Confused about choice of vector in a proof.

Main Question or Discussion Point

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 $|b\rangle = T|a\rangle$.

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fzero
Homework Helper
Gold Member
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.

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CompuChip
Homework Helper
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|.

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