In page 32 and page 33 of Boyd's convex optimization book, he said there exists a nonsingular matrix $A = (A_1, A_2) \in \mathbb{R}^{n \times n}$, such that
$AB = \begin{bmatrix}A_1 \\ A_2\end{bmatrix}B = \begin{bmatrix}I \\ 0\end{bmatrix}$
I'm not sure how $A$ can be nonsingular because as $B$ contains linearly independent vectors, to get a zero on the RHS, then an entire row of $A$ will b 0. And also, why such $A$ exist that $A_1 B = I$? $B$ is not even a square matrix.
Note that $B \in \mathbb{R}^{n \times k}$, with $k \leq n$, and $B$ is assumed to be full rank.
This means that the columns of $B$ are linearly independent. Therefore, we can choose $A_1 = B^{\dagger} \in \mathbb{R}^{k \times n}$ (the Moore-Penrose pseudoinverse of $B$). When $B$ has linearly independent columns we have
$$ B^{\dagger} B = \underbrace{(B^* B)^{-1} B^*}_{B^{\dagger}} B = I_{k}. $$
For $A_2$, we can choose any matrix in $\mathbb{R}^{(n - k) \times n}$ whose rows span the orthogonal complement of the column span of $B$.