Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

and

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

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

share|improve this question
1  
Yes, of course. Using adjoint in place of transpose. –  Berci Jul 10 '13 at 22:45
2  
By the way, I think you got your skew and sym operators mixed up. –  Joel Jul 10 '13 at 22:53
1  
@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

1 Answer 1

up vote 5 down vote accepted

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.