Let \(A\) be an \(m \times n\) matrix with entries in \(F\) and let \(T\) be the linear transformation from \(F^{n \times 1}\) into \(F^{m \times 1}\) defined by \(T(X) = AX\). Show that if \(m < n\) it may happen that \(T\) is onto without being non-singular. Similarly, show that if \(m > n\) we may have \(T\) non-singular but not onto.
Solution
If \(m<n\), take \[ A=\begin{bmatrix}I_m&0\end{bmatrix}, \] an \(m\times n\) matrix. Then \[ T(x_1,\ldots,x_n)=(x_1,\ldots,x_m), \] so \(T\) is onto. But \(T\) is not non-singular: for example, \(e_{m+1}\ne0\) and \(T(e_{m+1})=0\).
If \(m>n\), take \[ A=\begin{bmatrix}I_n\\0\end{bmatrix}, \] an \(m\times n\) matrix. Then \[ T(x_1,\ldots,x_n)=(x_1,\ldots,x_n,0,\ldots,0), \] so \(T\) is non-singular. It is not onto, since any vector whose last \(m-n\) coordinates are not all zero is outside its range.