# Kern(T) not the Ker(T)

I need some help understanding the following definition:

Definition: Let A$\in$Mn(ℂ) the complex vector space

C(A)={X$\in$Mn(ℂ) : XA=AX}

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

C(A,A*)={S$\in$Mn(ℂ) : SA=A*S}

H(A,A*)={H$\in$Mn(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$\in$Mn: 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$\in$Kern(T) means that P is an element of C(A,A*) which means that PA=A*P such that P is skew Hermitian

the defintion 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

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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.

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 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?

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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?

"S$\in$C(A,A*) implies S*$\in$C(A,A*)

this is done by one of his "standard propositions"

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

1) Kern(T)$\subseteq$iH(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$\in$iH(A,A*)

so Kern(T)$\subseteq$iH(A,A*)

Similarly let P$\in$iH(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