Question

25.Let \({\bf{T:}}{\mathbb{R}^{\bf{n}}} \to {\mathbb{R}^{\bf{m}}}\) be a linear transformation. Describe how to find a basis \(B\) for \({\mathbb{R}^n}\) and a basis \(C\) for \({\mathbb{R}^m}\) such that the matrix for \(T\) relative to \(B\) and \(C\) is an \(m \times n\) “diagonal” matrix.

Short answer

As \({\left( {T\left( {{{\bf{v}}_j}} \right)} \right)_C} = {\sigma _j}{{\bf{u}}_j}\). So, the matrix for \(T\) relative to \(B\) and \(C\) is an \(m \times n\) “diagonal” matrix.

Step-by-step solution

Write the matrix 

Consider the SVD for the standard matrix \(A\) of \(T\), that is \(A = U\sum {V^T}\). Let \(T:{\mathbb{R}^n} \to {\mathbb{R}^m}\) be a linear transformation.

And \(C = \left\{ {{{\bf{u}}_1},...,{{\bf{u}}_m}} \right\}\) be bases for \({\mathbb{R}^n}\) and \({\mathbb{R}^m}\).

Since the columns of V are orthogonal, \({V^T}{{\bf{v}}_j} = {{\bf{e}}_j}\).

Step 2: Compute the diagonal matrix

Find the matrix of \(T\) related to \(B\) and \(C\), find \(T\left( {{{\bf{v}}_j}} \right)\).

\(\begin{array}{c}T\left( {{{\bf{v}}_j}} \right) = A{{\bf{v}}_j}\\ = U\sum {V^T}{{\bf{v}}_j}\\ = U\sum {{\bf{e}}_j}\\ = {\sigma _j}U{{\bf{e}}_j}\\ = {\sigma _j}{{\bf{u}}_j}\end{array}\)

Which implies that, \({\left( {T\left( {{{\bf{v}}_j}} \right)} \right)_C} = {\sigma _j}{{\bf{u}}_j}\).

Thus, such that the matrix for \(T\) relative to \(B\) and \(C\) is an \(m \times n\) “diagonal” matrix.