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
$$x^TAy = y^TAx$$
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5 Answers
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$x^tAy$ is a scalar. So $x^tAy=(x^tAy)^t$. Can you continue from here?
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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.
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2$\begingroup$ This reply is underrated imho, this is the most accurate proof, since it involves directly the definition of quadratic form. $\endgroup$– Vexx23Oct 27, 2016 at 12:59
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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)
answered Sep 20, 2010 at 15:48
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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?
answered Sep 20, 2010 at 17:04
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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.