Relationship of eigenvalues if $\| Ax \| \le \| Bx\|$, where $A$ and $B$ are symmetric positive definite If symmetric positive definite matrices $A$ and $B$ in $\mathbb{R}^{n\times n}$ satisfy
$$
\| Ax \| \le \| Bx \|,
$$
for any $x\in\mathbb{R}^n$, then do we get $\lambda_i(A) \le \lambda_i(B)$ for all $i$, where $\lambda_i(A), \lambda_i(B)$ are the eigenvalues of $A$ and $B$ respectively? It seems intuitively the case, but I could be wrong.
 A: If $A$ and $B$ are symmetric positive semidefinite projection matrices, then each has a representation as $UDU^{*},$ where $U$ is unitary and $D$ is a diagonal matrix with ones and zeros along the diagonal. Then $\|Ax\|\leq \|Bx\|$ is equivalent to $x^{*}(B^{2}-A^{2})x\geq0$ for all $x\in\mathbb{R}^{n},$ and using the fact that $B^{2}=B,$ $A^{2}=A,$ this says that $A\preceq B$. Then in particular, for any vector $v$ in the subspace onto which $A$ projects (which is the span of the columns of $U_{A}$ corresponding to the diagonal entries of $D_{A}$ with value $1$), $\|Av\|=\|v\|,$ and since $\|Bv\|\leq\|B\|\|v\|\leq\|v\|,$ we see that $\|Bv\|=\|v\|,$ which means that the subspace onto which $A$ projects is contained in the subspace onto which $B$ projects.
Put differently, if $U_{A}=[U_{+},U_{0}],$ where $U_{+}$ are the columns of $U_{A}$ corresponding to diagonal entries of $D_{A}=1,$ and $U_{0}$ are the remaining columns of $U_{A}$ (note that there is no loss of generality here, since we have $U_{A}D_{A}U_{A}^{T}=(U_{A}P)(P^{T}D_{A}P)(U_{A}P)^{T}$ for any permutation matrix), then $B=U_{+}U_{+}^{T}+VV^{T},$ where $V$ is a matrix with orthonormal columns spanning $\mathrm{Ran}(B)\cap\mathrm{Ran}(U_{+})^{\perp}$, which is contained in $\mathrm{Ran}(U_{0})=\mathrm{Ran}(U_{+})^{\perp}.$ Conversely any matrix $B=U_{+}U_{+}^{T}+VV^{T},$ where $V$ is a matrix with orthonormal columns such that $\mathrm{Ran}(V)\subseteq\mathrm{Ran}(U_{0}),$ will satisfy $\|Ax\|\leq \|Bx\|$ for all $x\in\mathbb{R}^{n},$ since $\|Ax\|=\|U_{+}^{T}x\|,$ and $\|Bx\|=\|U_{+}^{T}x\|+\|V^{T}x\|\geq \|Ax\|.$ Thus, $$\|Ax\|\leq \|Bx\|\text{ for all }x\in\mathbb{R}^{n}\Leftrightarrow \mathrm{Ran}(A)\subseteq\mathrm{Ran}(B),$$ whenever $A$ and $B$ are symmetric positive semidefinite projection matrices, or equivalently, symmetric projection matrices.
