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## Main Question or Discussion Point

Hey guys, any hints on how to show that [tex]\frac{d}{dt}[/tex]

_{|t=0}[tex]det(I + tA) = tr(A) [/tex]? I did it for 2x2 but I can't figure out a generalization. Thanks- Thread starter jakey
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- #1

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Hey guys, any hints on how to show that [tex]\frac{d}{dt}[/tex]_{|t=0} [tex]det(I + tA) = tr(A) [/tex]? I did it for 2x2 but I can't figure out a generalization. Thanks

- #2

radou

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Did you try to use the definition of the determinant to conclude something?

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Which definition? Hmm, the co-factor expansion?Did you try to use the definition of the determinant to conclude something?

- #4

radou

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http://en.wikipedia.org/wiki/Leibniz_formula_for_determinants

So, after examining this definition (which may seem a bit complicated at the first glance, and if there is an easier way to prove it, it would be nice) you can conclude that the only permutation which generates a polynomial with a non-zero coefficient for

- #5

radou

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Oh yes, and note that the identity permutation is an even one.

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Now, for any polynomial, the t term is the symmetric of the sum of its roots; you may see this by looking at its factorization:

[tex]p\left(x\right)=a\left(x-r_1\right)\left(x-r_2\right)\cdots\left(x-r_n\right)[/tex]

So, when you differentiate [itex]p\left(-t)[/itex] and set [itex]t=0[/itex], what will be the only term left? And how is the trace of [itex]A[/itex] related to its eigenvalues?

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