Understanding the Relationship: Log, Traces, and Diagonalized Matrices

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Discussion Overview

The discussion revolves around the mathematical relationships involving logarithms, traces, and diagonalized matrices, specifically examining the equation \(\sum \log d_{j} = \text{tr} \log(D)\) and its implications. Participants explore the properties of eigenvalues in relation to these concepts.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions the validity of the relationship \(\sum \log d_{j} = \text{tr} \log(D)\) and expresses uncertainty about the implication that the log of a diagonal matrix corresponds to the log of each diagonal element.
  • Another participant asserts that the relationship is a fact about eigenvalues, stating that if \(\lambda\) is an eigenvalue of \(D\), then \(\log(\lambda)\) is an eigenvalue of \(\log(D)\).
  • It is noted that the relationship is often presented in reverse, where if \(\lambda\) is an eigenvalue of \(D\), then \(e^{\lambda}\) is an eigenvalue of \(e^D\), with the same eigenvector.
  • A participant acknowledges their previous misunderstanding and indicates they have resolved their confusion regarding a related question about the trace of transformed matrices.

Areas of Agreement / Disagreement

Participants express differing levels of confidence regarding the relationships discussed, with some asserting the validity of the eigenvalue properties while others remain uncertain about the initial logarithmic relationship.

Contextual Notes

The discussion does not resolve the initial participant's uncertainty regarding the logarithmic relationship, and assumptions about the properties of eigenvalues and matrix logarithms are not fully explored.

dm4b
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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
 
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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]
 
dm4b said:
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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Office_Shredder said:
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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