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Can someone please explain me how to prove the following formula?

$det(I +M) = \exp\;tr\;\ln(I+M)\;\,.$

Here $I$ and $M$ are a $2 \times 2$ identity matrix and an arbitrary $2 \times 2$ matrix, correspondingly.

Also, how to derive from the above formula the following one?

$− 2\,det\,M = tr(M^2) − (tr M)^2$

Do these formulae have counterparts of dimensions larger than $2 \times 2\,$?

Somewhere in the literature I saw the relation

$\ln\;det[M+Q] = \ln\;detM + Tr[M^{−1}\,Q] + O(Q^2)$

How to prove it?

Many thanks!

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First part follows the identity (see here) $$ \det (e^A) = e^{tr(A)}$$ where $A = \ln(I+M)$ in your example.

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I had better start with this, in any dimension. If $J$ is in Jordan Normal Form, we get $$ \det \exp (J) = e^{\operatorname{trace} J } \; . $$ Here we may separately consider each Jordan block, as $\exp (J)$ has the same block structure. In each lock, we have a scalar part $\lambda I$ and a nilpotent part $N.$ A these commute, finding the block $\exp (\lambda I + N) = \exp \lambda I \cdot \exp N = e^\lambda I \cdot \exp N = e^\lambda \exp N. $ We find $\exp N$ by a finite sum.

It follows that the same is true for any real or complex square matrix $A,$ as we can write $A = Q^{-1} J Q.$

The rest of your stuff quickly gets tricky. It is not the case that we can make sense out of $\log (I+M)$ for any square matrix $M.$ WE can expect to do so when all the entries of $M$ are quite small, as the exponential map takes a small neighborhood of the zero matrix to a small neighborhood of the identity matrix. For example, $\exp A$ is always invertible, the inverse being $\exp (-A).$ If your $I+M$ is not invertible, it also has no logarithm.

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