If $x_i \perp x_j$ then show that $x_j$ 's are linearly independent

Question

Let $x_1, x_2, ...,x_n \in \mathbb{H}$, $\mathbb{H}$ is the Hilbert space. $x_j \neq 0$ for every $j$. If $x_i \perp x_j$ for $i \neq j$ then show that $x_j$ 's are linearly independent.

I think I need to use the fact that $<\alpha x + \beta y , z> = \alpha<x,z> + \beta<y,z>$ but how?

Answer 1

Suppose $\;\sum\limits_{i=1}^na_ix_i=0\;$ , then for all $\;1\le k\le n\;$ :

$$\left\langle 0,x_k\right\rangle=\left\langle \sum\limits_{i=1}^na_ix_i\,,\,\,x_k\right\rangle=\sum\limits_{i=1}^na_i\langle x_i,\,x_k\rangle=a_k\langle x_k,\,x_k\rangle$$

End the proof now.

Answer 2

Always try a simpler case first when you are lost. Suppose $n=2$ and write $ax_1+bx_2=0$. Try to do the inner products with $x_1$ and $x_2$ on both sides. Then you see $a=b=0$. Can you generalize the argument?