Is the following Proof Correct? Please see the accompanying note for the notation used.

Proposition.Suppose $V$ is finite-dimensional and $P\in\mathcal{L}(V)$ is such that $P^2 = P$ and every vector in $\operatorname{null} P$ is orthogonal to every vector in $\operatorname{range}P$ . Prove that there exists a subspace $U$ of $V$ such that $P_U = P$.

Proof. Let $U = \operatorname{range}P$. Since $V$ is finite-dimensional then so is $U$ and by proposition $\textbf{6.47}$, $V = U\oplus U^{\perp}$. Now by $\textbf{6.50}$ and $\textbf{3.22}$ we have $$\dim V = \dim\operatorname{range}P+\dim(\operatorname{range}P)^{\perp}$$ $$\dim V = \dim\operatorname{null}P+\dim\operatorname{range}P$$ consequently $\dim\operatorname{null}P = \dim(\operatorname{range}P)^{\perp}$. This together with our hypothesis $\operatorname{null}P\subseteq(\operatorname{range}P)^{\perp}$ implies that $\operatorname{null}P = (\operatorname{range}P)^{\perp}$. In conclusion $V = \operatorname{range}P\oplus\operatorname{null}P$.

Now let $v\in V$ then $v = u+Pw$ where $u\in\operatorname{null}$ and $Pw\in\operatorname{range}P$ since $P^2 = P$ it follows that $P^2w = P(Pw) = Pw$ and evidently $Pu=0$ this together with the linearity of $P$ implies $P(u+Pw) = Pw$, in summary $P_{\operatorname{range}P} = P$.


Note: $P_U$ is the orthogonal projection of the vector space $V$ on the subspace $U$ where $V = U\oplus U^{\perp}$ where $P_Uv = P_U(u+w) = Pu$ where $u\in U$ and $w\in U^{\perp}$.

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    $\begingroup$ Don't reference propositions that we don't have unless you tell us what they are. $\endgroup$ – GFauxPas Jun 26 '18 at 22:33
  • $\begingroup$ @GFauxPas Will it be helpful if i were to add the statements of said propositions to the above post? $\endgroup$ – Atif Farooq Jun 26 '18 at 22:34
  • $\begingroup$ That’s a lot of “where”s. $\endgroup$ – amd Jun 26 '18 at 22:36
  • $\begingroup$ If $P$ is merely a bounded linear map satisfying $P^2=P$ then it is a projection onto its range, but not necessarily an orthogonal projection. $\endgroup$ – Lior Silberman Jun 28 '18 at 19:03

It seems correct, but no need to invoke dimensional arguments to show $V = \operatorname{range}P\oplus\operatorname{null}P$. Namely, for any $v \in V$ we have

$$v = \underbrace{Pv}_{\in \operatorname{range}P} + \underbrace{(v - Pv)}_{\in \operatorname{null}P}$$

so $V = \operatorname{range}P+\operatorname{null}P$. The assumption $\operatorname{range}P\perp\operatorname{null}P$ shows that the sum is orthogonal so $V = \operatorname{range}P\oplus\operatorname{null}P$.

Now proceed as you did.

  • $\begingroup$ Could you please explain how $v-Pv\in\operatorname{null}P$. Thanks $\endgroup$ – Atif Farooq Jun 26 '18 at 23:12
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    $\begingroup$ @AtifFarooq $P(v-Pv) = Pv - P^2v = Pv - Pv = 0$ because $P^2 = P$. $\endgroup$ – mechanodroid Jun 26 '18 at 23:14

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