambda$ and $- \lambda$ have the same algebraic or geometric multiplicity).

Let's prove first that this is necessary.

Assume such a $O$ exists and let $\lambda$ be a non zero eigenvalue of $H$ with eigenvector $v \neq 0$, i.e. $Hv = \lambda v$. Then, $HOv = - OHv = -\lambda Ov$ and $Ov \neq 0$ because $O$ is invertible. As for the dimensions, notice that $O_{| E_{\lambda}} : E_{\lambda} \rightarrow E_{-\lambda}$ is an isomorphism between the finite dimensional vector spaces $E_{\lambda}$ and $E_{-\lambda}$.

Now, let's prove that this is sufficient.

Let $\{\lambda_1, -\lambda_1, ..., \lambda_r, -\lambda_r\}$ be the set of non zero, distinct, eigenvalues of $H$. We have $$\Bbb C^n = E_{\lambda_1} \oplus E_{-\lambda_1} \oplus ... \oplus E_{\lambda_r} \oplus E_{-\lambda_r} \oplus \ker H$$ since $H$ is diagonalizable (and we omit $\ker H$ from the direct sum if it is equal to $\{0\}$).

For $\lambda_i$, let $\{v^1_{\lambda_i}, ..., v^{k_i}_{\lambda_i}\} \subset \Bbb C^n$ be a basis for the eigenspace $E_{\lambda_i}$ and let $\{w^1_{\lambda_i}, ..., w^{k_i}_{\lambda_i}\} \subset \Bbb C^n$ be a basis for the eigenspace $E_{-\lambda_i}$.

Define $O : \Bbb C^n \rightarrow \Bbb C^n$ to be the linear application such that $$Ov = v \ \forall v \in \ker H \\ Ov^j_{\lambda_i} = w^j_{\lambda_i} \\ Ow^j_{\lambda_i} = v^j_{\lambda_i}$$ for all $i = 1, ..., r$, for all $j = 1, ..., k_i$

Such an application is well defined, linear, invertible, and satisfies $O \circ O = I$.

Let $O_E \in \Bbb C^{n \times n}$ be the matrix of $O$ with respect to the canonical basis $\{e_1, ..., e_n\}$ of $\Bbb C^n$. Then $O_E$ is invertible and satisfies $O_E^2 = I$.

Furthermore, $HO_E = -O_EH$. Indeed, you just have to check the equality for a basis of $\Bbb C^n$. It is really easy to check with the basis of $\Bbb C^n$ consisting of the elements $v^j_{\lambda_i}$, $w^j_{\lambda_i}$ as above and any basis of $\ker H$ since those elements are eigenvectors of $H$ and $O$ is defined with them.

Other necessary conditions

If $H$ is any $\Bbb C^{n \times n}$ matrix satisfying $HO = -OH$ for an invertible $\Bbb C^{n \times n}$ matrix $O$, then $OHO^{-1} = -H$, so $$\det{H} = (-1)^n \det{H}$$ so either $\det H$ is zero, or $n$ is even.

Furthermore, $tr(OHO^{-1}) = tr(OO^{-1}H) = tr(H) = -tr(H)$, so $tr(H) = 0$

Application to the $3 \times 3$ hermitian case

For any $3 \times 3$ hermitian matrix $$H = \begin{pmatrix}a & b & c \\ \overline{b} & a & d \\ \overline{c} & \overline{d} & a \end{pmatrix}, \ a \in \Bbb R$$ we can show that the characteristic polynomial of $H$ is given by $$p_H(x) = x^3 - tr(H) x^2 + ((a^2 - \lvert d \rvert^2) + (a^2 - \lvert c \rvert^2) + (a^2 - \lvert b \rvert^2)) x - \det H$$

As mentioned, the relation $HO = -OH$ with $O$ invertible implies that $H$ has determinant and trace zero.

Assuming $H$ satisfies those two assumptions ($\iff$ $a = 0$ and at least one of the elements $b,c,d$ is zero as Kolja mentioned), the polynomial becomes $$p_H(x) = x(x^2 - (\lvert d \rvert^2 + \lvert c \rvert^2 + \lvert b \rvert^2))$$ Then, the condition of our result is always satisfied and in that case, Kolja showed explicitely the existence of the matrix $O$.

