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Exponential function and matrices

There's a proof here for real symmetric matrices. But I don't know if it extends to Hermitian matrices.

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user100928
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Just follow the proof there, replacing transpose with conjugate transpose. Note that the exponential function remains injective on the eigenvalues, because a Hermitian matrix has real eigenvalues.

Eric Auld
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  • But why is it that at the Wiki article on "Logarithm of a matrix." There's an added condition that the matrix has non-negative real eigenvalues. To quote:logarithm is not unique, but if a matrix has no negative real eigenvalues, then it has a unique logarithm whose eigenvalues lie all in the strip {z ∈ C | −π < Im z < π}. This logarithm is known as the principal logarithm. – user100928 Nov 11 '15 at 21:05
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    I think what's confusing you is that the exponential is injective on Hermitian matrices, but the logarithm is not single-valued. Note that these two are not incompatible...but the first condition implies that if you want to choose a Hermitian value for the logarithm, there is at most one. – Eric Auld Nov 11 '15 at 21:34
  • There's this exercise in Horn and Roger though. It asks me to prove that positive definite matrices have a unique Hermitian logarithm. And I'm just bothered because my proof did not use the positive definite hypothesis. – user100928 Nov 11 '15 at 21:54
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    Positive definite matrices are the only ones that can possibly have Hermitian logarithms. Said another way, whenever you exponentiate a Hermitian matrix you get a positive definite matrix. As an example, note that if you exponentiate a diagonal matrix of real numbers, you will get a diagonal matrix of positive numbers. – Eric Auld Nov 11 '15 at 22:13
  • I'm starting to get it. But what's wrong with this reasoning: if the matrix $A$ is Hermitian, then it has a logarithm that is a primary matrix function. A logarithm that is a polynomial of $A$, hence, Hermitian. – user100928 Nov 11 '15 at 22:31
  • Why is it a polynomial of $A$? – Eric Auld Nov 12 '15 at 00:10
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    No. I think my mistake was in assuming polynomial of a hermitian is hermitian. I just realized that works only for real polynomial. And that for example $p(x)=ix$ is skew-hermitian. Anyway thanks for your help. I'm correcting myself little by little. – user100928 Nov 12 '15 at 07:33
  • @user100928 I see. No problem. I am still curious why it would be a polynomial of $A$, if you have time to explain. – Eric Auld Nov 12 '15 at 15:50
  • The exponential of a matrix, because the exponential satisfies certain differentiability conditions, is, in a way, just determined through the exponential of the eigenvalues. So you can use an interpolation formula to get a polynomial that does what the exponential function does to the eigenvalues. The polynomial would depend on the matrix though. – user100928 Nov 12 '15 at 17:24