Let \(W\) be the space of all \(n \times 1\) column matrices over a field \(F\). If \(A\) is an \(n \times n\) matrix over \(F\), then \(A\) defines a linear operator \(L_A\) on \(W\) through left multiplication: \(L_A(X) = AX\). Prove that every linear operator on \(W\) is left multiplication by some \(n \times n\) matrix, i.e., is \(L_A\) for some \(A\).
Now suppose \(V\) is an \(n\)-dimensional vector space over the field \(F\), and let \(\mathfrak{B}\) be an ordered basis for \(V\). For each \(\alpha\) in \(V\), define \(U\alpha = [\alpha]_{\mathfrak{B}}\). Prove that \(U\) is an isomorphism of \(V\) onto \(W\). If \(T\) is a linear operator on \(V\), then \(UTU^{-1}\) is a linear operator on \(W\). Accordingly, \(UTU^{-1}\) is left multiplication by some \(n \times n\) matrix \(A\). What is \(A\)?
Solution
Let \(e_1,\ldots,e_n\) be the standard basis of the column space \(W=F^{n\times1}\). If \(L:W\to W\) is any linear operator, form the matrix \[ A=\begin{bmatrix}L(e_1)&\cdots&L(e_n)\end{bmatrix}. \] Then for \(X=\sum_jx_je_j\), \[ L(X)=\sum_jx_jL(e_j)=AX, \] so \(L=L_A\).
For the second part, the coordinate map \[ U:V\to W,\qquad U\alpha=[\alpha]_{\mathfrak B} \] is linear and bijective, since every coordinate column corresponds to exactly one vector of \(V\). Thus \(U\) is an isomorphism.
If \(T\) is a linear operator on \(V\), then \(UTU^{-1}\) is left multiplication by the matrix whose \(j\)th column is \[ UTU^{-1}(e_j)=U(T\alpha_j)=[T\alpha_j]_{\mathfrak B}. \] Therefore that matrix is precisely \[ A=[T]_{\mathfrak B}. \]