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Kern(T) not the Ker(T) 
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#1
Dec612, 03:47 AM

P: 84

I need some help understanding the following definition:
Definition: Let A[itex]\in[/itex]M_{n}(ℂ) the complex vector space C(A)={X[itex]\in[/itex]M_{n}(ℂ) : XA=AX} For A[itex]\in[/itex]M_{n}(ℂ) which is similar to A* we define the complex vector spaces: C(A,A*)={S[itex]\in[/itex]M_{n}(ℂ) : SA=A*S} H(A,A*)={H[itex]\in[/itex]M_{n}(ℂ): H is Hermitian and HA=A*H} [itex]\subset[/itex] C(A,A*) Define a map T:C(A,A*)→H(A,A*) by T(S)=[itex]\frac{1}{2}[/itex]S + [itex]\frac{1}{2}[/itex]S* As a map between real vector spaces, T is linear and Kern(T)={X[itex]\in[/itex]M_{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[itex]\in[/itex]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..._pp258283.pdf Thank you for any further comments 


#2
Dec612, 11:06 AM

Mentor
P: 18,230

A matrix P is an element of Kern(T) if [itex]P\in C(A,A^*)[/itex] and if [itex]T(P)=0[/itex].
So you know that [tex]PA=A^*P~\text{and}~\frac{1}{2}P+\frac{1}{2}P^*=0[/tex] The paper now claims that [tex]Kern(T)=iH(A,A^*)[/tex] and that these are exactly the skew Hermitian matrices. This is not a definition of Kern(T), but it is a theorem. 


#3
Dec612, 01:14 PM

P: 84

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=[itex]\frac{1}{2}[/itex](z+[itex]\overline{z}[/itex]) so if 0=[itex]\frac{1}{2}[/itex](z+[itex]\overline{z}[/itex]),then z is purely imaginary so if we look at P in the Kern(T) it's like saying P is skewHermitian which is analgous to a number being purely imaginary and if A=A* it's like saying z=[itex]\overline{z}[/itex],then z is real, so if A = A* and A=B+iC this implies that C=0 right? 


#4
Dec612, 06:32 PM

Mentor
P: 18,230

Kern(T) not the Ker(T)



#5
Dec812, 09:11 PM

P: 84

So let me start with this assertion he makes
"S[itex]\in[/itex]C(A,A*) implies S*[itex]\in[/itex]C(A,A*) this is done by one of his "standard propositions" for the proof of Kern(T)=iH(A,A*) 1) Kern(T)[itex]\subseteq[/itex]iH(A,A*) let P [itex]\in[/itex] Kern(T) then PA=A*P and [itex]\frac{1}{2}[/itex]P+[itex]\frac{1}{2}[/itex]P*=0 So P=P* It follows that P is skew hermitian and P[itex]\in[/itex]iH(A,A*) so Kern(T)[itex]\subseteq[/itex]iH(A,A*) Similarly let P[itex]\in[/itex]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 


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