Question

In Exercises 17–24, \(A\) is an \(m \times n\) matrix with a singular value decomposition \(A = U\Sigma {V^T}\) , where \(U\) is an \(m \times m\) orthogonal matrix, \({\bf{\Sigma }}\) is an \(m \times n\) “diagonal” matrix with \(r\) positive entries and no negative entries, and \(V\) is an \(n \times n\) orthogonal matrix. Justify each answer.

24. Using the notation of Exercise 23, show that \({A^T}{u_j} = {\sigma _j}{v_j}\) for \({\bf{1}} \le {\bf{j}} \le {\bf{r}} = {\bf{rank}}\;{\bf{A}}\)

Short answer

It is verified that \({A^T}{{\bf{u}}_j} = {\sigma _j}{{\bf{v}}_j}\).

Step-by-step solution

Write the matrix 

As from exercise 23,\(A = {\sigma _1}{{\bf{u}}_{\bf{1}}}{\bf{v}}_{\bf{1}}^{\bf{T}} + {\sigma _2}{{\bf{u}}_2}{\bf{v}}_2^{\bf{T}} + ... + {\sigma _r}{{\bf{u}}_r}{\bf{v}}_r^{\bf{T}}\).

Then since \(1 \le j \le r = {\rm{rank}}\;A\)

Step 2: Compute the notation

Find the product

\(\begin{array}{c}{A^T}{{\bf{u}}_j} = {\left( {{\sigma _1}{{\bf{u}}_{\bf{1}}}{\bf{v}}_{\bf{1}}^{\bf{T}} + {\sigma _2}{{\bf{u}}_2}{\bf{v}}_2^{\bf{T}} + ... + {\sigma _r}{{\bf{u}}_r}{\bf{v}}_r^{\bf{T}}} \right)^T}{{\bf{u}}_j}\\ = \left( {{\sigma _1}{{\bf{v}}_{\bf{1}}}{\bf{u}}_{\bf{1}}^{\bf{T}} + ... + {\sigma _r}{{\bf{v}}_r}{\bf{u}}_r^{\bf{T}}} \right){{\bf{u}}_j}\\ = \left( {{\sigma _j}{{\bf{v}}_j}{\bf{u}}_j^{\bf{T}}} \right){{\bf{u}}_j}\\ = {\sigma _j}{{\bf{v}}_j}\end{array}\)

Hence, \({A^T}{{\bf{u}}_j} = {\sigma _j}{{\bf{v}}_j}\).