How to show that a valid inner product on V is defined with the formula $[x, y] = \langle Ax, Ay\rangle$?

Let $A \in L(V,W)$ be an injection and $W$ an inner product space with the inner product $\langle \cdot,\cdot\rangle$. Prove that a valid inner product on $V$ is defined with the formula $[x, y] = \langle Ax, Ay\rangle$

$L(V, W)$ = The set of all linear mappings (linear operators) from V to W

To prove this, if I am correct, I need to show that the four properties of an inner products space apply on this formula:
1. $\langle x, y \rangle = \overline{\langle y,x\rangle }$
2. $\langle \alpha x, y\rangle = \alpha\langle x,y\rangle$
3. $\langle x+y,z\rangle = \langle x,z\rangle + \langle y,z\rangle$
4. $\langle x,x\rangle \space \ge 0 \space \space \forall x$
4.' $\langle x,x\rangle \space = 0 \Longleftrightarrow x=0$

4.

$[x, x] = \langle Ax, Ax\rangle , Ax \in W$, and since $W$ is an inner product space, $\langle Ax, Ax\rangle \space \ge 0$ implies $[x, x] \ge 0$.

4.'

Since A is an injection: $Ax = 0 \implies x = 0$ and since $W$ is an inner product space and $Ax \in W \implies \langle Ax, Ax \rangle = 0 \implies Ax = 0 \implies [x,x] = 0 \Longleftrightarrow x=0$

3.

$[x+y,z] = \langle A(x+y), Az\rangle = A$ is linear $= \langle Ax + Ay, Az\rangle = Ax, Ay, Az \in W$ and $W$ is an inner product space $= \langle Ax, Az\rangle + \langle Ay, Az\rangle = [x,z] + [y,z]$

2.

$[\alpha x, y] = \langle A(\alpha x), Ay\rangle = A$ is linear $= \langle \alpha (Ax), Ay \rangle = W$ is an inner product space $= \alpha \langle Ax, Ay\rangle = \alpha [x, y]$

1.

$[x, y] = \langle Ax, Ay \rangle = W$ is an inner product space $= \overline{\langle Ay, Ax\rangle } = \overline{[y, x]}$

But this seems to me a little to easy, did I maybe conclude something that can't be concluded so easily or maybe is my approach to prove this completely wrong?

• A LaTeX tip: < and > mean "less than" and "greater than", and produce spacing correct for that meaning only. When you want angle brackets, you need to use \langle and \rangle. May 18 '13 at 22:08

Yes, that's it all. So called 'routine verification'. $A$ embeds $V$ into $W$, and $[-,-]$ is just the inherited inner product from $W$, along this embedding.