# Positive definite matrix inequality

Can someone help me with the following proof involving positive definite matrices:

Suppose $X\succ 0$ positive definite. Show that $X-v{v^T}\succ 0$ if and only if ${v^T}X^{-1}v \le 1$.

-
Could you write the latex code correctly? – Alen Feb 2 '13 at 2:17
Is the statement for all v? – Inquest Feb 2 '13 at 2:30
@Inquest It was not specified in the question, but I believe yes. – xbl Feb 2 '13 at 2:32
@Alen I've reedited, but how do I flip the " $\prec$ " to go the other way? – xbl Feb 2 '13 at 2:35
$\succ$ is the symbol to use ( \succ ) – Alen Feb 2 '13 at 2:35

Here is another approach. I assume $X$ is symmetric.

To simplify life, all square matrices below are assumed symmetric.

Since $X>0$ it has a square root satisfying $X= (X^{\frac{1}{2}})^2$.

Note that if $B$ is invertible, then $A>0$ iff $BAB>0$. Also note that $I-u u^T>0$ iff $\|u\| <1$. To see the latter, note that $u$ is an eigenvector corresponding to the eigenvalue $1-\|u\|^2$, and all other eigenvalues are $1$.

Hence $X-v v^T >0$ iff $I-(X^{-\frac{1}{2}} v)(X^{-\frac{1}{2}} v)^T >0$ iff $\|X^{-\frac{1}{2}}v\|^2 < 1$.

Since $\|X^{-\frac{1}{2}}v\|^2 = v^T X^{-1} v$, we are finished.

-

I prefer copper.hat's approach, but it doesn't hurt to view the problem from other perspectives.

The assertion in the problem statement is obvious if $v=0$. So, assume $v\not=0$. Since $X$ is positive definite, it defines an inner product $\langle u,w\rangle=w^TXu$ on $\mathbb{R}^n$. Therefore, every nonzero vector $w\in\mathbb{R}^n$ can be written as $w=a(X^{-1}v)+bu$, where $(a,b)\not=(0,0)$ and $0\not=u\perp (X^{-1}v)$ w.r.t. the aforementioned inner product. That is, $0=\langle X^{-1}v,u\rangle=u^TX(X^{-1}v)=u^Tv$. So, \begin{align*} &\phantom{=} w^T(X - vv^T)w\\ &=(aX^{-1}v + bu)^T X (aX^{-1}v + bu) - (aX^{-1}v + bu)^T vv^T(aX^{-1}v + bu)\\ &=a^2 (v^T X^{-1}v)(1 - v^T X^{-1}v) + b^2 u^T Xu. \end{align*} Hence $X-vv^T$ is positive semidefinite if and only if $1 - v^T X^{-1}v\ge0$.

-