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 ${v^T}X^{-1}v \le 1$.
Thanks in advance.  
 A: 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.
A: 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 $1 - v^T X^{-1}v\ge0$.
