Confused about choice of vector in a proof.

[Luna=Luna]

This is probably going to be a very simple question, i just need justification for a seemingly simple step in a proof.

The statement is as follows:

An endomorphism T of an inner product space is {0} if and only if \langle b|T|a\rangle = 0 for all |a\rangle and |b\rangle.

Now it is obvious if T is 0 then \langle b|T|a\rangle = 0

For the converse proof if \langle b|T|a\rangle = 0 for all |a\rangle and b\rangle then T = 0, it starts by choosing |b\rangle = T|a\rangle.

Why is this valid, i guess a very naive reasoning would be doesn't this only prove it for the case that |b\rangle = T|a\rangle.

 

[fzero]

For the converse proof if \langle b|T|a\rangle = 0 for all |a\rangle and b\rangle then T = 0, it starts by choosing |b\rangle = T|a\rangle.

Why is this valid, i guess a very naive reasoning would be doesn't this only prove it for the case that |b\rangle = T|a\rangle.

In the converse we're already given that \langle b|T|a\rangle = 0 for all |a\rangle and b\rangle. We don't have to prove anything for all $|b\rangle$. We just consider $T|a\rangle$ and find that its norm must be zero for all $|a\rangle$. Therefore $T|a\rangle$ is the zero vector for all $|a\rangle$, hence $T$ is the zero map.

 

[CompuChip]

If the assumption in a proof is a "for all x in X" statement, you are always allowed to set x to any value you like. Often we are quite happy to take any allowed value ("let $x \in X$ arbitrary") as long as we know that it satisfies some condition, but you can always use that since the statement holds for all x in X, it holds for some particular convenient value.

Of course, if you are trying to prove a statement of the form "for all x in X", it is not sufficient to prove the statement for a single particular x. For example, you wrote

Now it is obvious if T is 0 then ⟨b|T|a⟩=0

Actually what you meant is

Let |a⟩ and ⟨b| be arbitrary. Now it is obvious if T is 0 then ⟨b|T|a⟩=0.

It would have been wrong to write

Let |a⟩ = 0 and ⟨b| = ⟨a|. Now it is obvious if T is 0 then ⟨b|T|a⟩=0.

Since that would not have been a statement about all |a⟩ and ⟨b|.