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Kern(T) not the Ker(T)

  1. Dec 6, 2012 #1
    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


    Thank you for any further comments
  2. jcsd
  3. Dec 6, 2012 #2
    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

    The paper now claims that


    and that these are exactly the skew Hermitian matrices. This is not a definition of Kern(T), but it is a theorem.
  4. Dec 6, 2012 #3
    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


    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
    Last edited: Dec 6, 2012
  5. Dec 6, 2012 #4
    But if A is hermitian,then C=0.

    OK, so you have the right analogue statements. But can you now prove the result for P directly?
  6. Dec 8, 2012 #5
    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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