Proof for pythagoras theorem

Question

Let $f,g$ orthogonals to each other.

$${\left\| {f + g} \right\|^2} = \left<f,f\right>+\left<g,f\right>+\left<f,g\right>+\left<g,g\right> = {\left\| f \right\|^2} + {\left\| g \right\|^2}$$

Can you please explain the second and third equalities (essentially the whole proof)?

Answer 1

The scalar product has is linear in both arguments so:

$$||f+g||^2 = \langle f+g , f+g\rangle = \langle f,f+g\rangle+\langle g ,f+g\rangle = \langle f,f\rangle+\langle g ,f\rangle + \langle f,g\rangle+\langle g ,g\rangle = (*)$$

$f$ and $g$ are orthogonal $\iff$ $\langle f,g \rangle = 0$ Because $\langle f,g \rangle = 0 = \langle g,f \rangle$ we can follow:

$$(*) = \langle f,f\rangle + \langle g ,g\rangle = ||f||^2 + ||g||^2$$

Answer 2

The key is to note that, by definition: $$ \| x\|^2= \langle x,x\rangle $$