Transformations of rank one

Proof. If \rho(A)=0 , then A=0 , and the statement is trivial. If \rho(A)=1 , that is, \mathcal{R}(A) is one-dimensional, then there exists in \mathcal{R}(A) a non-zero vector x_{0} (a basis in \mathcal{R}(A) ) such that every vector in \mathcal{R}(A) is a multiple of x_{0} . Hence, for every x , A x=y_{0} x_{0}, where the scalar coefficient y_{0} ( =y_{0}(x) ) depends, of course, on x . The linearity of A implies that y_{0} is a linear functional on \mathcal{V} . Let \mathcal{X}=\{x_{1}, \ldots, x_{n}\} be a basis in \mathcal{V} , and let (\alpha_{i j}) be the corresponding matrix of A , so that A x_{j}=\sum_{i} \alpha_{i j} x_{i}. If \mathcal{X}^{\prime}=\{y_{1}, \ldots, y_{n}\} is the dual basis in \mathcal{V}^{\prime} , then (cf. Section: Adjoints of projections , (2)) \alpha_{i j}=[A x_{j}, y_{i}]. In the present case \alpha_{i j}=[y_{0}(x_{j}) x_{0}, y_{i}]=y_{0}(x_{j})[x_{0}, y_{i}]=[x_{0}, y_{i}][x_{j}, y_{0}]; in other words, we may take \beta_{i}=[x_{0}, y_{i}] and \gamma_{j}=[x_{j}, y_{0}] .

Conversely, suppose that in a fixed coordinate system \mathcal{X}=\{x_{1}, \ldots, x_{n}\} the matrix (\alpha_{i j}) of A is such that \alpha_{i j}=\beta_{i} \gamma_{j} . We may find a linear functional y_{0} such that \gamma_{j}=[x_{j}, y_{0}] , and we may define a vector x_{0} by x_{0}=\sum_{k} \beta_{k} x_{k} . The linear transformation \tilde{A} defined by \tilde{A} x=y_{0}(x) x_{0} is clearly of rank one (unless, of course, \alpha_{i j}=0 for all i and j ), and its matrix (\tilde{\alpha}_{i j}) in the coordinate system \mathcal{X} is given by \tilde{\alpha}_{i j}=[\tilde{A} x_{j}, y_{i}] (where \mathcal{X}^{\prime}=\{y_{1}, \ldots, y_{n}\} is the dual basis of \mathcal{X} ). Hence \tilde{\alpha}_{i j}=[y_{0}(x_{j}) x_{0}, y_{i}]=[x_{0}, y_{i}][x_{j}, y_{0}]=\beta_{i} \gamma_{j}, and, since A and \tilde{A} have the same matrix in one coordinate system, it follows that \tilde{A}=A . This concludes the proof of the theorem. ◻

Proof. Since A\mathcal{V} = \mathcal{R}(A) has dimension \rho , we may find \rho vectors x_{1}, \ldots, x_{\rho} that form a basis for \mathcal{R}(A) . It follows that, for every vector x in \mathcal{V} , we have Ax = \sum_{i = 1}^\rho \xi_i x_i, where each \xi_{i} depends, of course, on x ; we write \xi_{i}=y_{i}(x) . It is easy to see that y_{i} is a linear functional. In terms of these y_{i} we define, for each i=1, \ldots, \rho , a linear transformation A_{i} by A_{i} x=y_{i}(x) x_{i} . It follows that each A_{i} has rank one and A=\sum_{i=1}^\rho A_{i} . (Compare this result with Section: Linear transformations , example (2).) ◻

Proof. Let \mathcal{R} and \mathcal{N} , respectively, be the range and the null-space of A , and let \{x_{1}, \ldots, x_{\rho}\} be a basis for \mathcal{R} . Let x_{\rho+1}, \ldots, x_{n} be vectors such that \{x_{1}, \ldots, x_{n}\} is a basis for \mathcal{V} . Since x_{i} is in \mathcal{R} for i=1, \ldots, \rho , we may find vectors y_i such that Ay_i = x_i ; finally, we choose a basis for \mathcal{N} , which we may denoted by \{y_{\rho + 1}, \ldots, y_n\} . We assert that \{y_{1}, \ldots, y_{n}\} is a basis for \mathcal{V} . We need, of course, to prove only that the y ’s are linearly independent. For this purpose we suppose that \sum_{i=1}^{n} \alpha_{i} y_{i}=0 ; then we have (remembering that for i=\rho+1, \ldots, n the vector y_{i} belongs to \mathcal{N} ) A\Big(\sum_{i=1}^{n} \alpha_{i} y_{i}\Big)=\sum_{i=1}^{\rho} \alpha_{i} x_{i}=0, whence \alpha_{1}=\cdots=\alpha_{\rho}=0 . Consequently \sum_{i=\rho+1}^{n} \alpha_{i} y_{i}=0 ; the linear independence of y_{\rho+1}, \ldots, y_{n} shows that the remaining \alpha ’s must also vanish.

A linear transformation P , of the kind whose existence we asserted, is now determined by the conditions P x_{i}=y_{i} , i=1, \ldots, n . Indeed, if i=1, \ldots, \rho , then P A y_{i}=P x_{i}=y_{i} , and if i=\rho+1, \ldots, n , then P A y_{i}=P 0=0 . ◻