0
$\begingroup$

Suppose that V is a vector space, and M is a subspace of V . A transformation $P : V \to V$ is called the projection of V onto M if

(i) there exists a subspace N such that every vector v ∈ V can be written uniquely as $v = x + y$ for some $x ∈ M$ and $y ∈ N$; and

(ii) P is given by $P(x + y) = x$, for all $x ∈ M$ and $y ∈ N$.

Question: Suppose that $P : V \to V$ is a linear transformation. Prove that P is a projection onto some subspace of V if and only if $P^2 = P$

So I can't seem to prove the reverse direction where I assume P^2 = P. Like what do we need to prove in order to show that it is a projection??

$\endgroup$
  • $\begingroup$ Does $V$ has finite dimension ? $\endgroup$ – user659895 May 13 at 9:39
  • $\begingroup$ yes v has a finite dimension $\endgroup$ – user673563 May 13 at 9:41
  • $\begingroup$ I think you should write "a projection of V onto M", rather than "the projection of V onto M", since the condition does not define a unique P. $\endgroup$ – Simon May 13 at 10:39
  • $\begingroup$ this was the way the question was worded. $\endgroup$ – user673563 May 13 at 11:05
  • $\begingroup$ Possible duplicate of Linear Algebra - Proving a projection onto a subspace is a linear transformation $\endgroup$ – user673563 May 14 at 3:32
1
$\begingroup$

Let $M$ be the range of $P$ and $N$ be the kernel. Then $x =Px+(x-P(x)), Px \in M$ and $x-Px \in N$ because $P(x-Px)=Px-P^{2}x=0$. Hence every vector is a sum of an element from $M$ and an element from $N$. Suppose $x+y=u+v$ where $x, u \in M$ and $y,v \in N$. Then we can write $x=Px',u=Pu'$ so $P(x-u)=P^{2}(x'-u')=P(x'-u')=x-u$. But $x-u=v-y$. Applying $P$ to both sides we get $x-u=P(v-y)=Pv-Py=0-0=0$. Hence $x=u$ and it is now obviuous that $v=y$. This proves (i). (ii) is easy: if $x =m+n, m \in M, n \in N$ then $Px=P(m+n)=Pm+0=Pm=m$ since $m =Pz$ for some $z$ which gives $Pm=P^{2}z=Pz=m$.

$\endgroup$
  • $\begingroup$ hi, why is x the sum of the range and kernel? $\endgroup$ – user673563 May 13 at 9:51
  • $\begingroup$ The equation $x=Px+(x-Px)$ is an identity and I have proved that the first term is in $M$ and the second term is in $N$. $\endgroup$ – Kavi Rama Murthy May 13 at 9:52
  • $\begingroup$ ah ok thank you $\endgroup$ – user673563 May 13 at 9:53
  • $\begingroup$ 𝑃(𝑥−𝑦)=𝑃^2(𝑥′−𝑢′) how did u get this? $\endgroup$ – user673563 May 13 at 10:03
  • $\begingroup$ $Px=P(Px')=P^{2}(x')$ and $Pu=P(Pu')=P^{2}(u')$ $\endgroup$ – Kavi Rama Murthy May 13 at 10:05
1
$\begingroup$

Hint: since $P^2=P$, we have $(1-P)^2=1-P$ and $P(1-P)=0$. Now, write any $v$ as $v = Pv + (1-P)v$.

$\endgroup$

Your Answer

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