Rank-One Lower Bound for the Inverse of a Positive Definite Matrix

Let $k\ge1$ be a natural number and let $J$ be a positive definite real matrix with $k$ rows and $k$ columns. Throughout, $\mathbb{R}^{k}$ is Euclidean space, $x\cdot y$ denotes the dot product, $Ax$ the matrix-vector product, $AB$ the matrix product and $A^{\top}$ the transpose. For $y\in\mathbb{R}^{k}$ let $y\otimes y$ be the real matrix with $k$ rows and $k$ columns and entries $(y\otimes y)_{\gamma\delta}=y^{\gamma}y^{\delta}$, which is symmetric because $y^{\gamma}y^{\delta}=y^{\delta}y^{\gamma}$. By Invertibility of Symmetric Positive Definite Matrices the inverse $J^{-1}$ exists and is symmetric positive definite. Let $z\in\mathbb{R}^{k}$ be nonzero.

Then the following hold.

1. (Positivity of the direction.) $z\cdot(Jz)>0$.

2. (Scalar form.) For every $x\in\mathbb{R}^{k}$,

$$(x\cdot z)^{2}\ \le\ \bigl(z\cdot(Jz)\bigr)\,\bigl(x\cdot(J^{-1}x)\bigr).$$

3. (Matrix form.) In the semidefinite order,

$$J^{-1}\ \succeq\ \frac{1}{z\cdot(Jz)}\,(z\otimes z).$$