Proof of Theorem CSNM

(⇒) Suppose $A$ is nonsingular. We wish to establish the set equality $\csp{A}=\complex{n}$. By Definition CSM, $\csp{A}\subseteq\complex{n}$. To show that $\complex{n}\subseteq\csp{A}$ choose $\vect{b}\in\complex{n}$. By Theorem NMUS, we know the linear system $\linearsystem{A}{\vect{b}}$ has a (unique) solution and therefore is consistent. Theorem CSCS then says that $\vect{b}\in\csp{A}$. So by Definition SE, $\csp{A}=\complex{n}$.

(⇐) If $\vect{e}_i$ is column $i$ of the $n\times n$ identity matrix (Definition SUV) and by hypothesis $\csp{A}=\complex{n}$, then $\vect{e}_i\in\csp{A}$ for $1\leq i\leq n$. By Theorem CSCS, the system $\linearsystem{A}{\vect{e}_i}$ is consistent for $1\leq i\leq n$. Let $\vect{b}_i$ denote any one particular solution to $\linearsystem{A}{\vect{e}_i}$, $1\leq i\leq n$.

Define the $n\times n$ matrix $B=\matrixcolumns{b}{n}$. Then \begin{align*} AB &=A\matrixcolumns{b}{n}\\ &=[A\vect{b}_1|A\vect{b}_2|A\vect{b}_3|\ldots|A\vect{b}_n]&&\knowl{./knowls/definition.MM.knowl}{\text{Definition MM}}\\ &=\matrixcolumns{e}{n}\\ &=I_n&&\knowl{./knowls/definition.SUV.knowl}{\text{Definition SUV}}\\ \end{align*}

So the matrix $B$ is a “right-inverse” for $A$. By Theorem NMRRI, $I_n$ is a nonsingular matrix, so by Theorem NPNT both $A$ and $B$ are nonsingular. Thus, in particular, $A$ is nonsingular. (Travis Osborne contributed to this proof.)