Application to the tridiagonal case

Let $H$ be a $n \times n$ tridiagonal real matrix $$H = \begin{pmatrix}a & b & 0 & 0 & ... & \\ c & a & b & 0 & ... & \\ 0 & c & a & b &... & \\ & & \ddots & \ddots & \ddots & & \end{pmatrix}, \ a, b, c \in \Bbb R$$

The eigenvalues of such a matrix are given by $$a + 2 \sqrt{bc} \cos \left(\frac{k \pi}{n+1}\right), k = 1, ..., n$$

If $b$ and $c \neq 0$, then all the eigenvalues are distinct. In that case, if we assume $a = 0$ (so that we have $tr(H) = 0$ as needed), then the condition of our result is always satisfied.

I don't think there is an easy result for the general case involving only the indices of the matrix.

  • 2
    $\begingroup$ Great proof ! Just one thing - I think it should be $O^2=O_E^2=\mathbb{1}$ (after defining $O$ and $O_E$). $\endgroup$ – Kolja Jun 4 '18 at 12:44
  • 1
    $\begingroup$ @L.K. If $v \in \ker H$ and $v \neq 0$, then $Hv = 0$ by definition of kernel. What about $Ov$ ? It certainly cannot be $0$ since $O0 = 0$ by linearity, so $Ov = 0 \Rightarrow v = 0$ by injectivity of $O$. But $HOv = -OHv = 0$, so $Ov \in \ker H$ and $Ov \neq 0$. So $O_{| \ker H} : \ker H \rightarrow \ker H$ is an isomorphism between the finite dimensional vector space $\ker H$ and itself such that $O \circ O = O$. $\endgroup$ – Desura Jun 5 '18 at 11:20
  • 1
    $\begingroup$ @L.K. Here, $O$ isn't really defined uniquely. When I construct it in the second part of my proof, I choose it to be the identity on $\ker H$ since it's easier (and always works if $\ker H = \{0\}$) but it's not necessarily true for every such $O$, i.e. $Ov = v$ is not necessarily true for $v \in \ker H$. $\endgroup$ – Desura Jun 5 '18 at 11:21
  • 1
    $\begingroup$ (It should be $O \circ O = I$. I made the same mistake again !) $\endgroup$ – Desura Jun 5 '18 at 11:28
  • 1
    $\begingroup$ @L.K. No, this is not possible. If $OH = -HO$ for an invertible $O$, the equation $\det H = (-1)^n \det H$ holds. In $n$ is odd, we necessarily have $\det H = 0$, so $H$ has not full rank and $\ker H \neq \{0\}$. $\endgroup$ – Desura Jun 5 '18 at 15:58

I just noticed that you wanted a general case. Instead I solved the case $n=3$. Nevertheless I think it will be helpful so I decided to leave the answer.

From the property $OH = -HO$ we can deduce that $$\det(HO)=\det(H)\det(O)=(-1)^3\det(O)\det(H)=\det(-HO)$$ Since $\det(O)\not=0$, we have $\det(H)=-\det(H)$. Since we're working over $\mathbb{R}$ it follows that $\det(H)=0$.

If $H$ is as follows $$ H =\begin{bmatrix}0 & a & c \\a & 0 & b \\ c & b & 0 \end{bmatrix} $$ then $\det(H)=2abc$. The determinant being zero tells us that at least one of $a,b,c$ must be $0$.

Since you already solved the case $c=0$, I believe you can finish up the other two cases ($a=0$ and $b=0$).

PS. I figured I could write them up quickly. For $d\in\{a,b,c\}$ take the matrix $O_d$ for the case $d=0$. $$ O_a =\begin{bmatrix}1 & 0 & 0 \\0 & 1 & 0 \\ 0 & 0 & -1 \end{bmatrix} $$ $$ O_b =\begin{bmatrix}-1 & 0 & 0 \\0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} $$ $$ O_c =\begin{bmatrix}1 & 0 & 0 \\0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix} $$

  • $\begingroup$ Very good observation, regarding vanishing of the det. Can we say something about the even case also? $\endgroup$ – L.K. May 29 '18 at 15:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.