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Let $V$ a finite vectorial space and $T:V\to V$ a linear transformation. Show that if $W$ is a subspace of $V$ invariant under $T$ so that $V={\rm{Im}}(T)\oplus W$ then $W=\ker T$.

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    $\begingroup$ Have you tried anything? $\endgroup$ – Brian Fitzpatrick Jul 11 '17 at 23:54
  • $\begingroup$ I tried with the range-nulity theorem but I not Have exit! $\endgroup$ – Diego1802 Jul 11 '17 at 23:56
  • $\begingroup$ for any $w\in W$, $w\in Im(T)$ by definition. But $W$ is $T$ invariant, what does it mean? $\endgroup$ – chan kifung Jul 11 '17 at 23:58
  • $\begingroup$ why if $w\in W$ then $w\in$ Im(T)$? $\endgroup$ – Diego1802 Jul 12 '17 at 0:03
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Suppose $w\in W$. Then $T(w)\in\DeclareMathOperator{image}{image}\image(T)$ and $T(w)\in W$ since $W$ is $T$-invariant. That is, $T(w)\in\image(T)\cap W$. But $\image(T)\cap W=\{\vec 0\}$ since $V=\image(T)\oplus W$. Thus $T(w)=\vec 0$. This shows that $W\subset \ker(T)$.

Now, the rank-nullity theorem implies $$ \dim\ker(T)=\dim(V)-\dim\image(T)=\dim(V)-\{\dim(V)-\dim(W)\}=\dim(W) $$ Hence $W=\ker(T)$.

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  • $\begingroup$ Shouldn't we leave a few gaps for the sake of OP's pleasure/curiosity? $\endgroup$ – user 1987 Jul 12 '17 at 0:10
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$\textbf{Hint: }$ Use Im$(T)\cap W = \{0\}$. Hence $T(v) = 0 \iff v \in $Im$(T)$$\cap W$. Use this to show the two containments.

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