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