Diagonalizability of a block-diagonal matrix I will very appreciate if someone would check my proof of the following problem:

Consider
  $$B=\begin{pmatrix}L&M\\O&N\end{pmatrix}\in\mathbb{C}^{2n\times 2n}$$
  for $L,M,N,O\in\mathbb{C}^{n\times n}$ such that $O$ is the zero
  matrix.
(a) Show that if $B$ is diagonalizable, then $L$ and $N$ must be
  diagonalizable.
(b) Show that if $L$ and $N$ are diagonalizable and do not share
  eigenvalues, then $B$ is diagonalizable.

My attempt is as follows:
(a) Suppose $B$ is diagonalizable, then there exists of a basis of its eigenvectors in $\mathbb{C}^{2n\times 2n}$. Note that $L$ is the restriction of $B$ to a subspace, spanned by first $n$ basis vectors, therefore the minimal polynomial of $L$ divides the minimal polynomial of $B$, which splits into a product of distinct linear factors, and so does the minimal polynomial of $L$, hence it's diagonalizable. Similarly for $N$.
(b) Suppose that $L$ and $N$ are diagonalizable and do not share eigenvalues. Assume to the contrary that $B$ isn't diagonalizable, then its minimal polynomial has a factor of $(t-\lambda)^k$ with $k\geq2$ for some eigenvalue $\lambda$. Then either a minimal polynomial of $L$ or $N$ has this factor and it contradicts diagonalizability of $L$ and $N$ respectively, or $L$ and $N$ share the same eigenvalue $\lambda$, that is a contradiction in any case.
I'm pretty sure that my approach is right, however, I'm not sure that I presented the solution rigorously enough. Is it obvious that $L$ can be viewed as a restriction of $B$ onto a subspace spanned by first $n$ basis vectors and therefore its minimal polynomial divides the minimal polynomial of $B$? Is it obvious that if $(t-\lambda)^k$ is one of the factors of the minimal polynomial of $B$, then it's either a factor in minimal polynomials of $L$ and $N$ or arises as a product of linear factors, corresponding to the same eigenvalue? If these statements aren't obvious, how to justify them?
 A: Problems with a:


*

*You should specify that $L$ is the restriction of $B$ to an invariant subspace.

*$N$ cannot be written as the restriction of $B$ to an invariant subspace.  However, $N^T$ can be written as a restriction of $B^T$ to an invariant subspace, so that gives you a workaround.  An alternative approach for both of these is to note that for any polynomial $p$ we have
$$
p(B) = \pmatrix{p(L) & Q\\ 0 & p(N)}
$$
for some matrix $Q$.  Thus, if $p(B) = 0$ it must be that $p(L) = 0$ and $p(N) = 0$.


Problem with b:


*

*You should prove that the minimal polynomial $m_B$ divides the product $m_L \cdot m_N$.  One approach to do so is to note that
$$
m_L(B)m_N(B) = \pmatrix{0 & Q\\ 0 & m_L(N)} \pmatrix{m_N(L) & R\\0 & 0} = \pmatrix{0&0\\0&0}
$$
by block-matrix multiplication.  Since $m_L(B)m_N(B) = 0$ it follows that $m_B$ divides $m_L \cdot m_N$.


To answer your other questions: it is clear that the minimal polynomial of a restriction will divide that of the original transformation so that's fine. Once you prove that $m_B$ divides the product $m_L \cdot m_N$, your answer for b is complete.
