Can Any Complex Matrix Be Decomposed into Hermitian and Skew-Hermitian Parts?

[ognik]

If C NOT Hermitian, show we can decompose C into $\frac{1}{2}\left( C + {C}^{\dagger} \right) +\frac{1}{2i}i\left( C- {C}^{\dagger} \right) $

I've managed to prove C = C a couple of times, EG taking Hermitian or conjugate of both sides, probably there is a bit of info I am not thinking of or missing altogether, a hint please?

 

[Ackbach]

If C NOT Hermitian, show we can decompose C into $\frac{1}{2}\left( C + {C}^{\dagger} \right) +\frac{1}{2i}i\left( C- {C}^{\dagger} \right) $

I've managed to prove C = C a couple of times, EG taking Hermitian or conjugate of both sides, probably there is a bit of info I am not thinking of or missing altogether, a hint please?

I'm guessing there's a bit more to this problem. That the RHS equals the LHS you've probably already shown, as indicated by your statement that you've proved "C = C a couple of times". I'm guessing the problem wants you to investigate the first matrix, $(C+C^{\dagger})/2$, and the second matrix, $(C-C^{\dagger})/2$, a bit more closely. What are their properties?

 

[ognik]

I had, but on second looks, it seems ridiculously easy...

Let $H(C)=\frac{1}{2}(C+C^{\dagger}) $, then $H^{\dagger}=\frac{1}{2}(C^{\dagger}+C) = H$

Similarly $S(C)^{\dagger}= -S$

I remember similar expressions for symmetric Matrices (same proof?) isn't there a similar formula for complex numbers?

...and why do they show the formula for C with $\frac{1}{2i}i$?

 

[Ackbach]

Exactly - you've got it. So, the moral of the story is that you can write any (square) matrix as the sum of an Hermitian matrix, and a skew-Hermitian matrix. These two kinds of matrices have special properties that are nice to work with in certain circumstances.

I have no idea why they didn't just write
$$C=\frac12\left(C+C^{\dagger}\right)+\frac12\left(C-C^{\dagger}\right).$$
Probably to confuse the newbie. ;)

 

[ognik]

This book excels at that suppose it gives the neurons more exercise. (Can't imagine why the Uni didn't use the latest edition...)

The other formula I was thinking of is $ z = \frac{1}{2}(z+z^*) + \frac{1}{2i}(z-z^*) $, real + imag component, even easier to prove

 

[Deveno]

This is my theory (as to the form of the decomposition):

$(C - C^{\dagger})^{\dagger} = -(C - C^{\dagger})$ is skew-Hermitian, whereas:

$(i(C - C^{\dagger}))^{\dagger} = (-i)(-(C - C^{\dagger})) = i(C - C^{\dagger})$ is Hermitian.