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

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?

• @Nat Sorry? I don't understand, it is just Mathematica 12.0.
– sky
Sep 26, 2020 at 7:28
• @Nat Oh, this is not the OCR, it is Chinese question mark. I forgot to close the Chinese input software.
– sky
Sep 26, 2020 at 12:37

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.

• Use MatrixLog instead. Sep 25, 2020 at 20:27
• @AccidentalFourierTransform: Great, thanks. Sep 25, 2020 at 20:38

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.

• It is not true that you can always diagonalise your matrix B. To correctly take the log of a general matrix, you should use the Jordan decomposition. So your proof holds, but only for diagonalisable matrices. Sep 25, 2020 at 9:30
• Well, since there can be no 0 eigenvalues, otherwise the $\ln$ would be ill-defined, te determinant will never be 0. I mean, for this procedure to work you implicitly assume the matrix to be diagonalisable. Sep 25, 2020 at 10:12
• Even if all eigenvalues are non-zero, or even non-negative, it is not guaranteed that your matrix will be diagonalisable. What you can assume always is that your matrix can be put in the Jordan decomposition form. However I agree that usually, to ensure positivity and reality of eigenvalues, we usually take the log of (positive definite) hermitian matrices, for which this proof works. Sep 25, 2020 at 15:19