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?

  • $\begingroup$ 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 $M$. $\endgroup$
    – Z Ahmed
    Aug 12 '19 at 4:29
  • $\begingroup$ "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. $\endgroup$ Aug 12 '19 at 14:14
  • $\begingroup$ 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. $\endgroup$ Aug 12 '19 at 14:20

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.


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