# How do I prove that $x^TAy = y^TAx$ if A is symmetric?

Ok this is for a HW but I'm not looking for a handout...just a hint to get me on the right track.

I have no idea where to begin proving this:

Show that if A is a symmetric matrix, then

$$x^TAy = y^TAx$$

$x^tAy$ is a scalar. So $x^tAy=(x^tAy)^t$. Can you continue from here?

If $A$ is symmetric then we know that $A_{ij} = A_{ji}$. If you understand that $x^T A y$ = $\sum_i\sum_j x_iA_{ij}y_j$ , then swapping the indices of $A$ should directly lead you to the answer.

• This reply is underrated imho, this is the most accurate proof, since it involves directly the definition of quadratic form. Oct 27, 2016 at 12:59

Prove that whenever $A$ and $B$ are matrices for which you can compute the product $AB$, then $$(AB)^t=B^tA^t$$.

Next apply $(\mathord-)^t$ to the left hand side of your equation, and compare the result to the right hand side, keeping in mind that both sides are actually numbers (well, $1$-by-$1$ matrices)

This condition can actually be used as an equivalent definition of $A$ being symmetric. The key point is that both sides are bilinear in $x$ and $y$, so it suffices to prove the result when $x$ and $y$ are basis vectors, say $e_i$ and $e_j$. What does the condition say then?

compute $[x_1, x_2]A[y_1 y_2]^T$ for a 2 x 2 symmetric matrix $A = [[a, b], [b, c]]$. you get $ax_1y_1 + b(x_1y_2 + x_2y_1) + cx_2y_2$. this expression is clearly symmetric in $x$ and $y$ i.e., does not change when $x$ and $y$ are exchanged.

for an n x n matrix you may want to use sigma notation.