# 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/… Commented Feb 19, 2016 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? Commented Feb 19, 2016 at 8:17
• The first two paragraphs of the section "Properties" are exactly the proof of your statement. Commented Feb 19, 2016 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 $$\det(I+tX)=(-t)^n\chi(-t^{-1})=\sum_{k=0}^n\sigma_k(X)t^k\text{.}$$