Determinant and symmetric function of eigenvalues

Question

Let $H$ be an $n$-dimensional vector space and $T$ a linear operator on it with eigenvalues $\lambda_{i_1},\dots,\lambda_{i_n}$. Let $I$ be the identity operator and $z\in\mathbb{C}$. How does one prove $$\det(I+zT)=\sum\limits_{k=0}^n z^k \sum_{i_1<\cdots<i_k}\lambda_{i_1}(T)\cdots \lambda_{i_k}(T)$$ The right sight came from the exterior product.

Answer 1

If we are allowed to assume that we know that the coefficients of the polynomial are symmetric polynomials of the eigenvalues (taken as distinct), then we can show this. Writing $z=-1/\lambda$, the right-hand side is $$ \det(I+zT) =z^n\det\left(T+\frac{1}{z}I\right) = \frac{(-1)^n}{\lambda^n}\det(T-\lambda I) =\frac{(-1)^n}{\lambda^n}(\lambda_1-\lambda)(\lambda_2-\lambda)\cdots(\lambda_n-\lambda)\,, $$ where $\lambda_n$ are the eigenvalues of $T$, taken as distinct. Then, \begin{align} \det(I+zT) &=\frac{(-1)^n}{\lambda^n}\sum_{k=0}^n\lambda^k(-1)^k e_{n-k}(\lambda_1,\dots,\lambda_n)\\ &=\sum_{k=0}^n\lambda^{k-n}(-1)^{k-n} e_{n-k}(\lambda_1,\dots,\lambda_n)\\ &=\sum_{k=0}^n(-\lambda)^{-k} e_{k}(\lambda_1,\dots,\lambda_n) \end{align} where $e_k(x_1,\dots,x_n)$ is the $k$'th elementary symmetric polynomial in the variables $x_1,\dots,x_n$. The symmetric polynomials have the alternative representation $$ e_k(x_1,\dots,x_n)=\sum_{1\leq j_1<\cdots<j_k\leq n}x_{j_1}\cdots x_{j_k}\,. $$ Finally, then, putting back $z$, we have \begin{align} \det(I+zT) &=\sum_{k=0}^n(-\lambda)^{-k} e_{k}(\lambda_1,\dots,\lambda_n)\\ &=\sum_{k=0}^nz^k e_{k}(\lambda_1,\dots,\lambda_n)\, \end{align} which is exactly the expression in the OP.

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To prove that the expansion of the product of $(\lambda_j-\lambda)$'s can be written in terms of the elementary symmetric polynomials, probably proof by induction would do it.