Eigenvalues of a matrix in a partial ordering

Question

If $A \succeq B$ for $A, B \in \mathbb{R}^{n \times n}$, then will $\lambda_{i}(A) \geq \lambda_{i}(B)$ for all $i \leq n$? Intuition says yes, but I don't know how to properly show this.

Answer 1

Since $A\succeq B$, then $\dfrac{x^TAx}{x^Tx} \ge \dfrac{x^TBx}{x^Tx}$ for all $x\in \mathbb{R}^n\setminus\{0\}$. In particular, for every subspace $U$ of $\mathbb{R}^n$ of dimension $i \le n$, $$\max_{x\in U\setminus\{0\}} \frac{x^TAx}{x^T x} \ge \max_{x\in U\setminus\{0\}} \frac{x^TBx}{x^Tx} \ge \min_{\dim V = i}\max_{x\in V\setminus\{0\}} \frac{x^TBx}{x^Tx}$$where the minimum is extended over all $i$-dimensional vector subspaces $V$ of $\mathbb{R}^n$. By the minimax principle, the latter expression equals $\lambda_i(B)$. Since $U$ is arbitrary, another application of the minimax principle yields $\lambda_i(A) \ge \lambda_i(B)$.