3
$\begingroup$

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 $P_{\operatorname{range}P} = P$.

$\blacksquare$


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 $w\in U^{\perp}$.

$\endgroup$
  • 3
    $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Could you please explain how $v-Pv\in\operatorname{null}P$. Thanks $\endgroup$ – Atif Farooq Jun 26 '18 at 23:12
  • 1
    $\begingroup$ @AtifFarooq $P(v-Pv) = Pv - P^2v = Pv - Pv = 0$ because $P^2 = P$. $\endgroup$ – mechanodroid Jun 26 '18 at 23:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.