Logs, Traces, Matrices

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  • #1
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Main Question or Discussion Point

Trying to make sense of the following relation:

[itex]\sum log d_{j} = tr log(D)[/itex]

with D being a diagonalized matrix.

Seems to imply the log of a diagonal matrix is the log of each element along the diagonal.

Having a hard time convincing myself that is true, though
 

Answers and Replies

  • #2
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One more:

if [itex]M = A^{-1}DA[/itex],

why is this true:

[itex]tr A^{-1}log(D)A=tr\ log (M)[/itex]
 
  • #3
Office_Shredder
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Trying to make sense of the following relation:

[itex]\sum log d_{j} = tr log(D)[/itex]

with D being a diagonalized matrix.

Seems to imply the log of a diagonal matrix is the log of each element along the diagonal.

Having a hard time convincing myself that is true, though
No, this is a fact about eigenvalues. If [itex] \lambda [/itex] is an eigenvalue of D, then [itex] \log(\lambda)[/itex] is an eigenvalue of [itex] \log(D)[/itex].

This is usually first presented in the other direction, that if [itex] \lambda[/itex] is an eigenvalue of D, then [itex] e^{\lambda}[/itex] is an eigenvalue of [itex] e^D[/itex]. (and the eigenvector is the same). It's very easy to see this by writing out the power series definition of eD and applying it to the eigenvector of D
 
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  • #4
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No, this is a fact about eigenvalues. If [itex] \lambda [/itex] is an eigenvalue of D, then [itex] \log(\lambda)[/itex] is an eigenvalue of [itex] \log(D)[/itex].

This is usually first presented in the other direction, that if [itex] \lambda[/itex] is an eigenvalue of D, then [itex] e^{\lambda}[/itex] is an eigenvalue of [itex] e^D[/itex]. (and the eigenvector is the same). It's very easy to see this by writing out the power series definition of eD and applying it to the eigenvector of D
ahh, thanks I should have known that.

I figured out the second one too, so no help needed on that one now.
 

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