If $A\in\mathcal M_n(\Bbb R)$ is nilpotent matrix such that $\ker(A)=\ker(A^T)$ then $A=0$.

Question

My try: since $\ker(A)=\ker(A^T)=(Im(A))^\perp$ and by the rank-nullity theorem we have $$\Bbb R^n=\ker(A)\oplus Im(A)$$ so if we write the matrix $A$ relative to a basis of $\Bbb R^n$ adapted to previous direct sum then we see that $A$ is similar to the block diagonal matrix $\operatorname{diag}(0,B)$ where $B$ is the matrix of the restriction of $A$ on the subspace $Im(A)$. And since $A$ is nilpotent then $A^n=0$ and so $B^n=0$. If $Im(A)\ne\{0\}$ then $B$ would be invertible and so $B^n\ne0$ which is a contradiction. Hence $Im(A)=\{0\}$ so $\Bbb R^n=\ker A$ and then $A=0$.

Is this proof Ok and are there other ways to prove it?

Answer 1

The basic idea is good but note that it is not the rank-nullity theorem that guarantess you that $\mathbb{R}^n = \operatorname{ker}(A) \oplus \operatorname{Im}(A)$ but the theorem that states that

$$ \mathbb{R}^n = W \oplus W^{\perp} $$

for all subspaces $W \subseteq \mathbb{R}^n$. If you take $W = \operatorname{Im}(A)$ and use your observation that $$W^{\perp} = \operatorname{Im}(A)^{\perp} = \ker(A^T) = \ker(A),$$ you'll get the direct sum decomposition.

Now, consider the operator $L_A \colon \mathbb{R}^n \rightarrow \mathbb{R}^n$ given by $L_A(x) = Ax$ and note that both $\ker(A) = \ker(L_A)$ and $\operatorname{Im}(A) = \operatorname{Im}(L_A)$ are both $L_A$-invariant subspaces. If $\dim \operatorname{Im}(A) = \dim \operatorname{Im}(L_A) > 0$ then $L_A|_{\operatorname{Im}(L_A)}$ is both invertible and nilpotent (as the restriction of a nilpotent operator to an invariant subspace), a contradiction.