# Proving a projection onto subspace if $P^2 = P$

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??

• Does $V$ has finite dimension ? – user659895 May 13 at 9:39
• yes v has a finite dimension – user673563 May 13 at 9:41
• 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. – Simon May 13 at 10:39
• this was the way the question was worded. – user673563 May 13 at 11:05
• Possible duplicate of Linear Algebra - Proving a projection onto a subspace is a linear transformation – user673563 May 14 at 3:32

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 $$Pm=P^{2}z=Pz=m$$.
• 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 $N$. – Kavi Rama Murthy May 13 at 9:52
• $Px=P(Px')=P^{2}(x')$ and $Pu=P(Pu')=P^{2}(u')$ – Kavi Rama Murthy May 13 at 10:05
Hint: since $$P^2=P$$, we have $$(1-P)^2=1-P$$ and $$P(1-P)=0$$. Now, write any $$v$$ as $$v = Pv + (1-P)v$$.