Why does $(A^T x) · y = x · (A y)$ hold?

Why does $(A^T x)· y = x ·(A y)$ hold?

The proof has to do with properties of transposes. I did a proof using coordinates (which was correct) but there is an infinitely easier way to do it.

A is an n by n matrix.

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If $x$ and $y$ are column vectors, then $x$ and $Ay$ are column vectors, so what does $x(Ay)$ mean? – Gerry Myerson Mar 15 '13 at 3:47
I edited the post. – iuppiter Mar 15 '13 at 3:48
Assuming the inner product is $u\cdot v=u^Tv$, just expand out $(A^Tx)^Ty$ and $x^T(Ay)$ and compare. – Rahul Mar 15 '13 at 4:18
This looks good but why does u⋅v=u^Tv? What property is this? – iuppiter Mar 15 '13 at 4:25
@orbis The usual dot product on $\mathbb{R}^n$ is defined by $\mathbf{u}\cdot\mathbf{v}=\sum_{i=1}^nu_iv_i$, which is precisely the result of $\mathbf{u}^T\mathbf{v}$ (I assume you know how to multiply two matrices). – user1551 Mar 15 '13 at 5:40

It doesn't...and I think there is some inner product and there're some assumptions on $\,A\,$, since if we denote by $\,\langle\,x,y\,\rangle\,$ the inner product of $\,x,y\,$ , then
$$\langle\,Ax\,,\,y\,\rangle=\langle\,x\,,\,A^*y\,\rangle$$
where $\,A^*\,$ is the adjoint of $\,A\,$ so if this is what is meant in your link and if $\, A^*=A^t\,$ then we're done (for example, if the linear space is real and we're working with an orthonormal basis...)
It doesn't matter, @orbis, since $\,(A^t)^t=A\,$ ... – DonAntonio Mar 15 '13 at 3:51