This question rises from the proof of the outer product Cholesky Factorization.

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

I have proved that the matrix $A$ is symmetric, which is easy, but I don’t know how to prove it is PSD. Any hints?

  • 3
    $\begingroup$ Take a look at the Schur complement. $\endgroup$ – A.Γ. Aug 17 at 9:08
  • $\begingroup$ Indeed, a Schur's complement of a PSD is itself PSD. $\endgroup$ – Jean Marie Aug 17 at 9:19
  • $\begingroup$ @A.Γ. Thanks! Got it. $\endgroup$ – UnbelieveTable Aug 17 at 12:08

By definition, $M$ is PSD, hence $$ \begin{bmatrix}x\\y\end{bmatrix}^TM\begin{bmatrix}x\\y\end{bmatrix}= \alpha x^2+2x\cdot q^Ty+y^TNy\ge 0,\qquad\forall x,y\tag{*} $$ Complete the squares in (*) $$ \alpha\left(x+\frac{1}{\alpha}q^Ty\right)^2+y^T\left(N-\frac{1}{\alpha}qq^T\right)y\ge 0,\qquad\forall x,y. $$ Take $x=-\frac{1}{\alpha}q^Ty$ to get $$ y^T\left(N-\frac{1}{\alpha}qq^T\right)y\ge 0,\qquad\forall y. $$


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