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, because $C_1=AB$ and $C_2=BA$. This is true as: for $P_i$ the permutation matrix exchanging rows $2i+1\leftrightarrow i+1$, 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 $PAP^*=\begin{pmatrix}0_{\lceil\frac{n}{2}\rceil}&A\\B&0_{\lfloor \frac{n}{2}\rfloor}\end{pmatrix}.$ The square block $0_m$ is a submatrix with all zeros of dimension $m$. Squaring the last identity gives your matrix $\begin{pmatrix}AB=C_1&0\\0&BA=C_2\end{pmatrix}.$

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 $\begin{pmatrix}AB=C_1&0\\0&BA=C_2\end{pmatrix}.$

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, because $C_1=AB$ and $C_2=BA$. This is true as: for $P_i$ the permutation matrix exchanging rows $2i+1\leftrightarrow i+1$, set $P=\prod_{i=1}^{\lfloor \frac{n-1}{2}\rfloor}P_i$; we get $PAP^*=\begin{pmatrix}0_{\lceil\frac{n}{2}\rceil}&A\\B&0_{\lfloor \frac{n}{2}\rfloor}\end{pmatrix}.$ The square block $0_m$ is a submatrix with all zeros of dimension $m$. Squaring the last identity gives your matrix $\begin{pmatrix}AB=C_1&0\\0&BA=C_2\end{pmatrix}.$

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, because $C_1=AB$ and $C_2=BA$. This is true as: for $P_i$ the permutation matrix exchanging rows $2i+1\leftrightarrow i+1$, 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 $PAP^*=\begin{pmatrix}0_{\lceil\frac{n}{2}\rceil}&A\\B&0_{\lfloor \frac{n}{2}\rfloor}\end{pmatrix}.$ The square block $0_m$ is a submatrix with all zeros of dimension $m$. Squaring the last identity gives your matrix $\begin{pmatrix}AB=C_1&0\\0&BA=C_2\end{pmatrix}.$