What is Kern and its relation to skew Hermitian matrices?

[BrainHurts]

I need some help understanding the following definition:

Definition: Let A\inM_n(ℂ) the complex vector space

C(A)={X\inM_n(ℂ) : XA=AX}

For A\inM_n(ℂ) which is similar to A* we define the complex vector spaces:

C(A,A*)={S\inM_n(ℂ) : SA=A*S}

H(A,A*)={H\inM_n(ℂ): H is Hermitian and HA=A*H} \subset C(A,A*)

Define a map T:C(A,A*)→H(A,A*) by T(S)=\frac{1}{2}S + \frac{1}{2}S*

As a map between real vector spaces, T is linear and Kern(T)={X\inM_n: X is skew Hermitian}=iH(A,A*)

I just want to make sure that my understanding is correct and what is "Kern" short for
To say that P\inKern(T) means that P is an element of C(A,A*) which means that PA=A*P such that P is skew Hermitian

the definition is from the paper I am reading it is by J. Vermeer on page 263

http://www.math.technion.ac.il/iic/ela//ela-articles/articles/vol17_pp258-283.pdf

Thank you for any further comments

 

[micromass]

A matrix P is an element of Kern(T) if P\in C(A,A^*) and if T(P)=0.
So you know that
PA=A^*P~\text{and}~\frac{1}{2}P+\frac{1}{2}P^*=0

The paper now claims that

Kern(T)=iH(A,A^*)

and that these are exactly the skew Hermitian matrices. This is not a definition of Kern(T), but it is a theorem.

 

[BrainHurts]

So I know we can look at the set of Hermitian matricies analogous to the real number (I think)

so let A be Hermitian. Then A can be written as A=B+iC where B and C are Hermitian.

If we look at that linear map it's like looking at the identity of complex numbers.

let z=x+iy

x=\frac{1}{2}(z+\overline{z})

so if 0=\frac{1}{2}(z+\overline{z}),then z is purely imaginary

so if we look at P in the Kern(T)

it's like saying P is skew-Hermitian which is analgous to a number being purely imaginaryand if A=A* it's like saying z=\overline{z},then z is real, so if A = A* and A=B+iC this implies that C=0
right?

 

[micromass]

So I know we can look at the set of Hermitian matricies analogous to the real number (I think)

so let A be Hermitian. Then A can be written as A=B+iC where B and C are Hermitian.

But if A is hermitian,then C=0.

If we look at that linear map it's like looking at the identity of complex numbers.

let z=x+iy

x=\frac{1}{2}(z+\overline{z})

so if 0=\frac{1}{2}(z+\overline{z}),then z is purely imaginary

so if we look at P in the Kern(T)

it's like saying P is skew-Hermitian which is analgous to a number being purely imaginary


and if A=A* it's like saying z=\overline{z},then z is real, so if A = A* and A=B+iC this implies that C=0
right?

OK, so you have the right analogue statements. But can you now prove the result for P directly?

 

[BrainHurts]

So let me start with this assertion he makes

"S\inC(A,A*) implies S*\inC(A,A*)

this is done by one of his "standard propositions"

for the proof of Kern(T)=iH(A,A*)

1) Kern(T)\subseteqiH(A,A*)

let P \in Kern(T)

then PA=A*P and \frac{1}{2}P+\frac{1}{2}P*=0

So P=-P*

It follows that P is skew hermitian and P\iniH(A,A*)

so Kern(T)\subseteqiH(A,A*)

Similarly let P\iniH(A,A*), let S=iP

Then SA=A*S such that S is skew Hermitian (if P is Hermitian, then iP is skew Hermitian)

this is the part that I'm getting stuck on