what is the eigen decomposition of the matrix $M = \left[ {\begin{array}{*{20}{c}} B&I\\ 0&B \end{array}} \right]$?

Question

I have the upper triangular matrix $M = \left[ {\begin{array}{*{20}{c}} B&I\\ 0&B \end{array}} \right]$ for which I want to write it as $M = V\Lambda {V^{ - 1}}$ where $\Lambda $ is a diagonal matrix. Is there any result that allows analytic answer for this question in matrix theory?

Answer 1

Your matrix is, in general, not diagonalizable.
If $B$ is non-diagnalizable than it is impossible because if it was possible it'd imply that $B$ is diagnalizable.
if $B$ is diagnalizable, $P^{-1}BP=D$, than $\begin{pmatrix}P^{-1} & 0 \\ 0 & P^{-1}\end{pmatrix}\begin{pmatrix}B & I \\ 0 & B\end{pmatrix}\begin{pmatrix}P & 0 \\ 0 & P\end{pmatrix}=\begin{pmatrix}D & I \\ 0 & D\end{pmatrix}$.
In that matrix, if $v$ is a vector with non-zero entries in it's upper half: $$v_i=a,(Av)_i=\lambda a,v_{i-n/2}=b,(Av)_{i+n/2}=\lambda b+a$$ Meaning they are not scaled by the same number, i.e. it's not an eigenvector. Meaning it doesn't have eigenvectors spanning $\mathbb R^n$, so not diagnalizable.