Show that an \(n\times m\) matrix \(A\)over a division ring \(D\) has an \(m\times n\) left inverse \(B\) (so \(BA = I_m\)) \(\iff \mathrm{rank} A = m\). Similarly, show \(A\) has a right \(m\times n\) inverse \(\iff \mathrm{rank} A = n\).
Solution
Regard \(A\) as the matrix of the right-\(D\)-linear map \[ T:D^m\longrightarrow D^n, \qquad x\longmapsto Ax. \] Its rank is the dimension of its image.
1. \(A\) has a left inverse if and only if \(\operatorname{rank}A=m\).
Suppose first that \(B A=I_m\). If \(Ax=0\), then \[ x=I_mx=BAx=0, \] so \(T\) is injective. Hence \[ \operatorname{rank}A=\dim_D D^m=m. \]
Conversely, if \(\operatorname{rank}A=m\), then \(T\) is injective. Choose a basis \(e_1,\ldots,e_m\) of \(D^m\). Then \(T(e_1),\ldots,T(e_m)\) is linearly independent in \(D^n\); extend it to a basis of \(D^n\). Define \(S:D^n\to D^m\) by \[ S(T(e_i))=e_i \] for \(1\le i\le m\), and send every added basis vector to \(0\). Then \(S\circ T=\operatorname{id}_{D^m}\). If \(B\) is the matrix of \(S\), then \(BA=I_m\).
2. \(A\) has a right inverse if and only if \(\operatorname{rank}A=n\).
Suppose \(AC=I_n\) for some \(m\times n\) matrix \(C\). Then the associated map \(T:D^m\to D^n\) is surjective, since for every \(y\in D^n\), \[ y=I_ny=ACy=T(Cy). \] Thus \(\operatorname{rank}A=n\).
Conversely, if \(\operatorname{rank}A=n\), then \(T\) is surjective. Choose a basis \(f_1,\ldots,f_n\) of \(D^n\), and choose \(v_i\in D^m\) with \(T(v_i)=f_i\). There is a unique linear map \(U:D^n\to D^m\) satisfying \(U(f_i)=v_i\). Then \(T\circ U=\operatorname{id}_{D^n}\). If \(C\) is the matrix of \(U\), then \(AC=I_n\).