From $\langle v,u_0\rangle=\langle v,u_1\rangle$ for all $v\in V$, show that $u_0=u_1.$

Question

$u_0$ and $u_1$ are vectors in the inner product space V with $\langle u_0,v\rangle=\langle u_1,v\rangle$ for all $v\in V$.

Show that $u_0=u_1$ (let $v=u_0-u_1$).

I'm would be glad to have a little pointer.

EDIT:

I get:

$\langle u_0-u_1,v=0\rangle$. If we let $v=u_0-u_1$:

$\langle v,v\rangle=0 \Rightarrow \vec{v}=\vec{0}$ and so $\vec{u_0}=\vec{u_1}=\vec{0}$

Is this correct?

Answer 1

Since $\langle u_0, v \rangle = \langle u_1,v \rangle$ we have $\langle u_0 - u_1, v\rangle = 0 $ for all $v \in V$. Now we can take $v = u_{0} - u_{1}$.

Answer 2

Hints:

$$\begin{align*}\langle u_0,v\rangle&=\langle u_1,v\rangle\iff \langle u_0-u_1,v\rangle=0\;\;\forall\,v\in V\\ \langle u,u\rangle&=0\iff u=0\end{align*}$$