If inner product of a vector and a transformation is zero, prove T(v) = 0

Question

For $T: V \rightarrow V$ being a linear mapping, are these valid steps and a valid conclusion?

Given that $\langle \textbf{u} | T(\textbf{v}) \rangle$ = 0 for all $u,v \in V$.
$$ \text{if what we're given is true then} \\ \langle T(\textbf{v}) | \textbf{u} \rangle = 0 = \langle \textbf{u} | T(\textbf{v})\rangle \\ \text{By linearity,} \langle T(\textbf{v})- \textbf{u} | T(\textbf{v}) - \textbf{u} \rangle = 0 \\ \text{hence,} T(\textbf{v}) = \textbf{u}$$.

Are these steps valid given our original statement?

How would we show, given our original statement, that this is only true IF $T(v) = 0$? \ \

Edit: Can someone check if this is correct as well? I can't see why it's wrong... $$ \langle u | T(v) \rangle = 0 \\ \text{add to both sides} \quad \langle u | T(v) \rangle + \langle T(v) | T(v) \rangle = \langle T(v) | T(v) \rangle \\ \langle u + T(v) | T(v) \rangle = \langle T(v) | T(v) \rangle \\ \therefore u + T(v) = T(v) \qquad \text{by the uniqueness property} \\ u = 0 \qquad (???)$$

Answer 1

I think you're misusing the (bi)linearity of the inner product, as I said in my comment.

Here is my solution:

Let $v \in V$.
We have: $T(v) \in V$ ($T$ is an endomorphism), so: $\langle T(v) | T(v) \rangle = 0$ (property of $T$) then: $T(v) = 0$ (the inner product is definite).

We've just proved that:If $T$ has the given property then $\forall v \in V, T(v) = 0$.
The converse is trivial.