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Dec 15, 2011 at 18:15 comment added darij grinberg Phil: Your proof using differentiation is another perfect example of how a property of a single matrix $A$ is proven by considering a whole family (actually, the one-parameter family $tA$).
Sep 8, 2011 at 14:16 comment added Dan Petersen Proving the density observation is nearly trivial. It's easy to see that a sufficient condition for diagonalizability is to have $n$ distinct eigenvalues, so the non-diagonalizable matrices are contained in the closed subvariety of $M_n(\mathbb C)$ defined by the vanishing of the discriminant of the characteristic polynomial. And clearly the complement of a hypersurface is dense.
Sep 6, 2011 at 1:08 comment added Phil Isett I'm not sure it's easier to do over the complex numbers. The function det(e^{tA}) solves $\frac{df}{dt}= \mbox{tr }A f$ with $f(0)=1$. The solution to the latter equation is unique since $\frac{d}{dt}(f e^{−t \mbox{tr }A} )=0$. Or you can just use the product definition $\mbox{\det } e^A =\lim e^{n \log \det(1+A/n)}$ and again differentiate $\det$ at the identity to conclude. Isn't proving this density observation you've quoted is a little harder than just differentiating $\det$ at the identity?
Sep 1, 2011 at 13:10 history answered Oscar Randal-Williams CC BY-SA 3.0