The Functional Determinants in Peskin and Schroeder (Eq.9.77)

Question

I'm working on the Eq.9.77 in Peskin (page 304):

To demonstrate this, we need only apply standard identities from linear algebra. First notice that, if a matrix $B$ has eigenvalues $b_i ,$ we can write its determinant as $$ \det{B} ~=~ \prod_{i}{b_i} ~~= \exp{\left[ \sum_{i}{\log{b_i}} \right]} ~=~ \exp{\left[ \operatorname{Tr}{\left(\log{B}\right)} \right]} \,, \tag{9.77} $$ where the logarithm of a matrix is defined by its power series.

However, I test the identity and find it doesn't work.

For $B=\{\{3, -2, 4\}, \{-2, 6, 2\}, \{4, 2, 3\}\} ,$ the left-hand side of the Eq.(9.77) is -98, and the right is 54.

Why does this happen? What are the conditions for this identity to be true? How to prove it?

Answer 1

The problem seems to be that Mathematica here evaluates ${\rm Log}[\text{matrix}]$ by taking ${\rm Log}$ of each matrix element, which is not the correct mathematical definition of the complex natural logarithm of a matrix.

Answer 2

Obviously, since we use the $\ln$ operation, none of the eigenvalues may be negative or 0.

It's fairly trivial to prove that this must hold: \begin{align} \det(B) &= b_1...b_n \\ &= e^{\ln(b_1)}...e^{\ln(b_n)} \\ &= \exp(\sum_i \ln(b_i)) \\ &= \exp[Tr(\ln B)] \end{align}

Since we can find the eigenvalues of $B$, we can also diagonalise it with the property that \begin{equation} \ln[diag(b_1,...,b_n)] = diag(\ln b_1,..., \ln b_n) \end{equation} Also remember that diagonalisation doesn't alter the trace due to the cyclic property: let $S$ be the matrix that diagonalises $B$ then
\begin{equation} Tr(B') = Tr(S^{-1}BS) = Tr(B S S^{-1}) = Tr(B) \end{equation} The same holds for the $\det$.
I am not familiar with your code, but I hope your $\log$ is nog the standard log with base 10, it has to be the natural one.