Is Loewner order equivalent to inequality of eigenvalues? I know from this previous question that if $A \geq B$ then $\lambda_k(A) \geq \lambda_k(B)$, where $\lambda_k$ denotes the $k$th ordered eigenvalue.
Now I would like to know if the reverse holds, i.e., if the Loewner order can be defined in terms of inequalities between eigenvalues.
Any book or reference where this issue is explained in detail would be appreciated.
 A: The reverse does not hold. To show this it can be noted that the inequality $A \geq B$ means that the eigenvalues of $A-B$ have to be greater or equal to zero. So the following matrices do satisfy the ordered eigenvalue inequalities
$$
A =
\begin{bmatrix}
4 & 0 \\ 0 & 2
\end{bmatrix}, \quad
B =
\begin{bmatrix}
1 & 0 \\ 0 & 3
\end{bmatrix}.
$$
Namely $\lambda_1(A)=4 > \lambda_1(B)=3$ and $\lambda_2(A)=2 > \lambda_2(B)=1$. However, the difference between the matrices is not positive semi-definite
$$
A - B =
\begin{bmatrix}
3 & 0 \\ 0 & -1
\end{bmatrix}.
$$
A: As noted in Kwin van der Veen's answer, the converse is not true as stated. However, order of the eigenvalues can be characterized in terms of the Löwner order:
Let $A,B\in M_n(\mathbb C)$ be hermitian matrices with eigenvalues $\lambda_1(A)\leq \dots\leq\lambda_n(A)$ and $\lambda_1(B)\leq\dots\leq\lambda_n(B)$ (counted with multiplicity, of course). There exists a unitary matrix $U\in M_n(\mathbb C)$ such that $A\leq U^\ast B U$ if and only if $\lambda_k(A)\leq\lambda_k(B)$ for all $k\in \{1,\dots,n\}$.
Since $B$ and $U^\ast B U$ have the same eigenvalues with the same multiplicities, one implication follows from the post linked by the OP. For the other implication note by the spectral theorem there exist unitary matrices $V,W\in M_n(\mathbb C)$ such that
$$
\begin{pmatrix}\lambda_1(A)&&\\&\ddots&\\&&\lambda_n(A)\end{pmatrix}=VAV^\ast,\;\begin{pmatrix}\lambda_1(B)&&\\&\ddots&\\&&\lambda_n(B)\end{pmatrix}=WBW^\ast.
$$
If $\lambda_k(A)\leq \lambda_k(B)$ for all $k\in\{1,\dots,n\}$, then $
VAV^\ast\leq WB W^\ast$, which implies $A\leq V^\ast W B W^\ast V$. Then $U=W^\ast V$ does the job.
