- #1
Demon117
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Homework Statement
Prove that for any matrix [itex]A[/itex], the following relation is true:
[itex]det(e^{A})=e^{tr(A)}[/itex]
The Attempt at a Solution
PROOF: Let [itex]A[/itex] be in Jordan Canonical form, then
[itex]A=PDP^{-1}[/itex]
where [itex]D[/itex] is the diagonal matrix whose entries are the eigenvalues of [itex]A[/itex]. Then,
[itex]e^{A}=Pe^{D}P^{-1}[/itex]
By application of the determinant operator we have
[itex]det(e^{A})=det(Pe^{D}P^{-1})=det(P)det(e^{D})det(P^{-1})=det(e^{D})[/itex]
Since the diagonal matrix has eigenvalue elements along its main diagonal, it follows that the determinant of its exponent is given by
[itex]det(e^{D})=e^{\lambda_{1}} \cdot e^{\lambda_{2}} \cdot \cdot \cdot e^{\lambda_{n}}[/itex]
By simple algebra the product of the exponents is the exponent of the sum, so
[itex]det(e^{D})=e^{\lambda_{1}+\lambda_{2}+ \cdot \cdot \cdot +\lambda_{n}}[/itex]
This of course is simply the exponent of the trace of [itex]D[/itex]. Therefore,
[itex]det(e^{A})=e^{tr(D)}[/itex]
Now, this is where I get messed up. My question is simply this: Because [itex]A[/itex] is similar to [itex]D[/itex], does it follow that the trace of [itex]D[/itex] is the same as the trace of [itex]A[/itex]? I heard this somewhere but I cannot verify this statement anywhere in my notes or textbooks. If this is a true statement, then the proof is complete.