Matrix positive definite Let $A$ and $B$ be both symmetric $ n \times n$ matrices, and $B \succ 0$;
$U$ be one $n \times q$ column orthogonal matrix ($n &gt; q$).
Assume $$ 0 \preceq U^{T} A U \preceq U^{T} B U,$$
do we have the following inequality $$UU^{T}AUU^{T} \preceq B.$$
Nope. Let 
$$
A=\begin{bmatrix}5 &amp; 3 \\ 3 &amp; 2\end{bmatrix},
\quad
B=\begin{bmatrix}5 &amp; 1 \\ 1 &amp; 1\end{bmatrix}
$$
(both are SPD). Let $U=[1,0]^T$. Then
$$
0\leq 5=U^TAU\leq U^TBU=5.
$$
But
$$
B-UU^TAUU^T=\begin{bmatrix}0 &amp; 1 \\ 1 &amp; 1\end{bmatrix},
$$
which is indefinite.
Yes.  Just premultiply by $U^T$ and postmultiply by $U$ after bringing $B$ over to the other side.  If $C \leq 0$ then $U^T C U\leq 0$ also.