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Matrix Norms

  1. Feb 27, 2010 #1
    Could someone please help me show that if A is Hermitian
    [tex]\left\|(A-\lambda I)^{-1}\right\|_{2}=\frac{1}{min_{\lambda_{i}\in\sigma(A)}|\lambda-\lambda_{i}|}[/tex]
    where [tex]\sigma(A)[/tex] denotes the eigenvalues of A.

    I have figured out how to solve the norm without an inverse, but the inverse confuses me a bit.
    Recall, that [tex]\left\|\cdot\right\|=\sqrt{r_{\sigma}(A^{*}A)}[/tex], which is to say the square root of the largest eigenvalue of [tex]A^{*}A[/tex].
  2. jcsd
  3. Feb 27, 2010 #2


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    If A is hermitian you can diagonalize it with the eigenvalues lying along the diagonal. Think about what your expressions look like with A in that form.
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