# Imaginary eigenvalues

Consider the matrix

$$A(\mu) = \begin{pmatrix} 0 & 1& 0 & 0 \\ -1 & -i\mu & 0 & i \\ 0 & 0 & 0 & 1 \\ 0 &i & -1 & i\mu \end{pmatrix}.$$

This matrix is for $$\mu \in \mathbb R$$ skew hermitian, i.e. all the eigenvalues are imaginary.

Let $$(\mu_i)_i$$ be a sequence of real numbers.

We consider the product

$$M=\prod_{i=1}^n A(\mu_i).$$

I claim the following two facts are true (observed numerically):

1.) If $$n$$ is odd, then all eigenvalues are imaginary (this is non-trivial for $$n\ge 3$$ since the matrix $$M$$ is in general not skew hermitian anymore)

2.) Show that the eigenvalues satisfy for $$n \in 2\mathbb N_0+1$$ that $$\lambda$$ is an eigenvalue of $$M$$ if and only if $$-\lambda$$ is. If you show this for one eigenvalue it will hold for all eigenvalues of $$M$$.

• Characteristic polynomial for odd $n$ is$$1+(3+\prod_{i=1}^n\mu_i^2)t^2+t^4$$ May 26, 2021 at 13:50
• @მამუკაჯიბლაძე Are you sure? I computed it for $n=3$ and got something completely different (the coefficient of $t^2$ is a messy polynomial in $\mu_1, \mu_2, \mu_3$) May 26, 2021 at 14:40
• Well unless Mathematica has a bug... :) May 26, 2021 at 14:42
• @AntoineLabelle OMG it seems I have a bug! I multiplied the matrices entrywise!! May 26, 2021 at 14:44
• Sorry my first comment is complete rubbish May 26, 2021 at 14:46

Define the unitary and Hermitian matrices $$U=\left( \begin{array}{cccc} 0 & 0 & -i & 0 \\ 0 & 0 & 0 & -i \\ i & 0 & 0 & 0 \\ 0 & i & 0 & 0 \\ \end{array} \right),\;\; V=\left( \begin{array}{cccc} 0 & 0 & -i & 0 \\ 0 & 0 & 0 & i \\ i & 0 & 0 & 0 \\ 0 & -i & 0 & 0 \\ \end{array} \right), \;\;U^2=I=V^2,$$ and note that, for $$\mu\in\mathbb{R}$$, $$UA(\mu)U=\bar{A}(\mu),\;\;VA(\mu)V=-A(\mu).$$

$$\bullet$$ Hence if $$\lambda$$ is an eigenvalue of $$\prod_{i=1}^n A(\mu_i)$$, then $$0=\overline{\det\bigl(\lambda I-\prod_i A(\mu_i)\bigr)}=\det\bigl(\bar{\lambda} I-\prod_i\bar{A}(\mu_i)\bigr)=\det\bigl(\bar{\lambda} I-\prod_i UA(\mu_i)U\bigr)=\det\bigl(\bar{\lambda} I-\prod_i A(\mu_i)\bigr)=0.$$ So the eigenvalues come in complex conjugate pairs: if $$\lambda$$ is an eigenvalue of $$\prod_i A(\mu_i)$$, then also $$\bar{\lambda}$$ is an eigenvalue. (This holds irrespective of whether $$n$$ is even or odd.)

$$\bullet$$ Similarly, if $$\lambda$$ is an eigenvalue of $$\prod_{i=1}^n A(\mu_i)$$ and $$n$$ is an odd integer, then $$0=\det\bigl(\lambda I-\prod_{i=1}^n VA(\mu_i)V\bigr)=\det\bigl(\lambda I-(-1)^n\prod_{i=1}^n A(\mu_i)\bigr)=\det\bigl(\lambda I+\prod_{i=1}^n A(\mu_i)\bigr)=0,$$ so the eigenvalues come in inverse pairs for odd $$n$$: if $$\lambda$$ is an eigenvalue then also $$-\lambda$$ is an eigenvalue. This proves property 2.

$$\bullet$$ Since $$\det A(\mu)=1$$ for any $$\mu$$, the product of the four eigenvalues of $$\prod_{i=1}^n A(\mu_i)$$ equals unity. This gives for odd $$n$$ the following three possibilities (with real $$c$$ and $$\phi$$):
A. $$\lambda_1=ic$$, $$\lambda_2=-ic$$, $$\lambda_3=i/c$$, $$\lambda_4=-i/c$$ (this is property 1),
B. $$\lambda_1=c$$, $$\lambda_2=-c$$, $$\lambda_3=1/c$$, $$\lambda_4=-1/c$$,
C. $$\lambda_1=e^{i\phi}$$, $$\lambda_2=-e^{i\phi}$$, $$\lambda_3=e^{-i\phi}$$, $$\lambda_4=-e^{-i\phi}$$.
The eigenvalues are either all four on the imaginary axis, or on the real axis, or on the unit circle.

• I think you also need to exclude the case of four eigenvalues $\lambda, -\lambda, \frac{1}{\lambda},-\frac{1}{\lambda}$ all real. May 26, 2021 at 15:51
• indeed, thanks for pointing this out. May 26, 2021 at 15:55