Let V be an inner product space. If $ x⊥y $, then show that

Question

Let $V$ be an inner product space. If $x_i ⊥ x_j $ when $i\neq j$, then show that $$\Bigg\Vert\sum_{i=0}^n x_i\Bigg\Vert^2\ =\sum_{i=0}^n \Vert x_i \Vert^2. $$

Answer 1

Here is a start, I'll do the special case $n=2$,

$$ ||x_1+x_2||^2=\langle x_1+x_2,x_1+x_2 \rangle= \langle x_1,x_1 \rangle + \langle x_1,x_2 \rangle + \langle x_2,x_1 \rangle + \langle x_2,x_2 \rangle. $$

Now, using the facts

i) $\langle x_1,x_1 \rangle=||x_1||^2$,

ii)$ \langle x_1,x_2 \rangle = \langle x_2,x_1 \rangle =0 $

the desired result follows.

Try to work out the general case.

Added:

$$\Bigg|\Bigg|\sum_{k=1}^{n}x_k \Bigg|\Bigg|^2 = \langle \sum_{k=1}^{n}x_k ,\sum_{m=1}^{n}x_m\rangle = \sum_{k=1}^{n}\sum_{m=1}^{n} \langle x_k, x_m \rangle = \sum_{k=1}^{n} \langle x_k, x_k \rangle =\sum_{k=1}^{n} ||x_k||^2, $$

since $ \langle x_k,x_m \rangle =0,$ for $k\neq m. $

Answer 2

Just use the definition of norm induced by the inner product,and the bilinearity or conjugate bilinearity of inner product