Question

8 Suppose V is finite-dimensional and P E L(V) is such that P2 = P. and || P v|| = || V || for every v E V. Prove that there exists a subspace U of V such that P = Pu.

Answer 1

8 Suppose V is finite-dimensional and P EL(V) is such that P2 = P. and ||PV| | for every v E V. Prove that there exists a subspace U of V such that P = Pu. finst, let us show that V = ker(P) = P(V) Let REV , then a can be written as x= x-P(x) + P(x) where P (2-pcx)) = P(a) – pº() = P(x) - PCx) =o => x-PCX) E ker(P) and PCR) E P (v) so that be wratten every as element sum of in is u elements can in Ker(p) and PCV). V = ker(P) + P(V)

To prove the above sum to be a dibeet - sum sum o we ne , we need to show that Kercp) OPCv2 = {o} Let Veker(8) pcv2 - P (v) = 0 v=P (4) for some е у → P (Plus) = P(V)=0 > pv (4)=0 > P(4)=0 vzo so that ker(p) o PCv2 = {o} so v = ker(p) @ PCV) Now we can see that p is a Projection on P(v) along ker(P). To show show that this projection is of thogonal projection, we need to

show that <x,y>zo txt ker(p) y EP (V). Now for ter ne ker(P) ,YEP(%) we see that P(xtky)= P(x) + PC+4)= 0 + tyrky so * Il y = 11 tyll = P atty) < llate yll = llall +2e Re (x,y) tt Ily in 2+ Re (acy) t lleis > 0 > Re (214) > - * 720 ak Re (ny) s Thus taking llei ttco to we get that Re (x,y)=0

now if <x,y) is beal then <x, y) = Re <alazzo So we are done. But if <a,y) complex then kx,y>l = eo Re (x, y) H = Relacedy) so we can start with xt ker(P) édy & P(X) and get the above besuet. le l<214) ) = 0 > <acy) = 0 # xe ker(p), YG PCV) So we have shown that V= ker(P) PCV) and <x,y)=0 taeker(p) YEP(V)

Therefore le u= P(u). P(V)