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Say I have $A \succcurlyeq 0$, $ \begin{bmatrix}A & \vec{b}_1 \\\ \vec{b}_1^T & t_1 \end{bmatrix} \succcurlyeq 0 $, and $ \begin{bmatrix}A & \vec{b}_2 \\\ \vec{b}_2^T & t_2 \end{bmatrix} \succcurlyeq 0 $.

Is $ \begin{bmatrix}A & [\vec{b}_1, \vec{b}_2] \\\ [\vec{b}_1, \vec{b}_2]^T & \operatorname{diag}([t_1, t_2]) \end{bmatrix} \succcurlyeq 0 $?

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No. For example $\begin{bmatrix} 1 & 1\\ 1 & 1 \end{bmatrix}\succeq 0$ and $\begin{bmatrix} 1 & 1&1\\ 1&1&0\\ 1&0&1\end{bmatrix}$ is not.

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