How can we show that there are not defective eigenvalues for this square block matrix of dimension $2d \times 2d $: \begin{bmatrix} A&B\\-B& 0 \end{bmatrix} where A, B are real matrices, $A =\text{diag}(-a,0, \dots,0,-a)$, $a>0$ while $B$ can be e.g. $$B= \begin{bmatrix} 2 &-1 &0 &0& \dots &0& 0 &0 \\ -1 & 3 & -1 &0& \cdots & 0 & 0&0 \\ 0 &-1&3&-1& \dots& 0 &0&0\\ \vdots &\vdots & \quad & \quad&\quad & \quad & \quad\\ \vdots &\vdots & \quad & \quad&\ddots & \ddots & \ddots\\ 0&0&0&0&\quad &-1&3&-1 \\ 0 &0& 0&0 & \dots & 0&-1 &2 \end{bmatrix}$$
Or is there a formula for the general form of the polynomial $\det (\lambda^2I-\lambda A+B^2)$? Can we say something about its roots? Or get some bounds on the eigenvalues? (Even though it seems that the argument must be somehow elementary, it seems quite difficult, and the matrix doesnt have any of the known forms to justify it automatically.)
Any help would be much appreciated.
