Show that $\Vert v\Vert^2= \langle v,AA' v\rangle \iff AA'=I$

Question

Show that $\Vert v\Vert^2= \langle v,AA' v\rangle \iff AA'=I$.

My thoughts: let $AA'=M$. Then, the above implies $$ \sum^n_{i=1}(1-M_{ii})v_i^2+\sum^n_{i=1}\sum^n_{j\neq i}v_iM_{ij}v_j=0. $$ How would one go about showing $M_{ii}=1$ and $M_{ij}=0$?

Answer 1

Let's prove $\|Av\| = \|v\|$ for all $v \in V$ iff $\langle Av, Aw\rangle = \langle v, w \rangle$ for all $v, w \in V$. This plus the hints should be enough for you to finish it.

If $\langle Av, Aw \rangle = \langle v, w \rangle $for all $v, w \in V$, then set $v = w$ to get $\|Av\|^2 = \langle Av, Av \rangle = \langle v, v \rangle = \|v\|^2$, so $\|Av\| = \|v\|.$

If $\|Av\| = \|v\|$ for all $v \in V$, then $\|A(v+w)\| = \|v+w\|$ for all $v, w \in V$. Notice

$$ \|A(v+w)\|^2= \langle A(v+w), A(v+w) \rangle = \langle Av + Aw, Av + Aw \rangle = \langle Av, Av\rangle + 2\langle Av, Aw \rangle + \langle Aw, Aw \rangle = \|v\|^2 + 2 \langle Av, Aw \rangle + \|w\|^2,$$ $$ \|v+w\|^2 = \langle v+w, v+w \rangle = \|v\|^2 + 2\langle v, w \rangle + \|w\|^2.$$ Setting these equal to each other and doing some algebra, we have

$$ 2 \langle Av, Aw \rangle = 2 \langle v, w \rangle \implies \langle Av, Aw \rangle = \langle v, w \rangle.$$

Edit: I'll also solve your problem in the comments. Fix $y$ and $z$. We claim $\langle x, y \rangle = \langle x, z \rangle$ for all $x$ iff $y = z$. The hard part is the implication, which we do now. If $\langle x, y \rangle = \langle x, z \rangle$, then $\langle x, y- z \rangle = 0$ for all $x$. In particular, $\langle y-z, y-z \rangle = \|y-z\|^2 = 0$. But recall $\|y-z\|^2 = 0$ if and only if $y - z =0$, or $y = z$.