Can we say that the matrix $\begin{bmatrix} A & A \\ 0 & A \end{bmatrix}$ is diagonalizable if and only if $A = 0$?

Solution 1:

Let $m$ be the minimal polynomial of $A$ and $M=\begin{pmatrix}A&A\\0&A\end{pmatrix}$. Note that $m(M)=\begin{pmatrix}m(A)&Am'(A)\\0&m(A)\end{pmatrix}$.

Thus $m^2(M)=0$ and, since $M$ is diagonalizable, $m(M)=0$. Suppose that $m$ has multiple roots; then, there is a polynomial $p$ that strictly divides $m$ and satisfies $p(M)=0$; that implies $p(A)=0$, a contradiction.

Finally, $m$ has simple roots and $m(M)=0$ implies that $Am'(A)=0$; since $m'(A)$ is invertible, $A=0$. $\square$

Solution 2:

The answer to your question is yes. Of course if $A = 0$ then $M$ is diagonalizable, but the trick is to prove that the converse holds.

One way to see this is as follows: a matrix $M$ of size $2n$ is diagonalizable if and only if $(M - \lambda I)^{2n}v = 0$ implies that $Mv = \lambda v$, i.e. that $(M - \lambda I)v = 0$.

Let $M$ be your matrix. For a vector $v = (x,y)$, we find (using block-matrix multiplication) that $$ Mv = \pmatrix{A&A\\0&A}\pmatrix{x\\y} = \pmatrix{A(x+y)\\Ay} $$ Now, let $y$ be an eigenvector of $v$ associated with eigenvalue $\lambda$. We note that taking $v = (0,y)$, we have $$ Mv = \pmatrix{Ay\\Ay} = \lambda \pmatrix{y\\y} \implies (M - \lambda I)v = \lambda \pmatrix{y\\y} - \lambda \pmatrix{0\\y} = \pmatrix{\lambda y\\0},\\ (M - \lambda I)^2 \pmatrix{0\\y} = (M - \lambda I) \pmatrix{\lambda y\\0} = 0. $$ So, $A$ cannot be diagonalizable if $A$ has any non-zero eigenvalues.

Now, suppose that $A$ has only zero eigenvalues. It follows that $M$ has only zero eigenvalues. However, if a matrix with zero as its only eigenvalues is diagonalizable, it must be the zero matrix. So, if $A$ has only zero eigenvalues and $M$ is diagonalizable, then $M = 0$ which in turn means that $A$ is zero.

So, the conclusion holds: if $M$ is diagonalizable, then $A$ must be zero.

Another way to see this is to note that your matrix can be expressed as $$ M = \pmatrix{1&1\\0&1} \otimes A, $$ where $\otimes$ denotes the Kronecker product, and then to use the properties of the Kronecker product. We could also use the Kronecker product to simplify notation in the proof outlined above.