Originally Posted by

**Mollier** __Problem statement:__

The Pythagorean theorem asserts that for a set of $\displaystyle n$ orthogonal vectors $\displaystyle {x_i}$,

$\displaystyle ||\sum^{n}_{i=1}x_i||^2 = \sum^{n}_{i=1}||x_i||^2$

(a) Prove this in the case $\displaystyle n=2$ by an explicit computation of $\displaystyle ||x_1 + x_2||^2$.

(b) Show that this computation also establishes the general case by induction.

-------------------------------------------------------------------------

__Attempt at solution:__

(a)

$\displaystyle ||x_1+x_2||^2 = ||x_1||^2+||x_2||^2$

$\displaystyle x^{T}_{1}x_1+2x^{T}_{1}x_2+x^{T}_2x_2 = x^{T}_{1}x_1 + x^{T}_{2}x_2$

Since the dot product of two orthogonal vectors is zero, we have:

$\displaystyle x^{T}_{1}x_1+x^{T}_2x_2 = x^{T}_{1}x_1 + x^{T}_{2}x_2$

(b)

Basis step,$\displaystyle n=1$: $\displaystyle ||x_1||^2 = ||x_1||^2$

I assume that this holds up to $\displaystyle k$ for $\displaystyle 1 \leq k < n$.

Now I show that it holds for $\displaystyle k+1$:

$\displaystyle ||x_1+x_2+\cdots+x_k+x_{k+1}||^2=||x_1||^2+||x_2|| ^2+\cdots+||x_k||^2+||x_{k+1}||^2$

By orthogonality $\displaystyle x_ix_j=0$ for $\displaystyle i \neq j$:

$\displaystyle x^{T}_{1}x_1 + x^{T}_{2}x_2 + \cdots + x^{T}_{k}x_k + x^{T}_{k+1}x_{k+1} = x^{T}_{1}x_1 + x^{T}_{2}x_2 + \cdots + x^{T}_{k}x_k + x^{T}_{k+1}x_{k+1}$

Since the statement is true for n=1, n=k and n=k+1, it is also true for n.