Lemma: The rank-nullity theorem holds over division rings.
Proof: A linear map \(\phi: D^m \to D^n\) induces a short exact sequence: \[ 0 \to \ker \phi \to D^m \mapsvia{\phi} \im\phi \to 0 \]
But every module over a division ring is free; in particular, \(\im \phi \leq D^n\) is a module over \(D\) and is thus free. So by a lemma in class, since the right-most term is a free module, this sequence splits and we have \[ D^m \cong \ker \phi \oplus \im \phi \]
and taking dimensions yields \[ m = \dim \ker(\phi) + \rank(\phi). \] \(\qed\)
- \(A \in M(n\times m, D)\) has a left inverse \(B \iff \rank(A) = m\):
\(\implies\): Suppose toward the contrapositive that \(\rank(A) < m\), so \(A\) has at least one pair of linearly dependent columns. So wlog write \[ A = [\vector a_1^t, \vector a_2^t, \cdots, \vector a_m^t] \] in block form with each \(\vector a_i\) a column vector, and we can assume that \(\vector a_1, \vector a_2\) are linearly dependent.
Now suppose such a left inverse \(B\) were to exist. Write it in block form as \[ B =[\vector b_1, \vector b_2, \cdots, \vector b_n]^t, \] so each \(\vector b_i\) is a row of \(B\).
Now if \(BA = I_m\) is to hold, noting that \((BA)_{ij} = \inner{\vector b_i}{\vector a_j}\), we must have
\[\begin{align*} I_{1,1} &= \inner{\vector b_1}{\vector a_1} = 1 \\ I_{1, 2} &= \inner{\vector b_1}{\vector a_2} = 0 \\ I_{1, 3} &= \inner{\vector b_1}{\vector a_3} = 0 \\ &\vdots \\ I_{2, 1} &= \inner{\vector b_2}{\vector a_1} = 0 \\ I_{2, 2} &= \inner{\vector b_2}{\vector a_2} = 1 \\ I_{2, 3} &= \inner{\vector b_2}{\vector a_3} = 0 \\ &\vdots .\end{align*}\]
But the claim is that this can not happen if \(\vector a_1, \vector a_2\) are linearly dependent. To see why, note that the linear dependence supplies elements \(d_1, d_2 \neq 0 \in D\) such that \(d_1 \vector a_1 + d_2 \vector a_2 = \vector 0\). But then taking inner products against, e.g. \(\vector b_1\) (that is, applying \(\inner{\vector b_1}{\wait}\) to everything in sight), we obtain
\[\begin{align*} d_1 \vector a_1 + d_2 \vector a_2 = \vector 0 \\ \implies \inner{\vector b_1}{d_1 \vector a_1} + \inner{\vector b_1}{d_2 \vector a_2} = \inner{\vector b_1}{\vector 0} = 0 \\ \implies d_1 \inner{\vector b_1}{\vector a_1} + d_2\inner{\vector b_1}{\vector a_2} = \inner{\vector b_1}{\vector 0} = 0 \\ \implies d_1 \inner{\vector b_1}{\vector a_1} + d_2\inner{\vector b_1}{\vector a_2} = 0 \\ \implies d_1 + d_2\inner{\vector b_1}{\vector a_2} = 0 \\ \implies \inner{\vector b_1}{\vector a_2} = -\frac{d_1}{d_2} \neq 0 ,\end{align*}\]
which contradicts \(\inner{\vector b_1}{\vector a_2} = 0\) as required by the previous equations.
\(\impliedby\): Suppose \(\rank(A) = m\), so \(A\) has \(m\) linearly independent columns – note that this is all of its columns.
Note: since row rank equals column rank, this also says that \(A\) has \(m\) linearly independent rows, so \(n\geq m\).
Viewing \(A\) as a representative of a map \(\phi: D^m \to D^n\), we find that \(\dim \im \phi = m \leq n\). In particular, from the rank nullity theorem, we have \[ m = \dim \ker \phi + \rank(\phi) = \dim \ker \phi + m \implies \dim \ker \phi = 0. \] So \(\ker A = \theset{\vector 0}\), and \(A\) represents an injective map \(f_A: D^m \to D^n\).
But any injective set map \(f: S_1\to S_2\) has a left-inverse \(g\) such that \(g\circ f = \id_{S_1}\). So \(f_A: D^m \to D^n\) as a set map has a left inverse \(g_B: D^n \to D^m\) satisfying \(g_B\circ f_A = \id_{D^m}\). But then taking the matrix associated to \(g_B\) yields a matrix \(B \in M(m\times n, D)\) such that \(BA = I_m\) as desired. \(\qed\)
- \(A\) has a right inverse \(B \iff \rank(A) = n\):
\(\implies\): By a similar argument, supposing that \(\rank A < n\) but \(AB = I_n\) for some \(B\), we find that \(A\) has at least two linearly dependent rows this time, say \(\vector a_1, \vector a_2\), whereas we obtain a system of equations of the form \(\inner{a_i}{\vector b_k} = \delta_{ik}\) where \(\vector b_i\) are now the columns of \(B\).
In a similar manner, the linear dependence forces, say, \(\inner{\vector a_2}{\vector b_1} \neq 0\), which is a contradiction.
\(\impliedby\): By another similar argument, we find that \(A\) represents a map \(f_A: D^m \to D^n\), and since \(\rank A = \dim \im A = n\), we find that \(A\) represents a surjective map \(f_A\). Surjective set maps have right inverses, so there is some \(g_B: D^n \to D^m\) such that \(f_A \circ g_B = \id_{D^n}\), and when translated to matrices this yields \(AB = I_{n}\). \(\qed\)
Solution
Fix the convention that \(D^m,D^n\) are right vector spaces over the division ring \(D\), and matrices act on column vectors by left multiplication.
For a \(D\)-linear map \[ \phi:D^m\to D^n, \] choose a basis of \(\ker\phi\) and extend it to a basis of \(D^m\). The images of the added basis vectors form a basis of \(\operatorname{im}\phi\). Hence \[ \boxed{m=\dim_D\ker\phi+\dim_D\operatorname{im}\phi}, \] which is rank-nullity.
Now let \(A\in M_{n\times m}(D)\).
If \(BA=I_m\) for some \(B\), then \(Av=0\) implies \[ v=BAv=0, \] so \(A:D^m\to D^n\) is injective. Rank-nullity gives \[ \operatorname{rank}A=m. \] Conversely, if \(\operatorname{rank}A=m\), then \(A\) is injective. Choose a basis of \(\operatorname{im}A\) and extend it to a basis of \(D^n\). Define a linear map \[ B:D^n\to D^m \] by sending \(A(e_i)\) back to \(e_i\) and sending the complementary basis vectors to \(0\). Then \[ BA=I_m. \] Thus \[ \boxed{A\text{ has a left inverse}\iff\operatorname{rank}A=m}. \]
Similarly, if \(AB=I_n\), then \(A\) is surjective, so \(\operatorname{rank}A=n\). Conversely, if \(\operatorname{rank}A=n\), then \(A\) is surjective. Choose a basis of \(\ker A\) and extend it to a basis of \(D^m\); the complementary basis maps to a basis of \(D^n\). Sending that latter basis back to its chosen preimages defines a linear section \[ B:D^n\to D^m \] with \[ AB=I_n. \] Hence \[ \boxed{A\text{ has a right inverse}\iff\operatorname{rank}A=n}. \]