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Let $A \in M_n(\mathbb{R})$ and let $B=A^TA \in M_n(\mathbb{R})$.

Prove that if A is invertible then B is positive definite.

So far I've done the following but I can't see how it helps. (https://i.stack.imgur.com/DXWBb.jpg)

Any help would be very much appreciated.

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    Well, $Q(x) >0$ unless $x=0$: isn't this the definition of positive definite? – 2017-01-10
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    @Thomas: you did it. – 2017-01-10
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    But if X was the zero vector then it would be positive semi definite not positive definite? – 2017-01-10
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    By definition, positive definite means that $x^TBx>0$ for all $x\ne 0$. You showed that $x^TBx=\|Ax\|^2$. Since $A$ is invertible, $Ax\ne0$ whenever $x\ne 0$, and hence $\|Ax\|^2\ne 0$ as well, which implies $\|Ax\|^2>0$. – 2017-01-10

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$B$ is positive definet if for every non zero $\mathbf v, \mathbf v^T B\mathbf v > 0$

$B = A^T A$

$\mathbf v^T B\mathbf v = \mathbf v^T A^T A \mathbf v = (A\mathbf v)^T(A\mathbf v)= \|A\mathbf v\|$

$\|A\mathbf v\| \ge 0$

If A is invertable, $\|A\mathbf v\| = 0$ only when $\mathbf v = \mathbf 0$