Understanding the Relationship: Log, Traces, and Diagonalized Matrices

  • Context: Graduate 
  • Thread starter Thread starter dm4b
  • Start date Start date
  • Tags Tags
    Matrices
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
3 replies · 2K views
dm4b
Messages
363
Reaction score
4
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
 
Physics news on Phys.org
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
 
Last edited:
  • Like
Likes   Reactions: 1 person
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.