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.

21. Justify the statement in Example 2 that the second singular value of a matrix \(A\) is the maximum of \(\left\| {A{\bf{x}}} \right\|\) as \({\bf{x}}\) varies over all unit vectors orthogonal to \({{\bf{v}}_{\bf{1}}}\), with \({{\bf{v}}_{\bf{1}}}\) a right singular vector corresponding to the first singular value of \(A\). (Hint: Use Theorem 7 in Section 7.3.)

Short answer

The given statement is verified.

Step-by-step solution

Show that vectors orthogonal to \({{\bf{v}}_{\bf{1}}}\)

As we know that the right singular vector \({{\bf{v}}_{\bf{1}}}\) is an eigenvector for the largest eigenvalue \({\lambda _1}\) of \({A^T}A\). Also, the second largest eigenvalue \({\lambda _2}\) is the maximum of \({{\bf{x}}^T}\left( {{A^T}A} \right){\bf{x}}\) overall unit vectors orthogonal to \({{\bf{v}}_{\bf{1}}}\) .

Show that overall unit vectors orthogonal to \({{\bf{v}}_{\bf{1}}}\)

Find\({{\bf{x}}^T}({A^T}A){\bf{x}}\).

\(\begin{array}{c}{{\bf{x}}^T}({A^T}A){\bf{x}} = {{\bf{x}}^T}{A^T}(A{\bf{x}})\\ = {(A{\bf{x}})^T}(A{\bf{x}})\\ = ||A{\bf{x}}|{|^2}\end{array}\)

Thus, the square root of \({\lambda _2}\), is the maximum of \(||A{\bf{x}}||\) overall unit vectors orthogonal to \({{\bf{v}}_{\bf{1}}}\).