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Rank(A)= Trace(AA*) ?

  1. Jun 21, 2011 #1
    Rank(A)= Trace(AA*) ??

    I have two questions and I hope it is acceptable...Seemingly unrelated, though I came to wonder about the first while thinking the second. Thanks

    1)Is this statement true? or is there a statement that relates Rank(A) and Trace(??)

    2) AB-BA=I (When does this identity hold if at all? Field can be closed or Z/Z2, or etc)
    Last edited: Jun 21, 2011
  2. jcsd
  3. Jun 21, 2011 #2
    Re: Rank(A)= Trace(AA*) ??

    Hi arthurhenry! :smile:

    The first one is not true. It is true that Rank(A)=Rank(AA*). But it isn't in general true that Rank(AA*)=Trace(AA*). For example, take

    [tex]A=\left(\begin{array}{cc} 2 & 0\\ 0 & 2\\ \end{array}\right)[/tex]

    Then AA* has rank 2, but the trace is 8.

    The second one is not true too. Take A the zero matrix and B an arbitrary matrix.
  4. Jun 21, 2011 #3
    Re: Rank(A)= Trace(AA*) ??

    Perhaps I was not clear, I will phrase it correctly:

    2)Does there exist matrices A and B such that AB-BA=I holds?

    In particular, what is the answer to the question in the case the field is Complex NUmbers and in the case the field is Z/Z2 ?

    1) Is rank(A) equal to Trace(A*A) ?

    not "is the rank(A*A) equal to Trace (A*A)?" As you have pointed out this one is definittely incorrect.
  5. Jun 21, 2011 #4
    Re: Rank(A)= Trace(AA*) ??

    Well, try it yourself. Take general 2x2-matrices and calculate AB-BA. Then solve the system to see whether they can equal I...

    We have that rank(A)=rank(A*A), so the same example applies.
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