Is it possible for $AA^T$ to be a nilpotent matrix if neither $A$ nor $A^T$ are?

Question

If $A$ is a square, non-nilpotent matrix with real-valued elements, and its transpose is $A^T$, then is it ever possible for $AA^T$ to be nilpotent? What if we allow complex-valued elements? Is $AA^H$ ever nilpotent (with $A^H$ being the conjugate transpose of the matrix)?

Answer 1

Observe that if $MM^H=0$ for some complex rectangular matrix $M$, we must have $M=0$. In fact, if $MM^H=0$, then $\langle M^Hx,\,M^Hx\rangle=\langle x,MM^Hx\rangle=0$ for every vector $x$. Therefore $M^Hx=0$ for all $x$ and in turn both $M^H$ and $M$ are zero.

Now, suppose $AA^H$ is nilpotent. Then $(AA^H)^{2^k}=0$ when $k$ is sufficiently large. Put $M=(AA^H)^{2^{k-1}}$, we obtain $MM^H=0$. Therefore $0=M=(AA^H)^{2^{k-1}}$. Continue this argument recursively, we finally arrive at $AA^H=0$. Hence $A=0$.

Note that the above argument still holds even if $M$ is a linear operator defined on an infinite dimensional inner product space. The notion of eigenvalue is not needed.

Answer 2

Calim : $A\in M_n(\Bbb{R}) $ not nilpotent implies $AA^{T}$ is not nilpotent.

Suppose, $AA^{T} $ is nilpotent .

$Spec(AA^{T})=\{0\}$

Then, $E_{0}(AA^{T})= \Bbb{R^n}$

$AA^{T} $ is a symmetric matrix, hence diagonalizable.

$\exists P\in Gl_n(\Bbb{R})$ such that

$P^{-1} (AA^{T}) P=\Lambda$

$\Lambda$ is a diagonal matrix with all eigenvalues $0$ , i.e $\Lambda=0$

Implies $AA^T=0$

Hence, $A=0$

And $A$ is a nilpotent matrix of degree of nilpotency $1$.

A contradiction.