Let $V$ be a $n$-dimensional complex vector space and $\phi:V\to V$ a linear mapping. Prove that $$V = \ker(\phi^n) \oplus \text{image}(\phi^n)$$

Here is my attempt:

Since $\phi^n$ is also a linear mapping of $V$ into $V$, we have that $$\dim V = \dim \ker(\phi^n) + \dim \text{image}(\phi^n).$$ We only need only to show that this sum is direct, in other words, that $$\ker(\phi^n) \cap \text{image}(\phi^n) = \{0\}.$$ since this would imply $$V = \ker(\phi^n) + \text{image}(\phi^n)$$

We let $v \in \ker(\phi^n) \cap \text{image}(\phi^n)$ be arbitrary and aim to show that $v=0$. $\ker(\phi^n)$ is the generalized eigenspace of $\phi$ for the eigenvalue $0$, so there is a $k \leq n$ such that $\phi^k(v) = 0$.

This is where I'm stuck. How do I proceed from here? Is there a different way to do this?

  • 1
    $\begingroup$ The Rank-Nullity Theorem gives that $\dim \ker T + \dim \textrm{image} T = n$ for any transformation $T: V \to V$. $\endgroup$ Sep 4, 2014 at 12:15
  • $\begingroup$ Recall also that dim$(V+W) = \text{dim}(V) + \text{dim}(W) - \text{dim}(V\cap W)$. $\endgroup$
    – fretty
    Sep 4, 2014 at 12:25
  • $\begingroup$ It will help to notice (using Jordan canonical form, say) that $\phi^{n}$ is diagonalizable. $\endgroup$ Sep 4, 2014 at 12:25
  • $\begingroup$ I still don't see the solution. :) Could you please elaborate on why $\phi^n$ is diagonalizable? Thank you! $\endgroup$
    – rehband
    Sep 4, 2014 at 12:38
  • $\begingroup$ Why is $V = \operatorname{ker}(\phi^n)+\operatorname{image}(\phi^n)$? This isn't true for a general linear map in place of $\phi^n$. $\endgroup$ Sep 4, 2014 at 12:50

2 Answers 2


We consider the chains $$V\subset\phi(V)\subset\cdots\subset\phi^n(V)\subset\cdots$$ and $$\dim(V)\geq\dim(\phi(V))\geq\cdots\geq\dim(\phi^n(V))\geq\cdots$$

  • If $\dim(\phi^n(V))=1$, it is easy to prove.

  • If $\dim(\phi^n(V))\geq2$, there exist $k\leq n$ such that $\phi^k(V)=\phi^{k+1}(V)=\cdots$. Actually, $\phi:\phi^n(V)\rightarrow\phi^n(V)$ is an isomorphism.

    • For any $a\in V$, there exists $b$ such that $\phi^n(a)=\phi^{2n}(b)$. Then we can prove $a=c+\phi^n(b)$ and $\phi^n(c)=0$.
    • We assume $a\in\ker(\phi^n)\cap\text{image}(\phi^n)$. That is to say, $\phi^n(a)=0$ and $\phi^n(b)=a$. If $a\ne0$, we get $\phi^{2n}(b)=0$ and $\phi^n(b)\ne0$ which will lead $\dim(\phi^n(V))>\dim(\phi^{2n}(V))$. That is a contraction.

Look at an another question Show that $V = \mbox{ker}(f) \oplus \mbox{im}(f)$ for a linear map with $f \circ f = f$.

  • In your question, $\phi:\phi^n(V)\rightarrow\phi^n(V)$ is an isomorphism. Also $\phi^n:\phi^n(V)\rightarrow\phi^n(V)$ is an isomorphism.

  • In another question, $f=id: f(V)\rightarrow f(V)$ is an isomorphism.

  • Lemma Let $f:V\rightarrow V$ be a linear map with $f:f(V)\rightarrow f(V)$ being an isomorphism. Then we have $$V=\ker(f)\oplus\text{image}(f)$$

  • $\begingroup$ Infintie thanks for the answer! I'm still trying to understand it :) $\endgroup$
    – rehband
    Sep 4, 2014 at 13:42

My answer is just a cleaned up version of the answer by gaoxinge.

$\newcommand{\inters}{\cap} $$\newcommand{\im}{\mathrm{im}} $The following chain of subspaces of the $n$-dimensional space $V\!$, $$ V\supseteq\phi V\supseteq\phi^2V\supseteq\cdots~, $$ cannot strictly decrease for ever, thus there is the least $k$ such that $\phi^kV=\phi^{k+1}V\!$, and then $\phi^kV=\phi^lV$ for all $l\geq k$. Set $U:=\phi^kV\!$. In the strictly descending chain $$ V\supset\phi V\supset\cdots\supset\phi^k V $$ the dimensions of subspaces decrease by at least one at each step, so we must have $k\leq n$. It follows that $\phi^{n+1}V=\phi^nV=\phi^kV=U$, so that $\phi\, U=U$, which means that the restriction of $\phi$ to $\phi_U\colon U\to U$ is an isomorphism. As you observed it suffices to prove that $\im(\phi^n)\inters\ker(\phi^n)=0$, so let us consider any $y$ in this intersection. Since $y\in\im(\phi^n)=\phi^nV=U$, and $\phi_U^ny=\phi^ny=0$, and $\phi_U^n\colon U\to U$ is an isomorphism, it follows that $y=0$, and we are done.


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