A hard exercise on endomorphisms and determinants The following exercise has been bugging me for some days, could someone help me with it ?
Let $E$ be a $\mathbb{C}$-vector space with dimension $n$ and $f\in\mathcal{L}(E)$
($\mathcal{L}(E)$ denotes the set of endomorphisms in E)
$\mathrm{tr}(f)$ is the trace of $f$, ie, the trace of $f$'s canonical matrix.
Show that, for $t$ small enough, $\displaystyle\exp\left(\sum_{k=0}^\infty\mathrm{tr}(f^k)\frac{t^k}k\right)=\det(\mathrm{Id}-t\cdot f)^{-1}$
 A: Recall that $$-\log(1-x)=\sum_{k=1}^{\infty}\frac{x^k}{k},$$
Using the fact that the exponential is the inverse of the logarithm we have,
$$\exp(\sum_{k=0}^{\infty}\frac{x^k}{k})=(1-x)^{-1}.$$
Next recall that $\exp(tr(f))=\det(\exp(f))$ when $f$ is a matrix.  This is a good time to broach how this goes.  The set of diagonable matrices is dense in the space of matrices, so if a formula involving continuous functions holds for diagonable matrices, it holds everywhere.  For diagonal matrices the formula is obviously true, and you can see that conjugating by a matrix doesn't do anything to the formula by looking term by term in the series.  This also means that the matrix logarithm and the matrix exponentials are inverses near the identity matrix.   By the way the convergence of these series is achieved using the Weierstrass $M$-test and the fact that if a series converges in norm, it converges.
By the same argument, substituting $tf$ in for $x$ we get,
$$\det(\exp(-\log(Id-t\cdot f)))=\exp(tr(-\log(Id-t\cdot f))).$$
For small $t$ , $Id-t\cdot f$ is near the identity, so the power series formula is valid,
$$(\det(Id-t\cdot f))^{-1}=\exp(tr(\sum_{k=1}^{\infty}\frac{t^kf^k}{k})).$$
Using the linearity of the trace we arrive at:
$$(\det(Id-t\cdot f))^{-1}=\exp(\sum_{k=1}^{\infty}\frac{t^ktr(f^k)}{k}).$$
