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## 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 [itex]T[/itex] of an inner product space is [itex]{0}[/itex] if and only if [itex]\langle b|T|a\rangle = 0[/itex] for all [itex]|a\rangle[/itex] and [itex]|b\rangle[/itex].

Now it is obvious if [itex]T[/itex] is [itex]0[/itex] then [itex]\langle b|T|a\rangle = 0[/itex]

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

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

The statement is as follows:

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

Now it is obvious if [itex]T[/itex] is [itex]0[/itex] then [itex]\langle b|T|a\rangle = 0[/itex]

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

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