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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.

enter image description here

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?

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

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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.

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    $\begingroup$ 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. $\endgroup$ – Frotaur Sep 25 at 9:30
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    $\begingroup$ 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. $\endgroup$ – JulianDeV Sep 25 at 10:12
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    $\begingroup$ 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. $\endgroup$ – Frotaur Sep 25 at 15:19

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