1
$\begingroup$

If the real symmetric matrix

$$M= \begin{bmatrix} \alpha & \textbf{q}^T \\ \textbf{q} & N \end{bmatrix} $$ is positive semidefinite with $\alpha > 0$, then the matrix $$N-\frac 1 \alpha \textbf{q} \textbf{q}^T$$ is also positive semidefinite.

This form reminds me of determinant and the fact that a matrix is PSD iff leading principal minors is non-negative. But I have no idea how to prove the proposition precisely.

  • 2
    https://en.wikipedia.org/wiki/Schur_complement#Schur_complement_condition_for_positive_definiteness_and_positive_semidefiniteness – 2017-02-09
  • 0
    Thanks, this helps a lot. I almost forgot this. – 2017-02-09

1 Answers 1

0

You need to prove that $$x^T \left(N-\dfrac1{\alpha}qq^T\right)x \geq 0$$ for all $x$. We know that since $M$ is positive semidefinite, $y^TMy \geq $ for all $y$. Now take $y=\begin{bmatrix}\beta\\x\end{bmatrix}$, where $\beta$ is a scalar such that $$y^TMy = x^T \left(N-\dfrac1{\alpha}qq^T\right)x$$

  • 0
    How to show such $\beta$ exists? – 2017-02-09
  • 0
    Why don't you multiply out and see what $\beta$ should be? It will turn out to be $-q^Tx/\alpha$. – 2017-02-09