proof that $det{e}^A=e^{\mathrm{tr}A}$

According to Schur decomposition the matrix $A$ can be written after a suitable change of basis as $A=D+N$ where $D$ is a diagonal matrix and $N$ is a strictly upper triangular matrix.

The formula we aim to prove

$$det{e}^A=e^{\mathrm{tr}A}$$

is invariant under a change of basis and thus we can carry out the computation of the exponential in any basis we choose.

By definition

$$e^A=\sum _{n=0}^{\mathrm{\infty }}\frac{A^n}{n!}$$(1)

By the properties of diagonal and strictly upper triangular matrices we know that both $DN$ and $ND$ will also be strictly upper triangular matrices and so will their sum.

Thus the powers of $A$ are of the form:

$A$$=$$(D+N)=D+N_1$(2)

$A^2$$=$$(D+N)(D+N)=D^2+N_2$(3)

$A^3$$=$$(D+N)(D^2+N_2)=D^3+N_3$(4)

$\mathrm{⋮}$(5)

$A^k$$=$$D^k+N_k$(6)

$\mathrm{⋮}$(7)

where all the $N_i$ matrices are strictly upper triangular.Explicitly, $N_2=D{N}_1+N_{1}D+N_1^2$ and by recursion $N_{n+1}=D{N}_n+N_{n}D+N_1{N}_n$.

Using equation 1 we can write

$$e^A=e^D+\tilde{N}$$(8)

where $\tilde{N}={\sum }_{n=1}^{\mathrm{\infty }}\frac{N_n}{n!}$ is strictly upper triangular and $e^D=\mathrm{diag}(e^{{\lambda }_1},\mathrm{\cdots },e^{{\lambda }_n})$, where $D=\mathrm{diag}({\lambda }_1,\mathrm{\cdots },{\lambda }_n)$.

$e^A$ will thus be an upper triangular matrix.Since the determinant of an upper triangular matrix is just the product of the elements in its diagonal, we can write:

$$det{e}^A=\prod _{i=1}^n{e}^{{\lambda }_i}=e^{{\sum }_{i=1}^n{\lambda }_i}=e^{\mathrm{tr}A}$$(9)