Let \(F\) be a field. Fix positive integers \(m,n\) and matrices \(A,B\in M_{m\times n}(F)\). Define \[ T:M_{n\times m}(F)\to M_{m\times n}(F), \qquad T(X)=AXB. \] Prove that if \(m\ne n\), then \(T\) is not invertible.
Solution
Regard a matrix in \(M_{a\times b}(F)\) as a linear map \(F^b\to F^a\).
1. If \(m<n\), then \(T\) is not injective.
Since \[ \operatorname{rank}A\leq m<n, \] the linear map \[ A:F^n\longrightarrow F^m \] has a nonzero kernel. Choose \(0\neq v\in\ker A\). Because \(m>0\), there is a nonzero matrix \(X\in M_{n\times m}(F)\) whose first column is \(v\) and whose remaining columns are zero. Then \[ AX=0, \] so \[ T(X)=AXB=0. \] Thus \(X\neq0\) lies in \(\ker T\), and \(T\) is not injective.
2. If \(m>n\), then \(T\) is not surjective.
Let \[ W\coloneqq\im A\subseteq F^m. \] Since \[ \dim W=\operatorname{rank}A\leq n<m, \] the subspace \(W\) is proper. For every \(X\in M_{n\times m}(F)\), the matrix \(T(X)=AXB\), viewed as a map \(F^n\to F^m\), satisfies \[ \im T(X)\subseteq\im A=W. \]
Choose \(y\in F^m\setminus W\). Since \(n>0\), let \(Y\in M_{m\times n}(F)\) have first column \(y\) and all remaining columns zero. Then \[ \im Y\nsubseteq W, \] so \(Y\neq T(X)\) for every \(X\). Hence \(T\) is not surjective.
3. If \(m\neq n\), then \(T\) is not invertible.
If \(m<n\), step 1 shows that \(T\) is not injective. If \(m>n\), step 2 shows that \(T\) is not surjective. Since \(m\neq n\), one of these two cases holds, so \(T\) cannot be invertible.
4. Q.E.D.
Step 3 is the desired conclusion.