Is it true: A real block diagonal matrix is diagonalizable if and only if each block is diagonalizable.

Question

We know this theorem and I know the proof for that.

Theorem. Define $A \in M_{n}(\mathbb{C})$ such that $$ A = \begin{bmatrix} A_{1} & & 0 \\ & \ddots & \\ 0 & & A_{k} \end{bmatrix} $$ where $A_{k} \in M_{n_{i}}$ are block matrices. Then, $A$ is diagonalizable if and only if each of $A_{k}$ is diagonalizable.

I am curious to know whether this theorem holds for real matrices or not?

Answer 1

Yes, this holds over the reals as well.

• A matrix can be diagonalized (over a given field) if and only if we can exhibit a basis of the relevant vector space consisting over eigenvectors of the given matrix.
• If all the blocks $A_i$ are diagonalizable, then clearly so is $A$.
• A column vector $[X_1,X_2,\ldots,X_k]^T$ (with component vector sizes matching the sizes of the blocks) is an eigenvector of $A$ if and only if all the components $X_i$ are eigenvectors of the block $A_i$, all belonging to the same eigenvalue. Therefore a basis of eigenvectors of $A$ also yields bases (for the relevant subspaces) consisting of eigenvectors of the block $A_i$ (a few details left as an exercise).