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

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

where $\displaystyle Tr$ is the trace.

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- Aug 18th 2009, 03:22 PMBruno 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 PMynj
- Aug 18th 2009, 03:40 PMBruno J.
$\displaystyle A$ is a matrix with complex entries.

- Aug 18th 2009, 03:46 PMynj
- Aug 18th 2009, 04:12 PMBruno J.
- Aug 18th 2009, 04:56 PMynj
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 PMBruno 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 PMynj
I have changed my proof

every matrix have eigenvalue, and my proof is not based on diagonalibility - Aug 18th 2009, 10:09 PMCaptainBlack
See here (2/3rd of the way down)

CB