Strict positive and conditionally positive augmented interpolation systems are invertible

In the strictly positive definite case, if $K_{X}a=0$, then $a^{\mathsf{T}}{K}_{X}a=0$, so [F1] forces $a=0$. Thus the linear map $a\mapsto K_{X}a$ has zero kernel.

In the augmented case, put $M=\left[\begin{array}{cc}{K}_X& P_X\\ P_X^{\mathsf{T}}& 0\end{array}\right]$ and suppose $M\left[\begin{array}{c}a\\ b\end{array}\right]=0$. The lower block gives $P_X^{\mathsf{T}}a=0$, and multiplying the upper block by $a^{\mathsf{T}}$ gives $a^{\mathsf{T}}{K}_{X}a+a^{\mathsf{T}}{P}_{X}b=0$. Since $a^{\mathsf{T}}{P}_X=(P_X^{\mathsf{T}}a{)}^{\mathsf{T}}=0$, strict conditional positivity [F2] forces $a=0$.

By [F3], the map $a\mapsto K_{X}a$ has rank $N$, so it is onto ${\mathbb{R}}^N$ as well as one-to-one. For the standard basis $e_1,\dots ,e_N$, choose $v_j$ with $K_X{v}_j=e_j$ and let $B$ have columns $v_j$. Then $K_{X}B=I_N$; for every $x$, injectivity of $K_X$ applied to $K_X(B{K}_{X}x-x)=0$ gives $B{K}_{X}x=x$. Thus $B$ is a two-sided inverse by [F5], and every system $K_{X}a=y$ has a unique solution.

With $a=0$, the upper block reduces to $P_{X}b=0$. The columns of $P_X$ are linearly independent by [F2], so $b=0$. The augmented square matrix therefore has zero kernel, and [F3] gives rank $N+J$ and surjectivity. For the standard basis of ${\mathbb{R}}^{N+J}$, choose the unique preimage of each basis vector and use those preimages as columns of a matrix $B$. Then $MB=I_{N+J}$; injectivity gives $BM=I_{N+J}$ by applying $M$ to $BMx-x$. Hence $M$ is invertible by [F5], and its system has a unique solution for every right-hand side.