$A \preceq MI_{d\times d}$ if and only if the eigenvalues of A are at most M

Question

Let $A \in R^{d\times d}$ be a symmetric matrix. Show that $A \preceq MI_{d\times d}$ if and only if the eigenvalues of A are at most M. Similarly, $mI_{d\times d} \preceq A$ if and only if the eigenvalues of A are at least m.