How do we prove by contradiction that, if $\langle x,y\rangle = \langle x,z\rangle$ for all $x$, then $y=z$?

Question

I must prove the statement below, I believe that I can do this by proof by contradiction. But am I not 100% sure how the contradiction statement should be.

Question Let $\langle \cdot ,\cdot \rangle$ be an inner product. Show that if $\langle x,y\rangle =\langle x,z\rangle$ for all $x$ then $y=z$

My attempt to provide a contradiction statement to prove incorrect:

Let $\langle x,y\rangle =\langle x,z\rangle$ for all $x$ be true but suppose $y\neq z$.

If I did it right then I can pick a value of $x$ that I can use to show why this contradiction will not work, hence proving that $y=z$.

Answer 1

Suppose that $\langle x,y\rangle = \langle x,z\rangle$ for every $x\in V$, where $y\in V$ and $z\in V$ are considered to be distinct. Then we can take $x = y - z$, apply the linearity of the inner product and its positive definiteness in order to conclude that: \begin{align*} \langle x,y\rangle = \langle x,z\rangle & \Rightarrow \langle x,y\rangle - \langle x,z\rangle = 0\\\\ & \Rightarrow \langle x, y - z\rangle = 0\\\\ & \Rightarrow \langle y - z, y - z\rangle = 0\\\\ & \Rightarrow y - z = 0 \Rightarrow y = z. \end{align*} which contradicts the assumption that $y\neq z$, and we are done.

Hopefully this helps!

Answer 2

$\langle x,y\rangle = \langle x,z\rangle$

Then $\langle x,y-z\rangle =0$

since this is true for all $x\in V$ , let $x=y-z$

Then $\langle y-z,y-z\rangle = 0$

Now by definitenes of inner product, we have $y-z=0$ i.e $y=z$