# To prove a matrix is PSD

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?

• Take a look at the Schur complement. – A.Γ. Aug 17 at 9:08
• Indeed, a Schur's complement of a PSD is itself PSD. – Jean Marie Aug 17 at 9:19
• @A.Γ. Thanks! Got it. – 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 $$y^T\left(N-\frac{1}{\alpha}qq^T\right)y\ge 0,\qquad\forall y.$$