# If $V_1=\ker(P)$, $V_2=\ker(I-P)$ and $V=V_1 \oplus V_2$ then prove $P^2=P$?

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.

• $P$ is equal to $0$ on $V_1$ and to $1$ on $V_2$. – Pierre-Yves Gaillard Oct 31 '11 at 11:03

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 $\rm P^2=P.$

• Thanks! that's very helpful – Freeman Oct 31 '11 at 14:39

$(P-P^2)v_1 = (I-P)Pv_1 = 0$ for $v_1\in V_1$, and $(P-P^2)v_2 = P(I-P)v_2 = 0$ for $v_2\in V_2$.

• That's clear, thank you. – Freeman Oct 31 '11 at 14:39

Hint: Let $x\in V$, and decompose $x=x_1+x_2$.

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

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 $P^2=P$.
• I take it you mean for $x_1$ to be in $V_1$, $x_2$ in $V_2$. – Gerry Myerson Oct 31 '11 at 11:35