I have to prove, that if ${\rm Im} (T) = \ker (T)$, then the transformation matrix is nilpotent.

How can I do this?

I know the Rank–nullity theorem:

If $T: V \to W$, then $\dim{\rm Im}(T) + \dim \ker (T) =\dim V$

In this case: $2 \dim{\rm Im} (T) = 2 \dim \ker (T) = \dim V$

I don't see how to prove that $T$ is nilpotent.


2 Answers 2


Notice that $T(T(v))=0$ for all $v\in V$. Hence $T$ is nilpotent.


Let's look at $T(T(v))$ for any vector $v$. Since $T(v)$ is in $Im\;T=Ker\;T$, applying $T$ on it will result in $0$ by definition of $Ker\;T$

  • $\begingroup$ Thanks! Is this true in the other direction? I mean if T is nilpotent, then im T = ker T? $\endgroup$
    – Nori
    May 15, 2020 at 6:42
  • $\begingroup$ @Nori The converse is false. Take the matrix $$ T =\pmatrix{0 & 1 & 0 \\ 0 & 0& 1 \\ 0& 0 &0}$$ Then the kernel has dimesion $1$ and the image has dimension $2$. $\endgroup$
    – Menezio
    May 15, 2020 at 7:22

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