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Show that $\exp(\mathrm{Tr}(X))=\det(\exp(X))$ where $X$ is a matrix using the concept of the Jordan normal form

I realised this formula by considering that: $\det(\exp(X))=\exp(\lambda_1) \times\cdots \times \exp(\lambda_n)=\exp(\lambda_1 + \cdots + \lambda_n)=\exp(\mathrm{Tr}(X))$

$\lambda_i$ are eigenvalues of $X$

I was unsure how to prove it using the Jordan normal form - could you help me please?

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    $\begingroup$ Prove the result for $X=J$, where $J$ is diagonalizable. Then show that diagonalizable matrices are dense in the space of $n\times n$ matrices, i.e. for any non-diagonalizable matrix, you can perturb each element slightly so that the result is diagonalizable. $\endgroup$ – Alex R. May 13 '16 at 19:47
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If $X$ is similar to $Y$ with $X = P^{-1}YP$ then

$$ \operatorname{Tr}(X) = \operatorname{Tr}(Y) \implies \exp(\operatorname{Tr}(X)) = \exp(\operatorname{Tr}(Y)), \\ \exp(X) = P^{-1}\exp(Y)P \implies \det(\exp(X)) = \det(\exp(Y)) $$

so we can assume that $X \in M_n(\mathbb{C})$ is in Jordan canonical form. In particular, $X$ is upper triangular with the eigenvalues $\lambda_1, \dots, \lambda_n$ (counted with multiplicities) on the diagonal. Then, by the definition of $\exp$, the matrix $\exp(X)$ is also upper triangular with the values $e^{\lambda_1}, \dots, e^{\lambda_n}$ on the diagonal and so

$$ \det(\exp(X)) = e^{\lambda_1} \dots e^{\lambda_n} = e^{\lambda_1 + \dots + \lambda_n} = \exp{\operatorname{Tr(X)}}. $$

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Here is another approach using differential equations:

Let $Y$ solve $\dot{Y} = XY$ subject to $Y(0) = I$. Note that the unique solution is given by $Y(t) = e^{tX}$.

Let $d(t) = \det Y(t)$, then $\dot{d}(t) = \det(Y(t)) \operatorname{tr} (Y^{-1}(t) X Y(t)) = \operatorname{tr} X \det Y(t) = \operatorname{tr} X d(t)$.

Hence $d(1) = d(0) e^{\operatorname{tr} X}$.

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