Does this mean the zero matrix is diagonalizable.

Question

Let $A \in M_{n}(\mathbb C)$ with $D = \begin{bmatrix} A & A \\ 0 & A \end{bmatrix} \in M_{2n}(\mathbb C)$. Show that $D$ is diagonalizable iff $A=0$.

This question perplexes me. Given that D is diagonalizable over $\mathbb C$ for example, does that not mean that by definition there are $2n$ unique eigenvalues. However if $D=0$ then $\chi_{D}=\lambda^{2n}$? Something does not add up.

Answer 1

"Given that D is diagonalizable over C for example, does that not mean that by definition there are 2n unique eigenvalues."

No. It can be only one eigenvalue. For example, $I$ (the identity matrix) is clearly diagonalizable (already diagonal), but has only the eigenvalue $1$.

Answer 2

A diagonalisable matrix can have repeated eigenvalues. A criterion for a matrix to be diagonalisable is that it's minimal polynomial has only simple roots, and the null matrix has $x$ as its minimal polynomial.

Answer 3

"Given that D is diagonalizable over $\Bbb C$ for example, does that not mean that by definition there are $2n$ unique eigenvalues."

No, what you mean to say is that there are $2n$ linearly independent (not unique) eigenvectors (not eigenvalues).