How do you prove that T and U are the same linear transformation on an inner product space V?

Question

Is it enough to show that $<T(x),y>$ = $<U(x),y>$ for any x and y in V?

Answer 1

Yes. This implies $$ \langle T(x)-U(x),y\rangle=0 $$ for all $x$ and $y$. Now, suppose $\langle v,y\rangle=0$ for all $y$. Then, in particular, $\langle v,v\rangle=0$, so $v=0$.

Thus you have $T(x)-U(x)=0$ for all $x$ and…

Answer 2

If you have a finite-dimensional space, you can choose bases to represent $T,U$ as matrices. Then $T,U$ are the same transformation iff there is some matrix $S$ with $STS^{-1}=U$. This last just tells you that $T,U$ are the same matrix represented in different bases.

An approach similar to that of egreg is this fix a non-zero vector $y$: if $<(T-U)(x),y>$=0 for all $x$, then the entire space is the orthogonal complement of $y$. This can only happen if $(T-U)(x)=0 \forall x, i.e., T(x)=U(x) \forall x \in V$