Constructing a Positive Definite Matrix

Question

I've recently been learning some linear algebra in preparation for university next year. I've just read a section on positive definite matrices and found a question that I have no idea how to answer.

We assume that $A$ is an $n\times n$ positive definite matrix. We also assume that $A$ is symmetric. We then take two $n$ dimensional vectors $x,y$ of real numbers such that $x^{T}y>0$. We then construct the matrix $$B=A-\frac{Axx^{T}A^{T}}{x^{T}Ax}+\frac{yy^{T}}{x^{T}y}$$

Is this matrix positive definite?

My initial thought was to try and see if $a^{T}Ba>0$ for some $n$ dimensional vector $a$ of real numbers, but I get pretty stumped with that. Does anyone have any hints?

Answer 1

Observe by Cauchy-Schwarz inequality, we have \begin{align} |\xi^TAx|^2 = |\xi^T A^{1/2}A^{1/2}x|^2 \leq |\xi^TA\xi||x^TAx| \end{align} Then it follows \begin{align} \xi^TB\xi = \xi^TA\xi -\frac{|\xi^TAx|^2}{x^TAx} +\frac{|\xi^Ty|^2}{x^Ty}\geq 0. \end{align} If $\xi$ is not perpendicular to $y$, we see that $\xi^TB\xi>0$. If $\xi$ is orthogonal to $y$, then we have \begin{align} \xi^TB\xi = \xi^TA\xi - \frac{|\xi^TAx|^2}{x^TAx} \end{align} which equals zero if and only if $\xi =\lambda x$. But since $\xi \perp y$, then $\xi$ can't be parallel to $x$. Hence it follows $B$ is indeed positive definite.

Answer 2

Let $z\neq 0$ be a vector $\in \Bbb{R}^n$. One has

$$(x+tz)^TA(x+tz)=x^TAx+t(z^TAx+xAz^T)+t^2z^TAz\geq 0 \forall t\in \Bbb{R}$$

The inequality being strict unless $\exists t_0 x=t_0z$. Now keep in mind that $A^T=A$ that yields $xAz^T=z^TAx$ The discriminant of the quadratic in $t$ is negative and we get

$$\left(z^TAx\right)^2-\left(x^TAx\right)\left(z^TAz\right)\leq 0$$

Dividing by $x^TAx\neq 0$ because $A$ is positive definite, keeping in mind that $(z^TAx)^T=z^TAx$ (it is a real)

$$\left(z^TAz\right)\geq {z^TAxx^TAz\over x^TAx}$$

Now ${z^Tyy^Tz\over x^Ty}\gt 0$ and we have proven that

$$z^TBz\gt 0$$