I have attempted this problem, and feel uneasy that my logic is not correct in my proof. Here is the question:
Let $V$ be a vector space over $\mathbb{R}$, and let $W$ be an inner product space over $\mathbb{R}$ with inner product $\langle \cdot,\cdot\rangle$. If $T:V \rightarrow W$ is linear, prove that $\langle x , y\rangle' = \langle T(x), T(y) \rangle$ defines an inner product on $V$ if and only if $T$ is one-to-one.
Here is my attempt at the proof.
$(\implies)$
Suppose $\langle x , y\rangle' = \langle T(x),T(y)\rangle$ defines an inner product on $V$, then $V$ is an inner product space, so if $\langle x,y \rangle' = \langle x,z \rangle'$ for all $x$, then $y=z$, and similarly $\langle T(x), T(y) \rangle = \langle T(x), T(z)\rangle $ and $T(y)=T(z)$. Therefore $T$ is one-to-one.
$(\impliedby)$
Suppose $T$ is one-to-one, and consider $T(x),T(y),T(z) \in W$.
Since $W$ is an inner product space, if $\langle T(x), T(y) \rangle = \langle T(x), T(z) \rangle$ for all $T(x)$, then $T(y)=T(z)$ which implies $y=z$, since $T$ is one-to-one.
Implies $\langle x,y \rangle' = \langle x,z \rangle'$
implies $y=z$.
If my proof is painfully incorrect, hints are greatly appreciated.
Best Answer
$(\Rightarrow)$ Let $T(x)=T(y)$
then $T(x)-T(y) =0$
which gives $T(x-y)=0$................by linearity
Now $\langle x-y , x-y\rangle' = \langle T(x-y), T(x-y) \rangle =0 $
hence $x-y=0$ i.e. $x=y$
Hence $T$ is one-one.
$(\Leftarrow)$ Now $\langle x , x\rangle' = \langle T(x), T(x) \rangle \geq 0$
Other linear properties of inner product satisfy because of linearity of $T$.
Only part remains to check is,
Let $x\in V$ s.t $\langle x , x\rangle' = 0$
then by definition of our inner product $ \langle T(x), T(x) \rangle =0$.
which gives $T(x)=0$
but as $T$ is one-one $x=0$.
Hence $\langle , \rangle'$ defines inner product on $V$.