Inner product space - simple proof

Question

This seems like a simple proof but I haven't been able to do it.

Problem: Suppose $V$ is an inner-product space with $u, v \in V$, with $v \neq 0$. Provided that $u_{perp} = u - u_{par}$ and that $u_{par} = \frac{<u, v>}{||v||^2}v$, prove that $<u_{perp}, u_{par}> = 0$.

My attempt:

Note that $<u_{perp}, u_{par}>$ = $<u - u_{par}, u_{par}>$ = $<u, u_{par}> - <u_{par}, u_{par}>$ = $<u, u_{par}>$

In a similar fashion, $<u_{perp}, u_{par}>$ = $<u_{perp}, u - u_{perp}>$ = $<u_{perp}, u> - <u_{perp}, u_{perp}>$ = $<u_{perp}, u>$

Therefore: $<u_{perp}, u_{par}>$ = $<u, u_{par}>$ = $<u_{perp}, u>$

$$\implies <u, u_{par}> = <u_{perp}, u>$$

$$\implies <u, u_{par}> - <u_{perp}, u> = 0$$

$$\implies <u - u_{perp}, u_{par} - u> = 0$$

$$\implies <u_{parr}, -u_{perp}> = 0$$

$$\implies -<u_{parr}, u_{perp}> = 0$$

$$\implies <u_{parr}, u_{perp}> = 0$$

$$\implies -<u_{perp}, u_{parr}> = 0$$

$$\implies <u_{perp}, u_{parr}> = 0$$

QED.

I can't help but think that I'm using wrong reasoning or my proof has some flaw. In fact, I'm pretty sure there is a flaw when expanding the inner products. In another attempt, I tried to use the fact that $||u_{perp} + u_{par}||^2 = ||u_{perp}||^2 + ||u_{par}||^2$ implies orthogonality, but the RHS here is $||u||^2$ whereas the LHS after expanding is $2<u_{par}, u_{par}> - 2<u, u_{par}> + ||u||^2$, which I'm unable to simplify further.

Any assistance or even hints are much appreciated.

Answer 1

$\langle u_{perp},u_{par} \rangle = \langle (u - \frac{\langle u,v \rangle}{\| v \|^{2}}\cdot v),\frac{\langle u,v \rangle}{\| v \|^{2}}\cdot v \rangle = \langle u,\frac{\langle u,v \rangle}{\| v \|^{2}}\cdot v \rangle - \langle \frac{\langle u,v \rangle}{\| v \|^{2}}\cdot v, \frac{\langle u,v \rangle}{\| v \|^{2}}\cdot v \rangle = \frac{\langle u,v \rangle}{\| v \|^{2}}\langle u,v \rangle - \frac{\langle u,v \rangle^{2}}{\| v \|^{4}}\langle v,v \rangle = \frac{\langle u,v \rangle}{\| v \|^{2}}\langle u,v \rangle - \frac{\langle u,v \rangle^{2}}{\| v \|^{4}}\|v\|^{2} = \frac{\langle u,v \rangle^{2}}{\| v \|^{2}} - \frac{\langle u,v \rangle^{2}}{\| v \|^{2}} = 0$