# Elementary symmetric polynomial related to matrices

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!

• True. See here, for example: en.wikipedia.org/wiki/… – Hans Lundmark Feb 19 '16 at 8:10
• @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? – C Weid Feb 19 '16 at 8:17
• The first two paragraphs of the section "Properties" are exactly the proof of your statement. – Hans Lundmark Feb 19 '16 at 11:32

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 $$\det(I+tX)=(-t)^n\chi(-t^{-1})=\sum_{k=0}^n\sigma_k(X)t^k\text{.}$$