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Given any matrix A, how can I determine whether the following system is neither positive semi definite nor negative semi definite?

\begin{pmatrix} I & A \\ A^T & 0 \\ \end{pmatrix}

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If $A$ is invertible, you can use the Schur complement

One can easily show that for $$X= \begin{pmatrix} A & B \\ B^T & C \\ \end{pmatrix}$$ $X$ is positive (semi-)definite if and only if both $A$ and $S$ are positive (semi-)definite where $S=C-B^TAB^{-1}$ is the Schur complement.

In your case, $$X=\begin{pmatrix} I & A \\ A^T & 0 \\ \end{pmatrix}$$

and $S=-A^TA^{-1}$.

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  • $\begingroup$ Thanks! Does the same logic hold for the negative case? (ie, X is negative semidefinite only iff both A and S are negative semidefinite?) $\endgroup$ – Theo Mar 17 '15 at 0:25
  • $\begingroup$ @Theo Yes. See mathoverflow.net/questions/118053/… $\endgroup$ – Nitish Mar 17 '15 at 0:32

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