# 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?

• My numerics indicate that the eigenvalues of this block matrix $B$ are nothing but the eigenvalues of $L$ and $N$, irrespective of the elements of the matrix $M$. Aug 12 '19 at 4:29
• "Similarly for $N$" won't do. $N$ is not the matrix of the underlying transformation $\beta$ on a subspace, but on the quotient space $V/U$ where $U$ is the (invariant) subspace spanned by the first $n$ basis vectors. Aug 12 '19 at 14:14
• I think you need to prove that $m_{B}(X)$ divides $m_{L}(X)m_{N}(X)$. As $m_{N}(B)$ pushes all vectors into the space of the first $n$ basis vectors this is easy to see. Aug 12 '19 at 14:20

• 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 $$p(N) = 0$$.
• 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.