# Decomposition an operator in terms of symmetric and anti-symmetric components

In linear algebra, we can write any operator as the sum of a symmetric and skew-symmetric parts:

$$A=A^{\mathrm{sym}}+A^{\mathrm{skew}}$$

where

$$A^{\mathrm{skew}}=\frac{1}{2}(A-A^T)$$

and

$$A^{\mathrm{sym}}=\frac{1}{2}(A+A^T)$$.

Can the same be done with any general (continuous) operator?

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Yes, of course. Using adjoint in place of transpose. – Berci Jul 10 '13 at 22:45
By the way, I think you got your skew and sym operators mixed up. – Joel Jul 10 '13 at 22:53
@Paul for what it's worth, this fails for higher order tensors. It is not the case that every tensor is expressed as a sum of a completely symmetric and completely antisymmetric tensor. One has to study representation theory to obtain a canonical decomposition. – James S. Cook Jul 10 '13 at 23:04

The answer is yes. This is a standard trick in Operator Theory. Provided that the operator $A$ is bounded (i.e. continuous) it has a bounded adjoint $A^*$. This would be the conjugate transpose of a matrix in finite dimensions.

We can decompose $A$ into a sum of a self adjoint operator and an anti-self adjoint operator by:

$$A = \frac{A + A^*}{2} + \frac{A - A^*}{2}.$$

This also holds for functions of a real variable in a similar way. For instance we can write any function $f: \mathbb{R} \to \mathbb{R}$ as a sum of a even and odd function by:

$$f(x) = \frac{f(x) + f(-x)}{2} + \frac{f(x) - f(-x)}{2}$$

In many instances we try to draw analogies between operators and real/complex numbers. This often happens through the spectral theorem.

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