# Kern(T) not the Ker(T)

by BrainHurts
Tags: kernt, kert
 P: 68 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/e..._pp258-283.pdf Thank you for any further comments
 PF Patron Sci Advisor Thanks Emeritus P: 15,673 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.
 P: 68 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?
PF Patron
 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?
 P: 68 So let me start with this assertion he makes "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