Verify Pythagorean Law for Normed Linear Space / Inner Product Space

Question

Given x $ = (1,1,1,1)^T$ and y $ = (8,2,2,0)^T$

1. Find vector projection p of x onto y

2. Compute $||$x$-$p$||_2$ , $||$p$||_2$ , $||$x$||_2$ (which are 2-Norms) and verify that the Pythagorean Law is satisfied

1. I know how to do part 1, and I found and confirmed p $=(\frac{4}{3}, \frac{1}{3}, \frac{1}{3}, 0)$

2. For part 2, I got

   $||$x$-$p$||_2 = \sqrt{2}$

   $||$p$||_2 = \sqrt{2}$

   $||$x$||_2 = 2$

   I don't quite understand how to verify the Pythagorean Law when it is only valid if x⊥y. We can check x and y are not orthogonal by performing the standard inner product of <x,y> which does not equal $0$.

Answer 1

Since $$p=\frac{\langle x,y\rangle}{\|y\|^2}y$$ we have \begin{align}\langle x-p, p\rangle&=\langle x,p \rangle -\|p\|^2 \\ &= \frac{\langle x,y\rangle}{\|y\|^2}\langle x,y \rangle -\frac{\langle x,y\rangle^2}{\|y\|^4}\|y\|^2\\ &=\frac{\langle x,y\rangle^2}{\|y\|^2} -\frac{\langle x,y\rangle^2}{\|y\|^2}\\ &=0\end{align}

Now you can verify Pythagorean law for $x-p$ and $p$.