Symmetric and Anti-symmetric Components

 

During various times in my math studies, I came across the following very similar looking equations.
For any complex number $z$, we can split it into a symmetric and antisymmetric part: $$\begin{align*} z & =\underbrace{\frac{z+\bar{z}}{2}}_{=:z_{1}}+\underbrace{\frac{z-\bar{z}}{2}}_{=:z_{2}}\\ \bar{z}_{1} & =z_{1}\\ \bar{z}_{2} & =-z_{2} \end{align*}$$ For any square matrix $A$, we can split it into a symmetric and antisymmetric part: $$\begin{align*} A & =\underbrace{\frac{A+A^{T}}{2}}_{=:A_{1}}+\underbrace{\frac{A-A^{T}}{2}}_{=:A_{2}}\\ A_{1}^{T} & =A_{1}\\ A_{2}^{T} & =-A_{2} \end{align*}$$ For any polynomial $f$, we can split it into an odd and even part: $$\begin{align*} f(x) & =\underbrace{\frac{f(x)+f(-x)}{2}}_{=:f_{1}(x)}+\underbrace{\frac{f(x)-f(-x)}{2}}_{=:f_{2}(x)}\\ f_{1}(-x) & =f_{1}(x)\\ f_{2}(-x) & =-f_{2}(x) \end{align*}$$
This prompts us to ask ourselves, is there a generalization of this?

Consider a vectorspace $V$. Let $g:V\rightarrow V$ be a function such that $$\begin{align*} g(x_{1}+x_{2}) & =g(x_{1})+g(x_{2})\qquad(\text{linearity})\\ g(cx) & =cg(x)\qquad\qquad(\text{linearity})\\ g^{2}(x) & =x \end{align*}$$ Then, given any $x\in V$, it can be split into a $g$-symmetric and $g$-antisymmetric component as follows: $$x=\underbrace{\frac{x+g(x)}{2}}_{=:x_{1}}+\underbrace{\frac{x-g(x)}{2}}_{=:x_{2}}$$ We can verify that $$\begin{align*} g(x_{1}) & =g\left(\frac{x+g(x)}{2}\right)=\frac{g(x)+g^{2}(x)}{2}=\frac{g(x)+x}{2}=x_{1}\\ g(x_{2}) & =g\left(\frac{x-g(x)}{2}\right)=\frac{g(x)-g^{2}(x)}{2}=\frac{g(x)-x}{2}=-x_{2} \end{align*}$$ Not only that, this decomposition is unique. If not, let $$x=x_{S}+x_{A}=\tilde{x}_{S}+\tilde{x}_{A}$$ be two such representations. Then $$0=\underbrace{(x_{S}-\tilde{x}_{S})}_{=:y_{S}}+\underbrace{(x_{A}-\tilde{x}_{A})}_{=:y_{A}}$$ Note that $$\begin{align*} g(y_{S}) & =g(x_{S}-\tilde{x}_{S})=g(x_{S})-g(\tilde{x}_{S})=x_{s}-\tilde{x}_{S}=y_{S}\\ g(y_{A}) & =g(x_{A}-\tilde{x}_{A})=g(x_{A})-g(\tilde{x}_{A})=-x_{A}+\tilde{x}_{A}=-y_{A} \end{align*}$$ Thus, $y_{S}$ is $g$-symmetric and $y_{A}$ is $g$-antisymmetric.

Now we have $$\begin{align*} 0 & =y_{S}+y_{A}\\ \implies0=g(0) & =g(y_{S})+g(y_{A})\\ 0 & =y_{s}-y_{A} \end{align*}$$ The first and third equations above imply $$y_{A}=0$$ and hence $$y_{S}=0$$ Thus $$\begin{align*} x_{S} & =\tilde{x}_{S}\\ x_{A} & =\tilde{x}_{A} \end{align*}$$ and the representation is unique.