Matrices with same eigenvalues

Question

This question is a more precise version of this question.

Let's assume we have the matrix

$$\left( \begin{array}{ccccc} 0 & a & 0 & 0 & 0 \\ f & 0 & b & 0 & 0 \\ 0 & e & 0 & c & 0 \\ 0 & 0 & d & 0 & r \\ 0 & 0 & 0 & g & 0 \\ \end{array} \right)$$

If we square it, we get the matrix

$$\left( \begin{array}{ccccc} a f & 0 & a b & 0 & 0 \\ 0 & a f+b e & 0 & b c & 0 \\ e f & 0 & b e+c d & 0 & c r \\ 0 & d e & 0 & c d+g r & 0 \\ 0 & 0 & d g & 0 & g r \\ \end{array} \right)$$

We see that this matrix decomposes into two submatrices

$$C_1:=\left(\begin{array}{cccc} a f & a b & 0 \\ e f & b e+c d & c r \\ 0 & d g & g r \\ \end{array}\right)$$ and $$C_2:=\left(\begin{array}{cc} a f+b e & b c \\ d e & c d+g r \\ \end{array}\right)$$ Now, one can check explicitly that the two submatrices are isospectral apart from one eigenvalue zero. I wonder if there is an abstract argument why this is so?

It would for instance follow if we can write the matrices as $C_1 = AB$ and $C_2= BA$, but I don't see how such a decomposition could work. In particular, this does not seem to be restricted to 5x5 matrices but holds for arbitrary matrices of the above form.

Answer 1

To answer say a previous question, call such tridiagonal matrix $T$; the matrices $T$ and $-T$ (of dimension $n$) are unitarily congruent (they have the same spectra), that is there is an 'alternating' $\pm 1$ diagonal matrix $U$ such that $U^*TU=-T$.
As you noticed when you square $T$ and rearrange it by a permutation $P$ the diagonal blocks have the same eigenvalues. This is true as: for $P_i$ the permutation matrix exchanging $2i+1\leftrightarrow i+1$, set $P=\prod_{i=1}^{\lfloor \frac{n-1}{2}\rfloor}P_i=P_{\lfloor \frac{n-1}{2}\rfloor}\cdots P_1$; we get $PTP^*=\begin{pmatrix}0_{\lceil\frac{n}{2}\rceil}&A\\B&0_{\lfloor \frac{n}{2}\rfloor}\end{pmatrix}.$ The square block $0_m$ is a submatrix with all zeros of dimension $m$. Squaring the last identity gives your matrix $\begin{pmatrix}AB=C_1&0\\0&BA=C_2\end{pmatrix}.$

Answer 2

Consider separately the even-numbered and the odd-numbered indices, place them in a vector $v=\{e,o\}$. The matrix $M$ only couples even-to-odd and odd-to-even, so acting on the vector $v$ it has the form ${0\;M_1}\choose{M_2\;0}$. The square is block diagonal, with matrices $M_1M_2$ and $M_2M_1$ that are isospectral.