 Orthogonal null-space completion of a square matrix

Let $A$ be a real $n\times n$ matrix of matrix rank $n-m$, and let the columns of $L,R$ be bases of $\ker A^{\top }$ and $\ker A$, respectively. Then $B=L{R}^{\top }$ has matrix rank $m$, $A^{\top }B=A{B}^{\top }=0$, and $M=A+B$ is an invertible matrix.

Proof. Since $R$ has independent columns, its Gram matrix $R^{\top }R$ is invertible, and $R^{\top }$ maps onto ${\mathbb{R}}^m$. Thus $\mathrm{im}B=\mathrm{im}L=\ker A^{\top }$ and $\mathrm{rank}B=m$. The basis conditions give the two zero products. In particular,If $M{M}^{\top }x=0$, taking its dot product with $x$ gives $\Vert A^{\top }x{\Vert }^2+\Vert B^{\top }x{\Vert }^2=0$. Hence $x\in \ker A^{\top }=\mathrm{im}B$ and simultaneously $x\in \ker B^{\top }=(\mathrm{im}B{)}^{\perp }$. Therefore $x=0$, and $M$ is invertible. The case $m=0$ is immediate.

For a common eigenvector of $A{A}^{\top }$ and $B{B}^{\top }$, the two eigenvalues have zero product and positive sum: exactly one is zero. Thus the added term fills the missing orthogonal output directions without interfering with the original ones.