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Kern(T) not the Ker(T)

by BrainHurts
Tags: kernt, kert
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BrainHurts
#1
Dec6-12, 03:47 AM
P: 84
I need some help understanding the following definition:

Definition: Let A[itex]\in[/itex]Mn(ℂ) the complex vector space

C(A)={X[itex]\in[/itex]Mn(ℂ) : XA=AX}

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

C(A,A*)={S[itex]\in[/itex]Mn(ℂ) : SA=A*S}

H(A,A*)={H[itex]\in[/itex]Mn(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]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[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..._pp258-283.pdf

Thank you for any further comments
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micromass
#2
Dec6-12, 11:06 AM
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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.
BrainHurts
#3
Dec6-12, 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 skew-Hermitian 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?

micromass
#4
Dec6-12, 06:32 PM
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P: 18,230
Kern(T) not the Ker(T)

Quote Quote by BrainHurts View Post
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=[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 skew-Hermitian 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?
OK, so you have the right analogue statements. But can you now prove the result for P directly?
BrainHurts
#5
Dec8-12, 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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