Inner Product Space on linear transformation on itself So $V$ is an inner product space and $T : V \to V$ is a linear map such that $$||T(v)|| = ||v||$$ for all $v \in V$. Prove that $$\langle T(v), T(w)\rangle = \langle v, w\rangle$$ for all $v,w \in V$.
However, I missed the class where they talked about this, and reading up in my book hasn't been completely helpful. I don't know what my first step should be in trying to prove this. 
 A: $$\langle v,v\rangle +\langle w,w \rangle +2\langle v,w \rangle=
 \langle (v+w),(v+w) \rangle $$
$$=\langle T(v+w),T(v+w) \rangle =\langle T(v),T(v) \rangle+2\langle T(v),T(w) \rangle  +\langle T(w),T(w) \rangle $$
So $$2\langle v,w \rangle=
 2\langle T(v),T(w) \rangle  $$
A: To solve this problem you will have to use linearity. This is pretty much how all basic linear algebra problems go. When in doubt, just mess around with what you're given.
To summarise what you might've missed (this is in fact most of what you need to know), for any scalars $\alpha,\beta \in \mathbb{R}$ (could be complex, but then you have to add a conjugate in one of the entries of the inner product -- a so-called "sesquilinear form"):
$$<\alpha v + w, u + \beta s> = \alpha <v, u> + \alpha\beta < v, s> + <w, u> + \beta < w, s>.$$
So the only thing you know is that $\|v\|^2 = <v, v> = <Tv, Tv>$. So let's use that on something related to what we need ($<v,w>$). Try 
\begin{align*}
\|v+w\|^2 &= \|T(v+w)\|^2 \\
&= <T(v+w), T(v+w)> \tag{by definition} \\
&= <Tv + Tw, Tv + Tw> \tag{by linearity of $T$} \\
&= <Tv,Tv> + <Tw, Tw> + 2<Tv, Tw>
\end{align*}
Now what happens if you expand $\|v+w\|^2 = <v+w, v+w>$ (hint: cancel stuff).
Note that if you were working in $\mathbb{R}^n$, you could think of this as
\begin{gather*}
<Tv,Tv> = (Tv)^t (Tw) = v^t T^t T v = v^t v,
\end{gather*}
so $T^t T = I$ ($T$ is orthogonal). $^t$ denotes the transpose.
A: Hint: $\langle T(v+w),T(v+w) \rangle=\langle v+w,v+w \rangle$.
A: $$||v+w||^{2} =||T(v+w)||^{2} = ||T(v)+ T(w)||^{2} = <T(v) + T(w), T(v) + T(w)> $$
$$ = \space <T(v), T(v)> + <T(v), T(w)> + <T(w), T(v)> + <T(w), T(w)>$$
$$ = \space ||T(v)||^{2} + 2\cdot Re(<T(v),T(w)>) +||T(w)||^{2} $$
$$ = \space ||v||^{2} + 2 \cdot Re(<T(v), T(w)>) + ||w||^{2} $$
Hence, $$||v||^{2} + 2\cdot Re(<v,w>) + ||w||^2 = ||v||^{2} + 2\cdot Re(<T(v), T(w)> + ||w||^{2} $$ implies $$Re(<v,w>)=Re<T(v),T(w)>$$
Now, this is equivalent to saying $$<v,w> = <T(v), T(w)>$$ if the inner product space is over R.
