Taking log of a matrix equation

Question

I have the following relation

$$\rho = \frac{1}{x}\exp(\log(\rho) + \delta),$$

where $x$ is a number and $\rho$ and $\delta$ are matrices of size $n\times n$. I would like to solve this for $\delta$. I take logs on both sides and get

$$\log\rho = \log\rho + \delta - \log x$$

What I have written here is dimensionally inconsistent since $x$ is a number and the other terms are matrices. But I'm not sure what's wrong since taking the log seems okay. Also it seems okay to use $\log(A/B) = \log(A) - \log(B)$ since $x$ is just a number and commutes with all the matrices.

What exactly is the right solution for $\delta$? Is it $\delta = x I$, where $I$ is the appropriately sized identity matrix? Or are there some technicalities that I'm missing?

Answer 1

It's true that, under some assumptions, $\log (a b)= \log a + \log b$ for scalars ("numbers") and $\log (A B)= \log A + \log B$ for commuting matrices. But it's obviously false that $\log(a B)=\log a + \log B$.

Hence, you must be careful. What you can write is

$$\log a B = \log aI B=\log(aI)+\log(B)$$

Further, consider $e^{tI}= \sum_n t^nI^n/n!=e^tI$ , hence $t I = \log (e^t I) \implies \log(aI)= \log(a)I$.

Then

$$\log \frac{1}{x} M =-\log(x)I + \log(M)$$

Answer 2

Your work looks correct, and it is perhaps the case (as it was in my math classes) that your notation is the standard one; but it is confusing. Notice that

$$\frac{1}{x}=\frac{1}{x}I$$

where $I$ is the $n\times n$ identity matrix. You can then distribute that $x^{-1}$ into the identity matrix to get a diagonal matrix with $x^{-1}$ as all of its elements. It becomes clear then that scalar multiples can always to be distributed into an identity matrix.

What you take away from this is that $x^{-1}$ is fine when we are working with multiplication, but once we perform operations like $\log$ that separate things into addition, $x^{-1}$ becomes a matrix with no change of notation, it is just understood that it is now a matrix. Which, of course, is confusing.