Let \(m<n\), and identify linear maps \(\mathbb R^m\to\mathbb R^n\) with real \(n\times m\) matrices. Let \(L\) be the set of maps of rank \(m\).
- Show that \(L\) is open.
- Show that there is a continuous assignment \[ A\longmapsto T(A)\in M_{m\times n}(\mathbb R) \] such that \[ T(A)A=I_m \] for every \(A\in L\).
Solution
1. A matrix \(A\in M_{n\times m}(\RR)\) lies in \(L\) if and only if\(A^T A\) is invertible.
For every \(v\in\RR^m\), \[ v^T A^T A v = \norm{Av}^2. \] Hence \[ A^T A v=0 \] implies \[ 0 = v^T A^T A v = \norm{Av}^2, \] so \(Av=0\). Conversely, \(Av=0\) immediately implies \(A^TAv=0\). Therefore \[ \ker(A^TA)=\ker(A). \] Since \(A\) has \(m\) columns, it has rank \(m\) exactly when \(\ker(A)=\{0\}\), and this is equivalent to invertibility of the \(m\times m\) matrix \(A^TA\).
2. The set \(L\) is open in \(M_{n\times m}(\RR)\).
By step 1, \[ L = \left\{ A\in M_{n\times m}(\RR): \det(A^TA)\neq0 \right\}. \] The map \[ A\longmapsto\det(A^TA) \] is a polynomial in the entries of \(A\), hence continuous. Therefore \(L\) is the inverse image of the open set \(\RR\sm\{0\}\) and is open.
3. Define, for \(A\in L\),\[ \boxed{ T(A)\coloneqq(A^TA)^{-1}A^T }. \]Then \(T(A)A=I_m\).
Step 1 makes \(A^TA\) invertible for every \(A\in L\), so the formula is well-defined. Directly, \[ T(A)A = (A^TA)^{-1}A^TA = I_m. \]
4. The assignment \(T:L\to M_{m\times n}(\RR)\) is continuous.
Transpose and matrix multiplication are polynomial operations in the matrix entries, hence continuous. Matrix inversion is continuous on \(\operatorname{GL}_m(\RR)\), since \[ B^{-1} = \frac{\operatorname{adj}(B)}{\det B} \] there. Thus \[ A \longmapsto A^TA \longmapsto (A^TA)^{-1} \longmapsto (A^TA)^{-1}A^T \] is continuous on \(L\).
5. Q.E.D.
Step 2 proves Part (1), while steps <1>3–<1>4 construct the continuous left inverse required in Part (2).