Let $M$ be an $n \times n$ block matrix defined as $$M = \begin{bmatrix} A & \mathbf{b} \\ \mathbf{b}^T & 1 \end{bmatrix}$$ where $A$ is an invertible and symmetric $(n-1) \times (n-1)$ matrix, and $\mathbf{b}$ is a column vector of size $(n-1)$. If $M$ is positive semidefinite, what is an equivalent expression in terms of $A$ and $\mathbf{b}$ for this condition to be satisfied?
My guess is that $A$ is positive semidefinite and $1 – \mathbf{b}^T A^{-1} \mathbf{b} \geq 0$. But I am not arriving to prove it. Can anyone clarify if it's true and why?
Best Answer
For invertible $A$ the proof is easy. It suffices to show the positive definitess of the matrix $$\begin{pmatrix} A^{-1/2} & 0\\ 0 & 1\end{pmatrix}\begin{pmatrix} A & b\\ b^T & 1\end{pmatrix}\begin{pmatrix} A^{-1/2} & 0\\ 0 & 1\end{pmatrix}=\begin{pmatrix} I & A^{-1/2}b\\ b^TA^{-1/2} & 1\end{pmatrix}$$ The last matrix is positive definite iff $I-b^TA^{-1}b>0.$ For nonivertible $A$ we can replace $A$ with $A+n^{-1}I$ and apply the limit procedure. It should work, I hope.