Calculating Derivatives and Traces to Solving for det(I + tA) = tr(A)

[jakey]

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

 

[radou]

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

 

[jakey]

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

Which definition? Hmm, the co-factor expansion?

 

[radou]

Some people have a thing about Wikipedia articles, but I don't believe it matters right now.

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 t is the identity permutation. After taking the derivative with respect to t of your determinant, and setting t = 0, everything vanishes except the coefficient of t, which equals Tr(A). This can be seen if you try to multiply (t a11 + 1)(t a22 + 1), or (t a11 + 1)(t a22 + 1)(t a33 + 1), etc. i.e. you will always have a term of the form t(a11 + a22 + a33 + ...) generated. All the other polynomials appearing in your sum of permutations are irrelevant, since they all vanish at t = 0, and the constant term is eliminated by taking the derivative itself.

 

[radou]

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

 

[JSuarez]

Note that \det \left(I+tA\right)=\det \left(I-\left(-t\right)A\right) is the characteristic polynomial of A evaluated at -t.

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:

p\left(x\right)=a\left(x-r_1\right)\left(x-r_2\right)\cdots\left(x-r_n\right)

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