# Connection between even/odd and symmetric/skew symmetric

I read awhile back that the set of continuous real valued functions from $\mathbb{R} \to \mathbb{R}$ has a direct sum decomposition into subspaces of strictly even and odd functions. Any such function $f$ could then be uniquely written in terms of even and odd parts by the formulas $\text{even}(x)=\dfrac{f(x)+f(-x)}{2}$ and $\text{odd}(x)=\dfrac{f(x)-f(-x)}{2}$ where $\text{even}(x)+\text{odd}(x)=f(x)$.

Recently I stumbled into a similar claim for $n\times n$ matrices, but instead of even and odd, the set of square matrices has a direct sum representation in terms of symmetric and skew symmetric parts. So given any square matrix $A$ we can write $B=\dfrac{A+A^T}{2}$ and $C=\dfrac{A-A^T}{2}$ as the symmetric and skew symmetric representations where $A=B+C$.

Does this mean that the notion of symmetric and skew symmetric have a connection to the notion of even and odd? Do they represent a generalization of even/odd, or is there another idea that generalizes these two ideas?

• You are very observant and this is something I had not thought of before! The connection is what happens when you change the sign of $x$ in $f(x)$. If it is even, there is no sign change and if it is odd the sign changes. Now in the case of matrices, what happens if you interchange $x$ and $y$ in $x^T M y$? If M is symmetric, then no change. If it is skew symmetric, the sign changes. Nice insight. I am going to use it! – user44197 Dec 31 '13 at 21:32

Let $V$ be any real vector space, and let $T$ be any operator on $V$ with $T$ not the identity but $T^2$ the identity. Given any $v$ in $V$, we have $v=u+w$ with $T(u)=u$ and $T(w)=-w$, namely, $u=(v+T(v))/2$ and $w=(v-T(v))/2$.