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.
