Prove that $a(u-u_{h},u-u_{h})\ge 0$

Question

Assume that $a$ is bilinear, symmetric and positive definite form, $u\in X$ and $u_{h}\in X_{h}\subset X$. I know the following fact: $$a(u-u_{h},u_{h})=0$$ Frm positive definiteness $a(u-u_{h},u-u_{h})>0$. I am trying to show that the equality applies: $$a(u-u_{h},u-u_{h})= 0$$ I am not able to go further, any hints or suggestions are welcome?

Answer 1

A bilinear, symmetric and positive-definite form is a scalar product: $a(v,v) \geq 0$ for every $v$, and $a(v,v)=0$ if and only if $v=0$.

Of course I am assuming that positive-definite is intended here in the strong sense: usually the condition "$a(v,v) \geq 0$ for every $v$" is called positive semidefinite.

Hence the equality means exactly that $u=u_h$.