Given two symmetric matrices A and B with same dimensions, if $\lambda_{max}(A)$ is biggest eigenvalue of A, $\lambda_{min}(B)$ is the smallest eigenvalue of B, does the following holds true:
$$A \leq B \Rightarrow \lambda_{max}(A)\leq \lambda_{min}(B) \ \ ?$$
Best Answer
$\lambda_{max}(A) \leq \lambda_{min}(B)$ is a sufficient property for the so-called "Loewner order".
See (https://en.wikipedia.org/wiki/Loewner_order)
But is is not a necessary condition. Here is a counterexample; take:
$$A=\pmatrix{1&1\\1&2} \ \ \text{and} \ \ B=\pmatrix{2&2\\2&3}.$$
We have: $A \leq B$ for the Loewner order because $B-A=\pmatrix{1&1\\1&1}$ is semi-positive-definite (with eigenvalues $0$ and $2$).
The spectra of $A$ and $B$ are resp.
$$\{0.3819,2.6180\} \ \ \text{and} \ \ \{0.4384, 4.5616\}.$$
but without having $\lambda_{max}(A) \leq \lambda_{min}(B).$