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Math Help - Show that skew-Hermitian is nonsingular

  1. #1
    Member Mollier's Avatar
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    Show that skew-Hermitian is nonsingular

    Problem statement:
    Let S be skew-Hermitian, i.e, S=-S^*.
    Show that I-S is nonsingular.

    Attempt:
    Well, I know that an arbitrary square matrix can be written as the sum of a Hermitian matrix and a skew-Hermitian matrix.
    This tells me that I-S is Hermitian, and since the determinant of a Hermitian matrix is real it is I-S is not singular.

    Now, I would like to express this in an more elegant manner using actual math
    Hope someone can give me a hint or two, thanks.
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  2. #2
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    Quote Originally Posted by Mollier View Post
    Problem statement:
    Let S be skew-Hermitian, i.e, S=-S^*.
    Show that I-S is nonsingular.

    Attempt:
    Well, I know that an arbitrary square matrix can be written as the sum of a Hermitian matrix and a skew-Hermitian matrix.
    This tells me that I-S is Hermitian, and since the determinant of a Hermitian matrix is real it is I-S is not singular.

    First, I-S is not hermitian in general, as you can easily check by using (A+B)^{*}=A^{*}+B^{*}. Second, even if it were then its determinant is real so...so what? It could be zero as zero is a real number! In fact, the zero matrix is Hermitian, and thus your argument doesn't work.

    Tonio


    Now, I would like to express this in an more elegant manner using actual math
    Hope someone can give me a hint or two, thanks.
    .
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  3. #3
    Member Mollier's Avatar
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    Some bad thinking on my part, thanks for the heads up.
    Here's another try:

    If  (I-S)y=0 for y \neq 0 it means that  I-S is singular.
    Then  Sy = 0 which says that  S has 1 as an eigenvalue.
    This contradicts the fact that skew-Hermitian matrices have pure imaginary eigenvalues.
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  4. #4
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    Quote Originally Posted by Mollier View Post
    Some bad thinking on my part, thanks for the heads up.
    Here's another try:

    If  (I-S)y=0 for y \neq 0 it means that  I-S is singular.
    Then  Sy = 0 which says that  S has 1 as an eigenvalue.
    This contradicts the fact that skew-Hermitian matrices have pure imaginary eigenvalues.


    Now this looks correct, but I think you meant (I-S)y=0\Longrightarrow Sy=y \Longrightarrow\,\,if\,\,\,y\ne 0\,\,\,then\,\,\,\lambda=1 is an eigenvalue of S, which is impossible since S is skew-Hermitian...nice!

    Tonio
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  5. #5
    Member Mollier's Avatar
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    Yes, I meant Sy=y. Was writing this up during work so I had to be quick
    Thanks!
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