Prove A is symmetric matrix iff $A$ is square and $x^T Ay = (Ax)^T y$ Prove A is a symmetric matrix iff $A$ is square and $x^T Ay = (Ax)^T y$. (for all $x,y \in \mathbb{R}^n$)
Going from the assumption that it is symmetric to the two conditions is fairly straightforward.
However, going the other way, I am stuck at proving $A^T = A$ using the second and condition, being stuck at $X^T (A-A^T)y=0$.
Note T is for transpose!
 A: The key part, alas missing, from $x^TAy=(Ax)^Ty$ is that this holds for all vectors $x,y$.  If $A-A^T$ had a nonzero entry, you could find an $x,y$ that would extract that entry.  
Hint: try standard basis vectors.
Double Hint: If $x=e_i, y=e_j$, then $x^TAy=(A)_{i,j}$.
A: First, lets remember the rules of transposing the product: $(AB)^T = B^T A^T$
Using this, we start with the giver equation $ x^TAy=(Ax)^Ty$ and apply the rule above which yields 
$x^TAy = x^TA^Ty$.
Note that the equation is true for all $x, y$, so the only way that is possible is if $A = A^T$.
A: Since you already accepted an answer I can show a more abstract method.
Consider $B_x=x^TA-(Ax)^T$, which is a row and so defines a linear map
$$\mathbb{R}^n\to\mathbb{R}$$
that, under your hypothesis, is the zero map. This means that $B_x=0$ for all $x$, because the rank of $B_x$ is zero. Now
$$B_x=x^TA-(Ax)^T=x^TA-x^TA^T=x^T(A-A^T)$$
so
$$B_x^T=(A-A^T)^Tx=(A^T-A)x$$
is the zero matrix for all $x$. Hence the linear map $\mathbb{R}^n\to\mathbb{R}^n$ associated to $C=A^T-A$ (the usual $x\mapsto Cx$) is the zero map and therefore $C$ has rank zero, so $C=0$.
