Prove $\mathrm{exp}(\mathrm{Tr}(A)) = \det(\mathrm{exp}(A))$ via maximal tori

Question

This is an exercise in Kris Tapp's little AMS booklet Matrix Groups for Undergraduates. I have reviewed the proof given earlier in the text but that proof relies on properties of exp and other aspects of commuting matrices, the definition of the derivative and other simpler observations. Ah! Commuting matrices must be one key here but I am afraid I haven't made enough of the connections. My feeling is that conjugation will be helpful as well, since all maximal tori are conjugates of each other. I believe I have all the parts of the jigsaw puzzle but have yet to fit the appropriate pieces. Maximal Tori is the name of the Chapter (#9) and they are, hence, a NEW topic, which again points toward their properties to be the glue to hold the pieces together.

Any suggestions?

Answer 1

I cannot imagine what maximal tori have to do with this. So here is the more or less standard proof. It depends on four facts:

• $\text{Tr}(AB)=\text{Tr}(BA)$ for all $A,B$.

• $\det AB=\det A\,\det B$ for all $A,B$.

• $\exp(JAJ^{-1})=J\exp(A)J^{-1}$ for all $A$ and all invertible $J$.

• The Schur decomposition: for any $A$, there exists a unitary $J$ such that $A=JTJ^{-1}$ with $T$ upper triangular.

The trace property is an easy exercise. Multiplicativity of $\det$ is not trivial, but it is a basic linear algebra fact. The property of the exponential follows easily from $(JAJ^{-1})^n=JA^nJ^{-1}$. The Schur decomposition is a basic fact in matrix analysis; it can be replaced here with the Jordan decomposition if you happen to know it better.

A quick analysis of the equality $A=JTJ^{-1}$ shows that $T$ contains the eigenvalues of $A$, counting multiplicities, in its diagonal. Call these $\lambda_1,\ldots,\lambda_n$.

Now \begin{align}\exp(\text{Tr}(A))&=\exp(\text{Tr}(JTJ^{-1})=\exp(\text{Tr}(TJ^{-1}J))\\ \ \\ &\tag1=\exp(\text{Tr}(T))=\exp(\lambda_1+\cdots+\lambda_n). \end{align} And \begin{align} \det(\exp(A))&=\det(\exp(JTJ^{-1}))=\det(J\exp(T)J^{-1})=\det(\exp(T))\\ \ \\ &\tag2=e^{\lambda_1}\cdots e^{\lambda_n}.\end{align} The only nontrivial step is that $\exp(T)$ has diagonal $e^{\lambda_1},\ldots,e^{\lambda_n}$. This follows by simply noticing that the diagonal of $T^n$ is $T_{11}^m,\ldots,T_{nn}^m$.

Now the equality betwen $(1)$ and $(2)$ is clear since $e^{a+b}=e^ae^b$.

Answer 2

The result clearly holds for $A$ diagonal, and thus it holds for $A$ diagonalizable by the fact that $\operatorname{tr}$ and $\det$ are both invariant under conjugation. But diagonalizable matrices are dense in $M_n(\mathbb{C})$ (in the usual metric), so the result holds for arbitrary $A$.