# An identity of Jacobi

• August 18th 2009, 03:22 PM
Bruno J.
An identity of Jacobi
Show that, if $A$ is any complex-valued matrix,

$\mbox{det }e^A = e^{Tr(A)}$

where $Tr$ is the trace.
• August 18th 2009, 03:38 PM
ynj
Quote:

Originally Posted by Bruno J.
Show that, if $A$ is any complex-valued matrix,

$\mbox{det }e^A = e^{Tr(A)}$

where $Tr$ is the trace.

where is the variable of A?
• August 18th 2009, 03:40 PM
Bruno J.
$A$ is a matrix with complex entries.
• August 18th 2009, 03:46 PM
ynj
Quote:

Originally Posted by Bruno J.
$A$ is a matrix with complex entries.

I am confused..how can you calculate the exponent of a matrix....
• August 18th 2009, 04:12 PM
Bruno J.
• August 18th 2009, 04:56 PM
ynj
Let $f$be a polynomial,if the eigenvalue of $A$is $\lambda_1,...,\lambda_n$, then the eigenvalue of $f(A)$is $f(\lambda_1),...,f(\lambda_n)$
we may look $e^A$as a polynomial series of $A$(limit)
note that: $\det A=\lambda_1\lambda_2...\lambda_n,Tr(A)=\lambda_1+\ lambda_2...+\lambda_n$
$e^{Tr(A)}=e^{\lambda_1}e^{\lambda_2}...e^{\lambda_ n}$
but $\{e^{\lambda_i}\}$is the eigenvalue of $e^A$(regard it as polynomial)
so $\det e^A=e^{\lambda_1}e^{\lambda_2}...e^{\lambda_n}=e^{ Tr(A)}$
• August 18th 2009, 05:20 PM
Bruno J.
Remember that not every matrix is diagonalizable!

In the case where the matrix is diagonal then it is easy to show that the identity holds because if $a$ is one of the diagonal entries, the corresponding entry of $e^A$ will be $e^a$.
• August 18th 2009, 05:53 PM
ynj
I have changed my proof
every matrix have eigenvalue, and my proof is not based on diagonalibility
• August 18th 2009, 10:09 PM
CaptainBlack
Quote:

Originally Posted by Bruno J.
Show that, if $A$ is any complex-valued matrix,

$\mbox{det }e^A = e^{Tr(A)}$

where $Tr$ is the trace.

See here (2/3rd of the way down)

CB