# An identity of Jacobi

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

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

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

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

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

where $\displaystyle Tr$ is the trace.

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

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

I am confused..how can you calculate the exponent of a matrix....
• Aug 18th 2009, 04:12 PM
Bruno J.
• Aug 18th 2009, 04:56 PM
ynj
Let $\displaystyle f$be a polynomial,if the eigenvalue of $\displaystyle A$is $\displaystyle \lambda_1,...,\lambda_n$, then the eigenvalue of $\displaystyle f(A)$is $\displaystyle f(\lambda_1),...,f(\lambda_n)$
we may look $\displaystyle e^A$as a polynomial series of $\displaystyle A$(limit)
note that:$\displaystyle \det A=\lambda_1\lambda_2...\lambda_n,Tr(A)=\lambda_1+\ lambda_2...+\lambda_n$
$\displaystyle e^{Tr(A)}=e^{\lambda_1}e^{\lambda_2}...e^{\lambda_ n}$
but $\displaystyle \{e^{\lambda_i}\}$is the eigenvalue of $\displaystyle e^A$(regard it as polynomial)
so $\displaystyle \det e^A=e^{\lambda_1}e^{\lambda_2}...e^{\lambda_n}=e^{ Tr(A)}$
• Aug 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 $\displaystyle a$ is one of the diagonal entries, the corresponding entry of $\displaystyle e^A$ will be $\displaystyle e^a$.
• Aug 18th 2009, 05:53 PM
ynj
I have changed my proof
every matrix have eigenvalue, and my proof is not based on diagonalibility
• Aug 18th 2009, 10:09 PM
CaptainBlack
Quote:

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

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

where $\displaystyle Tr$ is the trace.

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

CB