16
$\begingroup$

I have a matrix

$$ A= \begin{pmatrix} 0 & a & d & c\\ \bar a & 0 & b & d \\ \bar d & \bar b & 0 & a \\ \bar c & \bar d & \bar a & 0 \end{pmatrix} $$

As you can see, the matrix is always self-adjoint for any $a, b, c, d \in \mathbb C$.

But it has a funny property (that I found by playing with some numbers):

If $a,b,c$ are arbitrary real numbers and also $d$ is real, then the spectrum of $A$ is in general not symmetric with respect to zero. To illustrate this, we take $d := 2$, $a := 5$, $b := 3$, $c := 4$ then the eigenvalues are

$$\sigma(A):=\{10.5178, -6.54138, -3.51783, -0.458619\}$$

But once I take $d \in i \mathbb R$, the spectrum becomes immediately symmetric. In fact, $d := 2 i$, $a := 5$, $b := 3$, $c := 4$ leads to eigenvalues

$$\sigma(A)=\{-9.05607, 9.05607, -0.993809, 0.993809\}$$

Is there any particular symmetry that only exists for $d \in i\mathbb R$ that implies this nice inflection symmetry?

I am less interested in a brute-force computation of the spectrum than of an explanation of what symmetry causes the inflection symmetry.

$\endgroup$
2
  • $\begingroup$ Is it symmetric for all choices of $a$, $b$ and $c$ (real and or complex)? $\endgroup$
    – M. Winter
    Aug 5, 2020 at 10:01
  • $\begingroup$ @M.Winter good question, now I am tempted to say that it actually only works for $a,b,c$ real. $\endgroup$
    – Sascha
    Aug 5, 2020 at 10:09

2 Answers 2

37
$\begingroup$

For real $a,b,c$ and imaginary $d$ the matrix $A$ has chiral symmetry, meaning it anticommutes with a matrix $X$ that squares to the identity: $$X=\left( \begin{array}{cccc} 0 & 0 & 0 & -i \\ 0 & 0 & i & 0 \\ 0 & -i & 0 & 0 \\ i & 0 & 0 & 0 \\ \end{array} \right),\;\;XA+AX=0,\;\;X^2=I.$$ Hence the spectrum of $A$ has $\pm$ symmetry: $$\det (\lambda-A)=\det(\lambda X^2-XAX)=\det(\lambda+X^2A)=\det(\lambda+A),$$ so if $\lambda$ is an eigenvalue then also $-\lambda$.

$\endgroup$
3
  • $\begingroup$ how did you get $\det(\lambda I-A)=\det(\lambda X^2-XAX)$? $\endgroup$
    – vidyarthi
    Aug 5, 2020 at 11:50
  • 2
    $\begingroup$ Since $X^2=1$, you can multiply $\det(\lambda-A)$ with $\det X^2=(\det X)^2$; and since the product of determinants is the determinant of the matrix product, you have $\det(\lambda-A)=\det[X(\lambda-A)X]=\det(\lambda X^2-XAX)$. $\endgroup$ Aug 5, 2020 at 11:53
  • $\begingroup$ great! a useful trick in evaluating some determinants $\endgroup$
    – vidyarthi
    Aug 5, 2020 at 11:54
16
$\begingroup$

An equivalent trick : Let $J:= \operatorname{diag}(1,i,-1,-i)$. Then $J^*AJ=iB$ where $B$ is real and skew-symmetric. Hence the spectrum of $iB$ (thus that of $A$) comes by pairs $\pm\lambda$.

$\endgroup$

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.