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Let $A \succeq 0$ and $B \succeq 0$ be two semi-positive definte matrices. Assume $\{a_{1}, \cdots, a_{n}\}$ is the set of normalized orthogonal basises of matrix $(A-B)$. Let's we define $$ P\triangleq(a_{1}, \ldots,a_{m}), \qquad Q\triangleq(a_{m+{1}},\ldots, a_{n}),$$ in which $(a_{1}, \ldots,a_{m})$ correspnds to $(A-B)$'s nonnegative eigen-values, and $(a_{m+{1}},\ldots, a_{n})$ correspnds to $(A-B)$'s negative eigen-values.

Please proof or disprove

$$\begin{pmatrix}P^{T}BP & P^{T}AQ \\ Q^{T}A P& Q^{T}AQ \end{pmatrix}=\begin{pmatrix}P^{T}BP & P^{T}BQ \\ Q^{T}BP& Q^{T}AQ \end{pmatrix} \succeq 0$$

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  • $\begingroup$ Please share also what you already did on the subject and how far you got. $\endgroup$ – 5xum Feb 7 '14 at 9:25
  • $\begingroup$ I cannot proof its result. I just simulate some matrices that match this result $\endgroup$ – mewmew Feb 7 '14 at 11:24
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Disproving by a randomly generated example:

>> A = [5 6; 6 8];
>> B = [2 -1; -1 5];
>> [V D] = eig(A - B);
>> d = diag(D);
>> P = V(:,find(d >= 0));
>> Q = V(:,find(d < 0));
>> M = [P' * B * P, P' * A * Q; Q' * A * P, Q' * A * Q];
>> eig(M)

ans =

    3.3028
   -0.3028
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