An identity of Jacobi

**Bruno J.** · August 18th 2009, 03:22 PM

Show that, if $A$ is any complex-valued matrix,

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

where $Tr$ is the trace.

**ynj** · August 18th 2009, 03:38 PM

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?

**Bruno J.** · August 18th 2009, 03:40 PM

$A$ is a matrix with complex entries.

**ynj** · August 18th 2009, 03:46 PM

Originally Posted by Bruno J.

$A$ is a matrix with complex entries.

I am confused..how can you calculate the exponent of a matrix....

**Bruno J.** · August 18th 2009, 04:12 PM

Matrix exponential - Wikipedia, the free encyclopedia

**ynj** · August 18th 2009, 04:56 PM

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 $\det e^A=e^{\lambda_1}e^{\lambda_2}...e^{\lambda_n}=e^{ Tr(A)}$

**Bruno J.** · August 18th 2009, 05:20 PM

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 $e^a$ .

**ynj** · August 18th 2009, 05:53 PM

I have changed my proof
every matrix have eigenvalue, and my proof is not based on diagonalibility

**CaptainBlack** · August 18th 2009, 10:09 PM

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

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