Showing two matrix blocks are similar

Question

Let $A \in M_n$ and $B,C \in M_m$. Prove that if

$$H= \begin{bmatrix} A&0 \\ 0 & B \end{bmatrix}$$

is similar to

$$K = \begin{bmatrix} A&0 \\ 0 & C \end{bmatrix}$$

then $B$ is similar to $C$.

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I am not sure how I would do this I know that if $H$ is similar to $K$ then for some non-singular matrix $S$ then $S^{-1} H S=K$.

Answer 1

Using Jordan form: there exists matrices $P,Q,R$ such that $P^{-1}AP = J_A$, $Q^{-1}BQ = J_B$, and $RCR^{-1} = J_C$ are all in Jordan normal form.

We then note that $$ \pmatrix{P\\&Q}^{-1}\pmatrix{A\\&B} \pmatrix{P\\&Q} = \pmatrix{J_A\\& J_B}\\ \pmatrix{P\\&R}^{-1}\pmatrix{A\\&C} \pmatrix{P\\&R} = \pmatrix{J_A\\& J_C} $$ By the uniqueness of Jordan form (up to permutations of blocks), the two matrices on the right can only be similar if $J_B$ is similar to $J_C$, which is to say that $B$ is similar to $C$ as desired.

Answer 2

Since $H$ and $K$ are similar, $B$ and $C$ have same set of distinct eigenvalues. Also $B$ and $C$ have same generalized eigenspaces (same number of independent vectors) for distinct eigenvalues, i.e. $$ \operatorname{Rank}(\lambda_iI-B)^n=\operatorname{Rank}(\lambda_iI-C)^n $$ where $\lambda_i$ are distinct eigenvalues of $B$ and $C$. Hence $B$ and $C$ are similar.