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!