Is this a property of nilpotent operators?

Question

Let $A\in L(V,V)$ be a nilpotent operator (i.e. linear operator such that for some $m\in \mathbb N$ the condition $A^m=0$ holds) on $n$-dimensional linear space $V$.

Is it true that $A(\ker A^k)=\ker A^{k-1}$, where $k\geq2$ ?

Answer 1

Of course no; consider $$ A=\begin{bmatrix}0&1\\0&0\end{bmatrix} $$ Then $A^2=0$, so $A(\ker A^3)$ is the column space of $A$, which has dimension $1$, but on the other hand $\ker A^2$ has dimension $2$.

Answer 2

No. Take $A=\left[\begin{array}{cc}0 &1\\0 & 0\end{array}\right]$ and $k=2$.

$A(\ker A^2)$ is the image of $A$, which is the one-dimensional subspace spanned by $(0,1)$, and $\ker A^1$ is the one-dimensional subspace spanned by $(1,0)$.

Answer 3

One always has $A(\ker A^k)\subseteq\ker(A^{k-1})$ whenever $k>0$, by a very simple argument (that I shall omit). But the opposite inclusion is not always valid, as the LHS is contained in the image of $A$, but there is no reason that the RHS should be so (unless $k=1$, in which case RHS has dimension$~0$). Indeed for any given nilpotent operator $A$ on $V$ and any $k>1$, one may extend $A$ to $A'$ on $V\oplus W$, with the $A'$ acting on $W$ by the scalar $0$; then $W\not\subseteq A'(V\oplus W)\supseteq A'(\ker((A')^k)$, while $W\subseteq\ker((A')^{k-1})$, and it follows that $\ker((A')^{k-1})\not\subseteq A'(\ker((A')^k)$.