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)

Title	proof that $det{e}^A=e^{\mathrm{tr}A}$
Canonical name	ProofThatdetEAEoperatornametrA
Date of creation	2013-03-22 15:51:56
Last modified on	2013-03-22 15:51:56
Owner	cvalente (11260)
Last modified by	cvalente (11260)
Numerical id	7
Author	cvalente (11260)
Entry type	Proof
Classification	msc 15-00
Classification	msc 15A15
Related topic	SchurDecomposition