I am solving a problem, in the middle of which I reached the following block-tridiagonal matrix.
\begin{bmatrix} 0 & B_1 & &0 \\ B_1 & 0 & \ddots \\ &\ddots& \ddots&B_n \\ 0 & & B_n & 0 \end{bmatrix} Where $B_i$ is an invertible square matrix and size $0$ is equal to size $B_i$. Can I make this matrix similar to the following tridiagonal block matrix as shown below?
\begin{bmatrix} 0 & C & &0 \\ C & 0 & \ddots \\ &\ddots& \ddots&C \\ 0 & & C & 0 \end{bmatrix}
in which size $C$ and $0$ equals $B_i$. What about if size $B_i$ is 1?
This is not always true, even if the $B_i$ have size $1 \times 1$. For example, if $n = 3$ we have the following matrix and characteristic polynomial: $$ A = \begin{pmatrix} 0 & b_1 & 0 & 0 \\ b_1 & 0 & b_2 & 0 \\ 0 & b_2 & 0 & b_3 \\ 0 & 0 & b_3 & 0 \end{pmatrix}, \qquad p_A(\lambda) = \lambda^4 - (b_1^2 + b_2^2 + b_3^2)\lambda^2 + b_1^2 b_3^2. $$
Therefore:
If $(b_1,b_2,b_3) = (1,2,1)$, the characteristic polynomial is $\lambda^4 - 6\lambda^2 + 1$.
If $(b_1,b_2,b_3) = (c,c,c)$, the characteristic polynomial is $\lambda^4 - 3c^2\lambda^2 + c^4$.
Thus, for the matrices $$ \begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 2 & 0 \\ 0 & 2 & 0 & 1\\ 0 & 0 & 1 & 0 \end{pmatrix} \qquad \text{and} \qquad \begin{pmatrix} 0 & c & 0 & 0 \\ c & 0 & c & 0 \\ 0 & c & 0 & c\\ 0 & 0 & c & 0 \end{pmatrix} $$ to be similar, it is necessary that $c^2 = 2$ but $c^4 = 1$. This is impossible.