# proof that $\mathop{det}\displaylimits e^{A}=e^{\operatorname{tr}A}$

The formula we aim to prove

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

By definition

 $e^{A}=\sum_{n=0}^{\infty}\frac{A^{n}}{n!}$ (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:

 $\displaystyle A$ $\displaystyle=$ $\displaystyle(D+N)=D+N_{1}$ (2) $\displaystyle A^{2}$ $\displaystyle=$ $\displaystyle(D+N)(D+N)=D^{2}+N_{2}$ (3) $\displaystyle A^{3}$ $\displaystyle=$ $\displaystyle(D+N)(D^{2}+N_{2})=D^{3}+N_{3}$ (4) $\displaystyle\vdots$ (5) $\displaystyle A^{k}$ $\displaystyle=$ $\displaystyle D^{k}+N_{k}$ (6) $\displaystyle\vdots$ (7)

where all the $N_{i}$ matrices are strictly upper triangular. Explicitly, $N_{2}=DN_{1}+N_{1}D+N_{1}^{2}$ and by recursion $N_{n+1}=DN_{n}+N_{n}D+N_{1}N_{n}$.

Using equation 1 we can write

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

where $\tilde{N}=\sum_{n=1}^{\infty}\frac{N_{n}}{n!}$ is strictly upper triangular and $e^{D}=\operatorname{diag}(e^{\lambda_{1}},\cdots,e^{\lambda_{n}})$, where $D=\operatorname{diag}(\lambda_{1},\cdots,\lambda_{n})$.

$e^{A}$ 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:

 $\det e^{A}=\prod_{i=1}^{n}e^{\lambda_{i}}=e^{\sum_{i=1}^{n}\lambda_{i}}=e^{% \operatorname{tr}A}$ (9)
Title proof that $\mathop{det}\displaylimits e^{A}=e^{\operatorname{tr}A}$ ProofThatdetEAEoperatornametrA 2013-03-22 15:51:56 2013-03-22 15:51:56 cvalente (11260) cvalente (11260) 7 cvalente (11260) Proof msc 15-00 msc 15A15 SchurDecomposition