More precisely, suppose that $A$ is a square matrix of size $n$ and $B=\set{\vectorlist{x}{n}}$ is a basis of $\complex{n}$. Prove that $A$ is nonsingular if and only if $C=\set{A\vect{x}_1,\,A\vect{x}_2,\,A\vect{x}_3,\,\dots,\,A\vect{x}_n}$ is a basis of $\complex{n}$. (See also <acroref type="exercise" acro="PD.T33" />, <acroref type="exercise" acro="MR.T20" />.)

<solution contributor="robertbeezer">Our first proof relies mostly on definitions of linear independence and spanning, which is a good exercise. The second proof is shorter and turns on a technical result from our work with matrix inverses, <acroref type="theorem" acro="NPNT" />.<br /><br />

-<forwardimply /> Assume that $A$ is nonsingular and prove that $C$ is a basis of $\complex{n}$. First show that $C$ is linearly independent. Work on a relation of linear dependence on $C$,

+<implyforward /> Assume that $A$ is nonsingular and prove that $C$ is a basis of $\complex{n}$. First show that $C$ is linearly independent. Work on a relation of linear dependence on $C$,

<![CDATA[&&]]>\text{<acroref type="theorem" acro="MMSMM" />}

So we can write an arbitrary vector of $\complex{n}$ as a linear combination of the elements of $C$. In other words, $C$ spans $\complex{n}$ (<acroref type="definition" acro="SSVS" />). By <acroref type="definition" acro="B" />, the set $C$ is a basis for $\complex{n}$.<br /><br />

-<reverseimply /> Assume that $C$ is a basis and prove that $A$ is nonsingular. Let $\vect{x}$ be a solution to the homogeneous system $\homosystem{A}$. Since $B$ is a basis of $\complex{n}$ there are scalars, $\scalarlist{a}{n}$, such that

+<implyreverse /> Assume that $C$ is a basis and prove that $A$ is nonsingular. Let $\vect{x}$ be a solution to the homogeneous system $\homosystem{A}$. Since $B$ is a basis of $\complex{n}$ there are scalars, $\scalarlist{a}{n}$, such that

<![CDATA[\vect{x}&=\lincombo{a}{x}{n}]]>