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Let $L$ be a linear image from $\mathbb R^n$ to $\mathbb R^n$ that has $\mbox{Im}(L)=\mbox{Im}(L^2)$

Prove that $\mbox{Ker}(L) = \mbox{Ker}(L^2)$

I've been trying to get this for like two hours but with nothing really. I've tried anti-thesis and other methods but I can't seem to get this. The teacher gave us a hint telling to use dimension clause but I can't figure what else you can do with that but get the fact that $\dim\mbox{Ker}(L) =\dim\mbox{Ker}(L^2)$.

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  • $\begingroup$ if you can get dim Ker($L$) = dim Ker($L^2$) then you are done since clearly Ker($L$) $\subset$ Ker($L^2$) $\endgroup$ – hunter Jan 19 '14 at 11:08
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The dimension theorem:

$$\begin{align*}(1)&\;\;\dim\Bbb R^n=n=\dim\ker L+\dim\text{Im}\,L\\{}\\ (2)&\;\;\dim\Bbb R^n=n=\dim\ker L^2+\dim\text{Im}\,L^2\end{align*}$$

But $\;\dim \text{Im}\,L=\dim\text{Im}\,L^2\;$ , so

$$\dim\ker L=\dim\ker L^2\;\;\ldots \text{but also}\;\;\ker L\subset \ker L^2\;,\;\;\text{so}\;\ldots$$

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By assumption, $L : \mbox{Im}(L)\rightarrow \mbox{Im}(L)$ is onto. So this restriction has trivial kernel, which means $L(Lx)=0$ iff $(Lx)=0$. Equivalently, $\mbox{Ker}(L)=\mbox{Ker}(L^{2})$.

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  • $\begingroup$ How did you get that $\;L(Lx)=0\iff Lx=0\;$ ?? The direction $\;\Longleftarrow\;$ is trivial, but the other one... $\endgroup$ – DonAntonio Jan 19 '14 at 10:10
  • $\begingroup$ T.A.E. has explained this. Since $L|_{\mathrm{Im}(L)}$ is an endomorphism of a finite-dimensional vector space, which is onto, it is an isomorphism. Trivial kernel means exactly $L(Lx)=0 \Rightarrow Lx=0$. $\endgroup$ – Martin Brandenburg Jan 19 '14 at 10:12
  • $\begingroup$ @DonAntonio: What Martin said is it. In other words: if $Ly=0$ for some $y \in \mbox{Im}(L)$, then $y=0$; so $L(Lx)=0$ implies $(Lx)=0$. $\endgroup$ – DisintegratingByParts Jan 19 '14 at 10:17
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    $\begingroup$ No $x$ is arbitrary. $\endgroup$ – Martin Brandenburg Jan 19 '14 at 10:36
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    $\begingroup$ But $Lx \in \mathrm{Im} L$ :) $\endgroup$ – Ben Millwood Jan 19 '14 at 11:03
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If think you should look into this property :

Let $A$, $B$ be subspaces of a finite-dimensional vector space. Then if $A \subseteq B$ and $\dim A = \dim B$ we know we have $A =B$.

So here, two things to prove :

1) $\ker(L) \subset \ker(L^2)$ which is not that hard to prove (and is true for any linearimage $L$)

2) $\dim \ker(L) = \dim \ker(L^2)$ (which is only true here because of some assumption concerning $L$)

Then using the property I have highlighted you get your answer !

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