Which will be the largest and which the smallest eigenvalue of inverse of A?

Question

Let $A\in\mathbb R^{n\times n}$ be a symmetric positive definite matrix and $0 < \lambda_1 \le \lambda_2 \le \cdots \le\lambda_n$ be the eigenvalues of $A$.

Which will be the largest and which the smallest eigenvalue of inverse of $A$?

Thank you in advance.

Answer 1

Hint:

If $v$ is an eigenvector of $A$ for the eigenvalue $\lambda$ we have: $$ Av=\lambda v \iff A^{-1}Av=\lambda A^{-1}v \iff A^{-1}v=\frac{1}{\lambda} v $$ so if the eigenvalues of $A$ are $\{\lambda_1 \cdots \lambda_n\}$, the eigenvalues of $A^{-1}$ are the inverses.