Understanding the Relationship: Log, Traces, and Diagonalized Matrices

[dm4b]

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

 

[dm4b]

One more:

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

why is this true:

[itex]tr A^{-1}log(D)A=tr\ log (M)[/itex]

 

[Office_Shredder]

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 e^D and applying it to the eigenvector of D

 

[dm4b]

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 e^D 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.

 

[blue_raver22]

it does seem to work numerically.

I can understand your confusion about this relationship. Let me try to explain it in a more clear and intuitive way.

First, let's review some basic concepts. The logarithm function is the inverse of the exponential function, and it is commonly used in mathematics to simplify calculations involving very large or very small numbers. The trace of a matrix is the sum of its diagonal elements, and it is a useful tool in linear algebra for determining properties of a matrix. And a diagonalized matrix is a special type of matrix where all the non-diagonal elements are zero.

Now, let's look at the relationship between the sum of logarithms of the elements in a matrix (\sum log d_j) and the logarithm of a diagonalized matrix (tr log(D)). The key thing to understand here is that the logarithm of a diagonal matrix is not simply the logarithm of each element along the diagonal, but rather the logarithm of the determinant of the matrix.

In other words, if we have a diagonalized matrix D, with diagonal elements d_1, d_2, ..., d_n, then the logarithm of D is given by log(D) = log(d_1d_2...d_n) = log(d_1) + log(d_2) + ... + log(d_n). This is because the determinant of a diagonal matrix is simply the product of its diagonal elements.

Now, let's go back to the relationship in question. When we sum the logarithms of the elements in a matrix, we are essentially taking the logarithm of the product of those elements. And as we just saw, the logarithm of a diagonalized matrix is the sum of the logarithms of its diagonal elements. So, in a way, the relationship is saying that the logarithm of the product of the elements in a matrix is equal to the sum of the logarithms of the diagonal elements in a diagonalized matrix.

I hope this helps to clarify the relationship between the sum of logarithms and the logarithm of a diagonalized matrix. It may seem counterintuitive at first, but it is a fundamental concept in mathematics and has been proven to be true. If you're still having trouble understanding it, I suggest studying more about logarithms, determinants, and diagonal matrices to gain a deeper understanding.