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.

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