Demon117
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Homework Statement
Prove that for any matrix A, the following relation is true:
det(e^{A})=e^{tr(A)}
The Attempt at a Solution
PROOF: Let A be in Jordan Canonical form, then
A=PDP^{-1}
where D is the diagonal matrix whose entries are the eigenvalues of A. Then,
e^{A}=Pe^{D}P^{-1}
By application of the determinant operator we have
det(e^{A})=det(Pe^{D}P^{-1})=det(P)det(e^{D})det(P^{-1})=det(e^{D})
Since the diagonal matrix has eigenvalue elements along its main diagonal, it follows that the determinant of its exponent is given by
det(e^{D})=e^{\lambda_{1}} \cdot e^{\lambda_{2}} \cdot \cdot \cdot e^{\lambda_{n}}
By simple algebra the product of the exponents is the exponent of the sum, so
det(e^{D})=e^{\lambda_{1}+\lambda_{2}+ \cdot \cdot \cdot +\lambda_{n}}
This of course is simply the exponent of the trace of D. Therefore,
det(e^{A})=e^{tr(D)}
Now, this is where I get messed up. My question is simply this: Because A is similar to D, does it follow that the trace of D is the same as the trace of A? 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.