0
$\begingroup$

I've encountered with the following question while reading something about invariant polynomial in Chern-Weil theory:

For a matrix $X \in M(n;\Bbb{R})$ ,denote its eigenvalues by $\lambda_1,...,\lambda_n$,and the $n$ symmetric polynomials by $\sigma_1,...\sigma_n$,that is $$\sigma_1(X)=\lambda_1+...+\lambda_n$$ $$\sigma_2(X)=\lambda_1\lambda_2+...+\lambda_{n-1}\lambda_n=\sum_{i\lt_j}\lambda_i\lambda_j$$$$...$$ $$\sigma_n(X)=\lambda_1\lambda_2...\lambda_n$$

Is it true or false that $$\det(I+tX)=1+t\sigma_1(X)+t^2\sigma_2(X)+...+t^n\sigma_n(X)$$

I've checked the case when $n=2$ and $n=3$ ,but how can we prove it in general?

Much appreciated!

$\endgroup$
  • $\begingroup$ True. See here, for example: en.wikipedia.org/wiki/… $\endgroup$ – Hans Lundmark Feb 19 '16 at 8:10
  • $\begingroup$ @HansLundmark I know the fundamental theorem of symmetric polynomials but,I still don't know how to prove the above prop.,is it really useful for this problem? could you be more specific? $\endgroup$ – C Weid Feb 19 '16 at 8:17
  • $\begingroup$ The first two paragraphs of the section "Properties" are exactly the proof of your statement. $\endgroup$ – Hans Lundmark Feb 19 '16 at 11:32
0
$\begingroup$

You can actually just read this from the characteristic polynomial $$\chi(t)=\det(tI-X)=\prod_{k=1}^n(t-\lambda_k)=\sum_{k=0}^n(-1)^k\sigma_k(X)t^{n-k}\text{,}$$ because then your polynomial is $$\det(I+tX)=(-t)^n\chi(-t^{-1})=\sum_{k=0}^n\sigma_k(X)t^k\text{.}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.