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I am looking at a proof for why $A^TA$ is positive semi-definite when $A$ is $n\times n$ and it has this line. $$ v^TAA^Tv = A^Tv \cdot A^Tv ≥ 0. $$ I understand what $v^TAA^Tv$ means and the purpose of proving that it's nonnegative, etc... My problem is that I am a linear algebra novice and do not necessarily understand how the first part $v^TAA^Tv$ is equivalent to $A^Tv \cdot A^Tv$. I know that $a^Tb = a \cdot b$, but something else is going on, no? Appreciate any help!

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By AT, do you mean $A^{T}$, the transpose of $A$? – Isaac Solomon Apr 18 '12 at 6:51
yes sorry, lost my formatting. The T's are all ^T's – user29373 Apr 18 '12 at 6:53

For any column vector $v$, we have $v^tA^tAv=(Av)^t(Av)=(Av)\cdot (Av)\geq 0$, therefore $A^tA$ is positive semi-definite. In particular, if $A$ is a nonsingular matrix, then $A^tA$ is positive definite.

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The other thing that is going on is that $(AB)^t=B^tA^t$, and $(A^t)^t=A$, applied to $A^tv$

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