# Math Help - Inner product space and linear transformation proof

1. ## Inner product space and linear transformation proof

Hi guys.
I need to prove the following:

Let $T:V\rightarrow V$ be a linear transformation.
Let $ImgT$ be the span of a set $B$, and $V$ be the span of a set $S$.
I need to prove that if for every $u\in S,v\in B:=0$ then $T=0$.

I'm not sure where to start, I need some guidance please.

2. ## Re: Inner product space and linear transformation proof

Hey Stormey.

For your problem can you show that if u is non-zero (in general) then since <u,v> = 0 implies u = 0 or v = 0, then if v is not necessarily zero then u = 0.

if u = Tx and x is not necessarily zero, then you need to show T = 0 but for all x.

So the proof now becomes showing that T = 0 satisfies Tx = 0 for all valid vectors in x. You could try proof by contradiction by assuming that T is non-zero and find a value for some realization of x' where Tx = 0 but Tx' != 0 for x' != x. This will prove the statement that T must be zero in order for Tx = 0 for all valid x in the vector space.

3. ## Re: Inner product space and linear transformation proof

Hi chiro. thanks for the help.

Originally Posted by chiro
Hey Stormey.

For your problem can you show that if u is non-zero (in general) then since <u,v> = 0 implies u = 0 or v = 0, then if v is not necessarily zero then u = 0.
but that just means that u and v are orthogonal, it can't tell that either is 0.

Originally Posted by chiro
if u = Tx and x is not necessarily zero, then you need to show T = 0 but for all x.

So the proof now becomes showing that T = 0 satisfies Tx = 0 for all valid vectors in x. You could try proof by contradiction by assuming that T is non-zero and find a value for some realization of x' where Tx = 0 but Tx' != 0 for x' != x. This will prove the statement that T must be zero in order for Tx = 0 for all valid x in the vector space.
what's 'x' here? vector in V? in S?

4. ## Re: Inner product space and linear transformation proof

When I say <u,v> = 0 I mean when this holds for all possible v where v can vary. I should have been more clear on that.

Also x is in set S.

OK, thanks!