# Trace and derivatives

1. Mar 30, 2010

### 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

2. Mar 30, 2010

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

3. Mar 30, 2010

### jakey

Which definition? Hmm, the co-factor expansion?

4. Mar 30, 2010

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.

5. Mar 30, 2010

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

6. Mar 30, 2010

### 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?