Let S be any subspace of C^I. Show that if Q is invertible and Q(S) = S then Q^-1(S) = S.?

Question

C^I contains all vectors w for which the complex conjugate transpose of A * w = 0.

Q is a Hermitian matrix.

Hint: If Qt=Qs, then t=s

I was just explaining what C^I is... let me reword it... it's a subspace containing all vectors w for which A*w = 0
A* is the complex conjugate transpose of A

the nullspace of A* is the subspace C^I, containing all the vectors w for which A*w = 0

Q is Hermitian, which means it is equal to it's complex conjugate transpose. Q = Q*.

What you wrote (Modulo) may work... but it doesn't talk about s^perpendicular though... so i'm not so sure if that could be used... i'm wondering where the prof was going with the hint that was given...

Answer 1

I don't see where this A or A* enters in. I am right to just treat S as a subspace and not worry about where it comes from?

Hermitian means self adjoint, so that Q=Q*

That hint simply means that Q is injective, which you already know because it's invertible.

All right, here goes. You're showing that S is an invariant subspace under Q and Q^-1.

Let
v e S, w e S^perp
Qv e S
=0=
=
,
here is where the A enters, I think. Since this inner product is zero, and we want to show that Q^-1v e S then somehow we must work in Qw e S^perp. I think that that's where the nature of A and S must enter in.


Maybe
v e S, w e S^perp
=0= -> Qw e S^perp
returning to the last line above:
=0
-> Q^-1v e S
That should work, dont' you think?

Answer 2

Well, your question is confusedly phrased.
So I hereby blame any error on you. :
But if QS=S for all S,
then Q must be the identity matrix. (It is well known
in group theory there can only be one left-identity...)
Hence Q^-1 is also the identity.