Let $A \in M \left( n\times n, \mathbb{C} \right)$ then
$$ -\frac{d}{dt} \log \det(E_n-tA) = \sum_{m=1}^\infty \mathrm{trace}(A^m)t^{m-1} $$
How do I show that the equation is true? Do I need something like jordan normal form?
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One can conjugate $A$ to put it in triangular form with diagonal elements $a_1,\ldots,a_n$ say. Then $$\det(A-tI)=(a_1-t)\cdots(a_n-t)$$ and logarithmic differentiation gives $$-\frac d{dt}\det(A-tI)=\frac1{t-a_1}+\cdots+\frac1{t-a_n}.$$ This doesn't seem to correspond to your RHS though.
ADDED IN EDIT
If all $|a_i|<1$ then $$\sum_1^\infty\text{trace}(A^m)t^{m-1}=\sum_i\sum_1^\infty a_i^mt^{m-1}=\frac{a_1}{1-a_1t}+\cdots+\frac{a_n}{1-a_nt}$$ which isn't the LHS. So, what should the LHS be?
Angina Seng
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I also got to this point but it is a exercise in university so I guess, it must be true. – fpmoo Jun 14 '17 at 06:16
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Why must it be true? – Angina Seng Jun 14 '17 at 06:21
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It says: Proof that the following equation is correct. – fpmoo Jun 14 '17 at 06:25
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Okay. Sorry you are right. – fpmoo Jun 14 '17 at 06:27
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Denote $B=E_n - tA$ and use the formula for the derivative of $\log\det$ $$ d_t(\log\det B_t) =\operatorname{trace}\left(B_t^{-1}\;dB_t\right), $$
Insert $dB_t=-A$ and adapt the Neumann series formula for the inverse (replace $A$ with $tA$): $$(I-A)^{-1}=I+A+A^{2}+A^{3}+\ldots$$
rych
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