# Algebra (Artin) Ch.6 Problem 6.10

Problem statement: Let $M$ be a matrix made up of two diagonal blocks: $M = \begin{bmatrix} A & 0 \\ 0 & D \end{bmatrix}$. Then $M$ is diagonalizable if and only if $A$ and $D$ are diagonalizable.

Solution: Let $A, D$ be $n \times n$ matrices. Then $M$ is an $2n \times 2n$ matrix. Let $T$ be the linear operator and $(u_1,..,u_n,u'_1,..,u'_n)$ be the basis of the vector space $F^{2n}=V$ w.r.t the matrix $M$. Denote $Sp({u_1,..,u_n})$ by W and $Sp({u'_1,..,u'_n})$ by W’. Then W and W’ are T-invariant subspaces. Also, $V$ is the direct sum of $W$ and $W'$.

The if part:

Let $M$ be diagonalizable. Then there is a basis of eigenvectors of the vector space $F^{2n}$. Let the basis be $(v_1,..,v_{2n}).$ Then $v_i = w_i + w'_i$ for some $w_i \in W$ and $w'_i \in W'$. Then $T(v_i)=\lambda_i v_i = \lambda_i w_i + \lambda_i w'_i = T(w_i) + T(w'_i)$.

Then $T(w_i)-\lambda_i w_i + T(w'_i)-\lambda_i w'_i = 0$. Since W and W’ are T-invariant, $T(w_i)=\lambda_i w_i, T(w'_i)=\lambda_i w'_i$. Then the nonzero $w_i s$ and $w'_i s$ are eigenvectors of W and W’ respectively. Also, since V is the direct sum of W and W’, and $(v_1,..,v_{2n})$ is a basis of V, $(w_1,..,w_n)$ generates W and $(w'_1,..,w'_n)$ generates W’. Then we can obtain a basis of eigenvectors for W and W’ by dropping the dependent vectors in $(w_1,..,w_n)$ and $(w'_1,..,w'_n)$. This essentially means that A and D are diagonalizable.

The only-if part:

As $A, D$ are diagonalizable, let $P^{-1}AP$ and $Q^{-1}DQ$ be diagonal matrices where $P,Q$ are invertible. Then $\begin{bmatrix} P & 0 \\ 0 & I_n \end{bmatrix}$ $\begin{bmatrix} I_n & 0 \\ 0 & Q \end{bmatrix}$ diagonalizes $M$.