Let \(p\), \(m\), and \(n\) be positive integers and \(F\) a field. Let \(V\) be the space of \(m \times n\) matrices over \(F\) and \(W\) the space of \(p \times n\) matrices over \(F\). Let \(B\) be a fixed \(p \times m\) matrix and let \(T\) be the linear transformation from \(V\) into \(W\) defined by \(T(A) = BA\). Prove that \(T\) is invertible if and only if \(p = m\) and \(B\) is an invertible \(m \times m\) matrix.
Solution
1. \((\Leftarrow)\) If \(p = m\) and \(B\) is an invertible \(m \times m\) matrix, then \(T\) is invertible:
1.1. Since \(B\) is invertible, there exists \(B^{-1} \in M_m(F)\) such that \(B^{-1} B = B B^{-1} = I_m\).
definition of an invertible matrix.
1.2. Define the linear map \(S: W \to V\) by \(S(C) = B^{-1} C\).
matrix multiplication is linear.
1.3. For any \(A \in V = M_{m \times n}(F)\):\[ S(T(A)) = S(BA) = B^{-1}(BA) = (B^{-1} B) A = I_m A = A. \]
associativity of matrix multiplication.
1.4. For any \(C \in W = M_{m \times n}(F)\):\[ T(S(C)) = T(B^{-1} C) = B(B^{-1} C) = (B B^{-1}) C = I_m C = C. \]
associativity of matrix multiplication.
1.5. Thus \(S = T^{-1}\), so \(T\) is an invertible linear transformation.
3 and 4.
2. \((\Rightarrow)\) If \(T\) is invertible, then \(p = m\):
2.1. The vector spaces have dimensions \(\dim_F V = mn\) and \(\dim_F W = pn\).
dimension of matrix spaces over a field.
2.2. An invertible linear map between finite-dimensional vector spaces requires \(\dim V = \dim W\).
isomorphism preserves dimension.
2.3. Thus \(mn = pn\).Since \(n \ge 1\), dividing by \(n\) yields \(p = m\).
cancellation in \(\mathbb{Z}\).
3. \((\Rightarrow)\) If \(T\) is invertible, then \(B\) is an invertible \(m \times m\) matrix:
3.1. From 2, \(p = m\), so \(B\) is a square \(m \times m\) matrix.
2.
2.2. Let \(v \in F^m\) be any vector such that \(B v = 0\).
setup for injecting nullspace of \(B\).
2.3. Construct the matrix \(A \in M_{m \times n}(F) = V\) whose first column is \(v\) and whose remaining columns are all \(0\).
matrix construction.
2.4. Compute \(T(A) = BA\): The columns of \(BA\) are \(B\) applied to the columns of \(A\), so \(\operatorname{col}_1(BA) = Bv = 0\) and \(\operatorname{col}_j(BA) = B(0) = 0\) for \(j \ge 2\).Thus \(T(A) = 0\).
column-by-column matrix multiplication.
2.5. Since \(T\) is invertible, it is injective, so \(T(A) = 0 \implies A = 0\).
injectivity of \(T\).
2.6. In particular, the first column \(v\) of \(A\) is \(0\), so \(\ker(B) = \{0\}\).
\(A = 0 \implies \operatorname{col}_1(A) = 0\).
2.7. For a square matrix over a field, having trivial kernel is equivalent to invertibility.Thus \(B\) is invertible.
Rank–Nullity Theorem for square matrices.
4. Conclusion: \(T\) is invertible if and only if \(p = m\) and \(B\) is an invertible \(m \times m\) matrix.
1, 2, and 3.
Q.E.D.