How to show that if $x^T x = 0$, then $x=0$ for any column vector $x$?

Question

I did the proof and got it wrong, it'd be really helpful if someone could point me in the right direction or show me how to do it. The $T$ in $(x)^Tx=0$ means transpose; I wasn't sure if that was clear in the title.

Answer 1

Let $$x=\left( \begin{matrix} x_1 \\ x_2 \\ . \\ .\\ .\\ x_n \end{matrix}\ \right). $$ Then $x^Tx$ = $x_1^2+x_2^2+...+x_n^2$.

Since for all real $x_i$, $x_i^2 = 0$ if $x_i=0$ and $x_i^2>0$ otherwise,

$x^Tx=0$ implies $x_i=0$ for all $i$, i.e., $x=\mathbf 0.$