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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!

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  • $\begingroup$ True. See here, for example: en.wikipedia.org/wiki/… $\endgroup$ Commented Feb 19, 2016 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
    Commented Feb 19, 2016 at 8:17
  • $\begingroup$ The first two paragraphs of the section "Properties" are exactly the proof of your statement. $\endgroup$ Commented Feb 19, 2016 at 11:32

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

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