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

Claim 1. Since $J$ is positive definite and $z$ is nonzero, $z\cdot(Jz)>0$ by the definition of positive definiteness.

Two elementary identities. Let $M$ be a real matrix with $k$ rows and $k$ columns and let $u,v\in\mathbb{R}^{k}$. By the index formulas for the matrix-vector product and the dot product,

from which two facts follow. First, the expression is linear in $u$ for fixed $v$ and linear in $v$ for fixed $u$. Second, if $M$ is symmetric, so that $M_{\delta\gamma}=M_{\gamma\delta}$ by the definition of the transpose, then exchanging the names of the two summation indices gives $u\cdot(Mv)=v\cdot(Mu)$.

Claim 2. Fix $x\in\mathbb{R}^{k}$. By claim 1 the real number $\theta=(x\cdot z)/\bigl(z\cdot(Jz)\bigr)$ is well defined; put $y=x-\theta\,Jz$. The matrix $J^{-1}$ is symmetric positive definite, hence positive semidefinite, so $y\cdot(J^{-1}y)\ge0$. Expanding by the bilinearity just recorded,

By the defining property of the inverse, $J^{-1}J=I_{k}$ with $I_{k}$ the identity matrix, and the matrix product satisfies $(J^{-1}J)z=J^{-1}(Jz)$ by the associativity of the index sums; hence $J^{-1}(Jz)=z$. The second term is therefore $-\theta\,(x\cdot z)$ and the fourth is $\theta^{2}\,(Jz)\cdot z=\theta^{2}\,z\cdot(Jz)$, the dot product being symmetric. For the third, the second identity above applied to the symmetric matrix $J^{-1}$ gives $(Jz)\cdot(J^{-1}x)=x\cdot\bigl(J^{-1}(Jz)\bigr)=x\cdot z$, so the third term is $-\theta\,(x\cdot z)$ as well. Consequently

the last equality by the choice of $\theta$. Multiplying by $z\cdot(Jz)>0$ gives claim 2.

Claim 3. Put

a difference of symmetric matrices formed entrywise, hence symmetric. For $x\in\mathbb{R}^{k}$ the index formulas give

so that, by the linearity in the matrix argument evident from the same formula and by claim 2,

Thus $M$ is positive semidefinite, which by the definition of the semidefinite order is exactly claim 3.