# Logs, Traces, Matrices

Trying to make sense of the following relation:

$\sum log d_{j} = tr log(D)$

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

One more:

if $M = A^{-1}DA$,

why is this true:

$tr A^{-1}log(D)A=tr\ log (M)$

Office_Shredder
Staff Emeritus
Gold Member
Trying to make sense of the following relation:

$\sum log d_{j} = tr log(D)$

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 $\lambda$ is an eigenvalue of D, then $\log(\lambda)$ is an eigenvalue of $\log(D)$.

This is usually first presented in the other direction, that if $\lambda$ is an eigenvalue of D, then $e^{\lambda}$ is an eigenvalue of $e^D$. (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:
1 person
No, this is a fact about eigenvalues. If $\lambda$ is an eigenvalue of D, then $\log(\lambda)$ is an eigenvalue of $\log(D)$.

This is usually first presented in the other direction, that if $\lambda$ is an eigenvalue of D, then $e^{\lambda}$ is an eigenvalue of $e^D$. (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.