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Proof of normal matrix criterion

  1. Jul 30, 2008 #1
    I need to proof:
    A in normal matrix if and only if [tex]trace(A^*A)=|t_1|^2+...|t_n|^2[/tex]
    where [tex]t_1,...,t_n[/tex] are the characteristic roots of A.

    I have a problem only with the second direction:
    [tex]trace(A^*A)=|t_1|^2+...|t_n|^2[/tex] --> A is normal.

    can you help me ?
     
  2. jcsd
  3. Jul 30, 2008 #2

    morphism

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    Presumably A is a complex matrix, so we can put it into Schur form, i.e. we can find a unitary matrix Q and an upper triangular matrix T such that A=Q-1TQ. Notice that trace(A*A)=trace(T*T). So if A is not normal, then <blank>.
     
  4. Jul 31, 2008 #3
    thanks.

    note [tex]T=[\alpha_{ij}][/tex]
    so [tex]trace(T^*T)=trace(A^*A)=|t_1|^2+...|t_n|^2[/tex]
    hence
    [tex]\sum_{\substack{
    0\leq\i\leq n \\
    0\leq\j\leq n
    }} |\alpha_{ij}|^2
    =|t_1|^2+...|t_n|^2[/tex]
    because of the eigenvalues are the diagonal entries of T we have
    [tex]\sum_{\substack{
    0\leq\i\leq n \\
    0\leq\j\leq n \\
    j\ne i
    }} |\alpha_{ij}|^2
    =0[/tex]
    hence for [tex]i\ne j[/tex] we have [tex]\alpha_{ij}=0[/tex]

    so T is diagonal matrix. A=Q^-1TQ and hence A is normal.
     
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