So, I want to prove the following statement:
Let A be diagonal block matrix with the blocks $ A_{1},...A_{r}$, then $xI-A=\left(\begin{array}{cccc} xI-A_{1}\\ & xI-A_{2}\\ & & \ddots\\ & & & xI-A_{r} \end{array}\right) $
Proof:
For each $1\leq i\leq r $ let $ n_{i}xn_{i} $ be the size of $ A_{i}$ . and let $ \sum_{i=1}^{n}n_{i}=n $ be the size of $A $. let $ n_{0}=1 $. and let $ I_{nxn} $. So:
$ [xI]_{lm}=\begin{cases} x & l=m\\ 0 & l\neq m \end{cases} $ and by definition :
$ [xI-A]_{lm}=[xI]_{lm}-[A]_{lm} $
now for any $ 0\leq i\leq r-1 $ we have for any $ n_{i}\leq l,m\leq n_{i+1} $
$ [xI-A]_{lm}=[xI]_{lm}-[A_{n_{i}}]_{lm}=\begin{cases} x-[A_{n_{i}}]_{lm} & l=m\\ -[A_{n_{i}}]_{lm} & l\neq m \end{cases}=[xI-A_{n_{i}}]_{lm} $
so for any $ 1\leq l,m\leq n\ $ we have : $ [xI-A]_{lm}=[xI]_{lm}-[A]_{lm}=\begin{cases} [xI-A_{n_{i}}]_{lm} & \exists\left(0\leq i\leq1:n_{i}\leq l,m\leq n_{i+1}\right)\\ 0 & else \end{cases} $
Thus:
$ xI-A=\left(\begin{array}{cccc} xI-A_{1}\\ & xI-A_{2}\\ & & \ddots\\ & & & xI-A_{r} \end{array}\right) $
I feel like something's missing. I guess the usual way would be to prove it by induction. but im not sure how to do it. I'll be glad to see some more ways to prove this statement, and to hear some of your thoughts.