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True, false questions about matrices

  1. Apr 7, 2009 #1
    could you help me with some true false questions about matrices?

    If [tex]\mathbf{A}=\mathbf{B}-\mathrm{i}\mathbf{C}[/tex] is hermitian matrix [tex]\mathbf{B},\mathbf{C}[/tex] are real, then [tex]\mathbf{B},\mathbf{C}[/tex] are anti-symmetric matrices. True? False?

    My solution
    If [tex]\mathbf{A}[/tex] is hermitian, then [tex]\mathbf{A}^{\mathrm{H}}=\mathbf{A}[/tex] so [tex](\mathbf{B}-\mathrm{i}\mathbf{C})^{\mathrm{H}}=\mathbf{B}^{\mathrm{H}}-(\mathrm{i}\mathbf{C})^{\mathrm{H}}=\mathbf{B}^{\mathrm{T}}+i\mathbf{C}^{\mathrm{T}}[/tex]. It implies the fact [tex]\mathbf{B}^{\mathrm{T}}+i\mathbf{C}^{\mathrm{T}}=\mathbf{B}-\mathrm{i}\mathbf{C}[/tex]. [tex]\mathbf{B}[/tex] is symmetric and [tex]\mathbf{C}[/tex] is anti-symmetric. FALSE

    If A is diagonalizable and its eigenvalues are [tex]\{\lambda_1,\lambda_2,\cdots,\lambda_n\}[/tex], then [tex]\prod_{k=1}^{n}(x-\lambda_k)=0[/tex] has n different solutions

    My solution
    Matrix is diagonalizable, then [tex]\lambda_i\ne\lambda_j[/tex] for [tex]i\ne j[/tex]. So polynomial has n different soln's. TRUE

    Thank you very much for your help...
  2. jcsd
  3. Apr 7, 2009 #2


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    Science Advisor

    You have a much bigger problem that whether this statement is true of false! What makes you think that if a matrix is diagonalizable, then [itex]\lambda_i\ne\lambda_j[/itex]?

    The identity matrix is diagonalizable because it is diagonal. What are its eigenvalues?

    (An n by n matrix is diagonalizable if and only if it has n independent eigenvectors. It doesn't matter what the eigenvalues are.)
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