Proof of 2 Matrix identities (Traces, Logs, Determinants) I am working through a derivation in someone's thesis at the moment to understand an important result, but I am more than a bit rusty on matrices. Could anyone give me some tips on these identities? They are stated without proof and I'm having a hard time finding a derivation online.
Below, X is a matrix and E is a scalar, and X is a function of E.
1) $Tr(X' X^{-1}) = \frac{d}{dE} Tr(ln(X))$
When I first saw this I thought it would be the same as treating X as a scalar, then by the definition of the ln function the above would be true. Is the fact that there is a trace and that X is a matrix important in the derivation? 
He does mention that "" $Tr(log X)' = Tr[X' X] = Tr[X X']$ "" but I think this was probably a typo, since an expression of the form $Tr[X' X]$ does not appear in his calculation. 
2) $Tr(ln(X)) = ln(det(X))$
This one I am a bit stuck on, I would guess that it has something to do with the definitions of the trace and determinant but not sure where to go from there. I haven't done anything with matrices in about 3 years, and I'm a physicist, so keep it basic :)
EDIT
OK here is my working for proof 1 using Robert's guide below:
$$ X' = \frac{d}{dE} \sum_{n=0}^{\infty} \frac{L^n}{n!} $$
Using the chain rule, $$ X' = \sum_{n=0}^{\infty} \frac{n L' L^{n-1}}{n!} = \sum_{n=0}^{\infty} \frac{L' L^{n-1}}{(n-1)!} = \sum_{n=1}^{\infty} \frac{L' L^{n-1}}{n!}$$
Here is the bit I don't quite follow, regarding the introduction of the dummy j which seems to cancel later on in the calculation without being used.
$$ \sum_{j=0}^{n-1} \sum_{n=1}^{\infty} \frac{L' L^j L^{n-1 - j}}{n!} $$
Now using this expression for X':
$$ X' X^{-1} = X' e^{-L} = \sum_{j=0}^{n-1} \sum_{n=1}^{\infty} \frac{L' L^j L^{n-1 - j}}{n!} e^{-L} $$
Here is where I find a problem now, since 
$$ \sum_{j=0}^{n-1} (\sum_{n=1}^{\infty} \frac{L^{n-1}}{n!}) L' e^{-L} $$
Now my part in the brackets in that last expression isn't $e^L$ so doesn't cancel nicely. I am pretty sure I am missing something with commutativity and when you introduced the sum over j!? EDIT 2 Just realised my last step on the chain rule, changing the sum from n=0 to n=1 doesn't make much sense.
 A: 1) is a bit tricky if $X'$ and $X$ don't commute.  $\log(X)$ is a matrix $L$ such that $\exp(L) = X$.  Now 
$$\dfrac{d}{dE} X = \dfrac{d}{dE} \sum_{n=0}^\infty \dfrac{L^n}{n!} =
\sum_{n=1}^\infty \sum_{j=0}^{n-1} \dfrac{L^j L' L^{n-1-j}}{n!} $$
and so
$$ X' X^{-1} = X' \exp(-L) = \sum_{n=1}^\infty \sum_{j=0}^{n-1}\frac{L^j L' L^{n-1-j}}{n!} \exp(-L)$$
but since $\text{Tr}(AB) = \text{Tr}(BA)$ and $L$ commutes with $\exp(-L)$,
$$ \text{Tr}(X' X^{-1}) = \sum_{n=1}^\infty \sum_{j=0}^{n-1} \dfrac{\text{Tr}\left(L^j L' L^{n-1-j} \exp(-L) \right)}{n!} = \sum_{n=1}^\infty  \dfrac{\text{Tr}\left( L'L^{n-1} \exp(-L)\right)}{(n-1)!} = \text{Tr}(L' \exp(L) \exp(-L)) = \text{Tr}(L')$$
2) If $X$ has eigenvalues $\lambda_j$ (counted by algebraic multiplicity), $\log(X)$ has eigenvalues $\log(\lambda_j)$.  Then $\text{Tr}(\log(X)) = \sum_j \log(\lambda_j) = \log(\prod_j \lambda_j) = \log(\det(X))$.  Actually there is a question of which branches of the logarithm to use when there are non-positive eigenvalues, so it is more accurate to say that $\text{Tr}(\log(X))$ is one of the branches of $\log(\det(X))$.
