Proof of Theorem RNNM

(1 $\Rightarrow$ 2) Theorem CSNM says that if $A$ is nonsingular then $\csp{A}=\complex{n}$. If $\csp{A}=\complex{n}$, then the column space has dimension $n$ by Theorem DCM, so the rank of $A$ is $n$.

(2 $\Rightarrow$ 3) Suppose $\rank{A}=n$. Then Theorem RPNC gives \begin{align*} \nullity{A}&=n-\rank{A}&&\knowl{./knowls/theorem.RPNC.knowl}{\text{Theorem RPNC}}\\ &=n-n&&\text{Hypothesis}\\ &=0 \end{align*}

(3 $\Rightarrow$ 1) Suppose $\nullity{A}=0$, so a basis for the null space of $A$ is the empty set. This implies that $\nsp{A}=\set{\zerovector}$ and Theorem NMTNS says $A$ is nonsingular.