If $A$ is a square matrix, then $A=P+Q$ for $P$ symmetric, $Q$ antisymmetic. This is what's to be shown.

If $A$ is a square matrix, then $A$ can be taken apart uniquely into two matrices $P+Q$, such that $P$ is symmetric and $Q$ is antisymmetric.

This is my work so far:
$A$ can always be taken apart into some $P+Q$.
Since $Q$ is to be antisymmetric, it must follow that $Q+Q^t = 0$. Since $Q$ and $Q^t$ have the same diagonal entries, all of them must be $0$. So $P$ must have the same diagonal entries as $A$. This is all good, but I'm stuck here. I'm trying to work my way backwards, trying to see how the elements in $P$ and $Q$ must be related to each other, but I just cant seem to find any rhyme or reason in the elements off the diagonal. 
Intuition tells me that if $P$'s elements are the same as $A$, and $Q$'s are all $0$, then this should be enough to prove uniqueness. Any tips?
I don't think this should be too hard, so it's kinda frustrating getting stuck! :(
This is not homework, or classwork of any kind, this is for my own pleasure.

Hints:
If $A=P+Q$ with $P$ symmetric and $Q$ antisymmetric, then $A^\top=P^\top+Q^\top=P-Q$.
Solve for $P$ in terms of $A$, then solve for $Q$ in terms of $A$.

The only blip to remember with this reasoning is that it doesn't work in a fields of characteristic $2$ (where $2$ isn't a unit.) In those cases, "symmetric" and "antisymmetric" mean the same thing, and this decomposition can't happen.
