Let \(T\) be a linear operator on \(F^n\), let \(A\) be the matrix of \(T\) in the standard ordered basis for \(F^n\), and let \(W\) be the subspace of \(F^n\) spanned by the column vectors of \(A\). What does \(W\) have to do with \(T\)?
Solution
The columns of \(A\) are precisely the vectors \(T(e_1),\ldots,T(e_n)\), where \(e_1,\ldots,e_n\) is the standard basis of \(F^n\). Hence \[ W=\operatorname{span}\{T(e_1),\ldots,T(e_n)\}. \] Every vector \(x=\sum_j x_je_j\) satisfies \[ T(x)=\sum_j x_jT(e_j), \] so \(\operatorname{range}T\subseteq W\). The reverse inclusion is immediate because every \(T(e_j)\) lies in the range. Therefore \[ W=\operatorname{range}T. \] In particular, the column rank of \(A\) is the rank of \(T\).