Making sense of the formula $\operatorname{Det} (I+M )= e^{\operatorname{Tr} \ln (I+M)}$, especially in the infinite dimensional cases

Question

$\DeclareMathOperator\Det{Det}\DeclareMathOperator\Tr{Tr}$In physics literature dealing with quantum field theory, the formula \begin{equation} \Det(I+M) = e^{\Tr \ln(I+M)} \end{equation} appears frequently. Moreover, in the so-called infinitesimal cases, \begin{equation} \ln(I+M) \simeq M \end{equation} is used in analogy with the Taylor series for the usual logarithm.

What is worse, $M$ can be not only unbounded but singular as well, containing $\delta(x-y)$ for example and yielding \begin{equation} \Tr\delta(x-y)=\delta(x-x). \end{equation}

I searched everywhere in order to find relevant materials and make sense of the outrageous formulae, especially the "Mathematical Methods of Modern Physics" series by Reed & Simon. However, I cannot find reference to start with.

Could anyone please help me?

Answer 1

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Q: How to make sense of the outrageous formula $\ln\operatorname{Det}M=\operatorname{Tr}\ln M$ ?

A: You might first consider a diagonalizable $M$, and then since both determinant and trace do not change if you evaluate them in a basis where $M$ is diagonal, the identity follows immediately.

If $M$ is not diagonalizable, decompose it into Jordan blocks and use the identity $$\ln{{1\;1}\choose{0\;1}}={{0\;1}\choose{0\;0}}.$$