10
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ The Rank-Nullity Theorem gives that $\dim \ker T + \dim \textrm{image} T = n$ for any transformation $T: V \to V$. $\endgroup$ – Travis Sep 4 '14 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 '14 at 12:25
  • $\begingroup$ It will help to notice (using Jordan canonical form, say) that $\phi^{n}$ is diagonalizable. $\endgroup$ – Andrew D. Hwang Sep 4 '14 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 '14 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 $\phi^n$. $\endgroup$ – Santiago Canez Sep 4 '14 at 12:50
3
$\begingroup$

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 $$V=\ker(f)\oplus\text{image}(f)$$

$\endgroup$
  • $\begingroup$ Infintie thanks for the answer! I'm still trying to understand it :) $\endgroup$ – rehband Sep 4 '14 at 13:42
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.