403/503 – Determinants

Here is a problem that you may enjoy thinking about. Given an n\times n matrix A, define a new n\times n matrix e^A by the power series


This means, of course, the matrix whose entries are the limit of the corresponding entries of the sequence of matrices \sum_{k=0}^n\frac{1}{k!}A^k as n\to\infty.

(This limit actually exists. Those of you who have seen Hilbert spaces should see a proof easily: Recall we defined the norm \|A\| of A as \sup_{\|v\|=1}\|Av\|, where in this supremum \|v\| and \|Av\| denote the usual norm  (of v or Av, respectively) in {\mathbb C}^n defined in terms of the usual inner product. One checks that a series of vectors \sum_{k=0}^\infty v_k converges (in any reasonable sense) in a Banach space if it converges absolutely, i.e., if \sum_{k=0}^\infty \|v_k\| converges. Since \|A^k\|\le\|A\|^k, the series defining e^A clearly converges absolutely.)

The matrix e^A is actually a reasonable object to study. For example, the function f(t)=e^{tA}v_0 is the unique solution to the differential equation f'(t)=Af(t), f(0)=v_0. Here, v_0\in{\mathbb C}^n is a fixed vector.

Note that, for any A, the matrix e^A is invertible, since e^Ae^{-A}=I, as a direct computation verifies.

Anyway, the problem: Show that for any matrix A, we have \det(e^A)=e^{{\rm tr}(A)}. Note this is not completely unreasonable to expect: A direct computation shows that if v is an eigenvector of A with eigenvalue \lambda, then e^Av=e^\lambda v, so the formula is true whenever A is diagonalizable.


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