91影视

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

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