Let $A$ and $B$ be both symmetric $ n \times n$ matrices, and $B \succ 0$; $U$ be one $n \times q$ column orthogonal matrix ($n > q$). Assume $$ 0 \preceq U^{T} A U \preceq U^{T} B U,$$ do we have the following inequality $$UU^{T}AUU^{T} \preceq B.$$
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Nope. Let $$ A=\begin{bmatrix}5 & 3 \\ 3 & 2\end{bmatrix}, \quad B=\begin{bmatrix}5 & 1 \\ 1 & 1\end{bmatrix} $$ (both are SPD). Let $U=[1,0]^T$. Then $$ 0\leq 5=U^TAU\leq U^TBU=5. $$ But $$ B-UU^TAUU^T=\begin{bmatrix}0 & 1 \\ 1 & 1\end{bmatrix}, $$ which is indefinite.
Algebraic Pavel
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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.
JPi
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I'm not vert undertand your expression. Could you clarify more detali? – mewmew Feb 07 '14 at 03:22
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But my question is juse reverse. Knowing $U^{T}CU \leq 0$, we need proof $C\leq 0$. – mewmew Feb 07 '14 at 05:50