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.

23. Let \(U = \left( {{u_1}...{u_m}} \right)\) and \(V = \left( {{v_1}...{v_n}} \right)\) where the \({{\bf{u}}_i}\) and \({{\bf{v}}_i}\) are in Theorem 10. Show that \(A = {\sigma _1}{u_1}v_1^T + {\sigma _2}{u_2}v_2^T + ... + {\sigma _r}{u_r}v_r^T\).

Short answer

It is verified that \(A = {\sigma _1}{{\bf{u}}_{\bf{1}}}{\bf{v}}_{\bf{1}}^{\bf{T}} + {\sigma _2}{{\bf{u}}_2}{\bf{v}}_2^{\bf{T}} + ... + {\sigma _r}{{\bf{u}}_r}{\bf{v}}_r^{\bf{T}}\).

Step-by-step solution

Write the matrix 

It is given that, \(U = \left( {{{\bf{u}}_{\bf{1}}}...{{\bf{u}}_{\bf{m}}}} \right)\) and \(V = \left( {{{\bf{v}}_{\bf{1}}}...{{\bf{v}}_{\bf{n}}}} \right)\)

Therefore, \({V^T} = \left( {\begin{array}{*{20}{c}}{{\bf{v}}_1^T}\\{{\bf{v}}_n^T}\end{array}} \right)\)

Step 2: Compute the matrix A from SVD:

From theorem 10,\(U\sum = \left\{ {\left( {{\sigma _1}{{\bf{u}}_1}...{\sigma _r}{{\bf{u}}_r}} \right)} \right\}\)where\(r\)is the rank of the matrix\(A\), thus find\(A = U\sum {V^T}\).

\(\begin{array}{c}A = U\sum {V^T}\\ = \left\{ {\left( {{\sigma _1}{{\bf{u}}_1}...{\sigma _r}{{\bf{u}}_r}} \right)} \right\}\left( {\begin{array}{*{20}{c}}{{\bf{v}}_1^T}\\{{\bf{v}}_n^T}\end{array}} \right)\\ = {\sigma _1}{{\bf{u}}_{\bf{1}}}{\bf{v}}_{\bf{1}}^{\bf{T}} + {\sigma _2}{{\bf{u}}_2}{\bf{v}}_2^{\bf{T}} + ... + {\sigma _r}{{\bf{u}}_r}{\bf{v}}_r^{\bf{T}}\end{array}\)

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