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?
<and>mean "less than" and "greater than", and produce spacing correct for that meaning only. When you want angle brackets, you need to use\langleand\rangle. – Zev Chonoles May 18 '13 at 22:08