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Let $V$ be a finite dimensional vector space over the reals and $P:V \rightarrow V$ be a linear transformation. Let

$V_1=\ker(P)$

$V_2=\ker(I-P)$

and $V=V_1 \oplus V_2$

How would you prove $P^2=P$? I've been trying to show that $\mathrm{Im}(P)=\ker(I-p)$ as that would be equivalent.

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  • $\begingroup$ $P$ is equal to $0$ on $V_1$ and to $1$ on $V_2$. $\endgroup$ – Pierre-Yves Gaillard Oct 31 '11 at 11:03
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By definition, $ \rm V_1 = \{ x\in V: P(x) = 0 \} $ and $\rm V_2 = \{ x\in V : P(x) = x \} .$

Let $\rm v_1 \in V_1 $ and $\rm v_2 \in V_2$. Then

$$\rm (P^2-P)(v_1 + v_2) = P^2(v_1) + P^2(v_2) - P(v_1) - P(v_2) .$$

Since $\rm v_2 \in V_2$ we have $\rm P^2 (v_2) = P( P(v_2) ) = P(v_2) $. We know $\rm v_1 \in V_1$ so $\rm P(v_1) = 0 $ and $\rm P^2(v_1) = P (P(v_1))=P(0) = 0$ as well. Thus $$\rm (P^2-P)(v_1 + v_2) =0 $$ so $\rm P^2=P.$

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  • $\begingroup$ Thanks! that's very helpful $\endgroup$ – Freeman Oct 31 '11 at 14:39
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$(P-P^2)v_1 = (I-P)Pv_1 = 0$ for $v_1\in V_1$, and $(P-P^2)v_2 = P(I-P)v_2 = 0$ for $v_2\in V_2$.

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  • $\begingroup$ That's clear, thank you. $\endgroup$ – Freeman Oct 31 '11 at 14:39
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Hint: Let $x\in V$, and decompose $x=x_1+x_2$.

You should then apply $P$ twice, and see that, indeed, $P(P(x))=P(x)$.

Edit: A more complete answer.

Let $x\in V$, and let $x_1\in V_1$ and $x_2\in V_2$ be the unique elements such that $x=x_1+x_2$.

By definition of $V_1$ and $V_2$, $P(x_1)=0$ and $x_2-P(x_2)=0$, so \begin{align*} P(x)&= P(x_1) + P(x_2)\\ &= 0 + x_2\\ P(P(x))&= P(0) + P(x_2)\\ &=x_2, \end{align*} so we have that $$\forall x\in V, P^2(x)=P(x),$$ which proves indeed that $P^2=P$.

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  • $\begingroup$ I take it you mean for $x_1$ to be in $V_1$, $x_2$ in $V_2$. $\endgroup$ – Gerry Myerson Oct 31 '11 at 11:35
  • $\begingroup$ @Gerry: yes indeed, but seeing this is homework, the above was more intended as a hint than a definitive answer. :) $\endgroup$ – FelixCQ Oct 31 '11 at 13:01
  • $\begingroup$ I take your point. I made a judgement that the hint was just a little too elliptical to be helpful. I might have been wrong. $\endgroup$ – Gerry Myerson Oct 31 '11 at 21:52
  • $\begingroup$ @Gerry: Judging from the votes -- and the comments from the OP -- it seems you were right! I will try to be less elliptical in the future. $\endgroup$ – FelixCQ Oct 31 '11 at 23:00

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