Question

Question: Find an SVD of each matrix in Exercises 5-12. (Hint: In Exercise 11, one choice for \(U = \left( {\begin{array}{*{20}{c}}{ - \frac{1}{3}}&{\frac{2}{3}}&{\frac{2}{3}}\\{\frac{2}{3}}&{ - \frac{1}{3}}&{\frac{2}{3}}\\{\frac{2}{3}}&{\frac{2}{3}}&{ - \frac{1}{3}}\end{array}} \right)\). In Exercise 12, one column of U can be \(\left( {\begin{array}{*{20}{c}}{\frac{1}{{\sqrt 6 }}}\\{\frac{{ - 2}}{{\sqrt 6 }}}\\{\frac{1}{{\sqrt 6 }}}\end{array}} \right)\).)

5. \(\left( {\begin{array}{*{20}{c}}{ - 2}&0\\0&0\end{array}} \right)\)

Short answer

The singular value decomposition of \(A\) is \(A = \left( {\begin{array}{*{20}{c}}{ - 1}&0\\0&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}2&0\\0&0\end{array}} \right)\left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)\).

Step-by-step solution

The Singular Value Decomposition

Consider \(m \times n\) matrix with rank \(r\) as \(A\). There is a \(m \times n\) matrix \(\sum \) as seen in (3) wherein the diagonal entriesin \(D\) are the first \(r\) singular valuesof A, \({\sigma _1} \ge \cdots \ge {\sigma _r} > 0,\) and there arises an \(m \times m\) orthogonal matrix \(U\) and an \(n \times n\) orthogonal matrix \(V\) such that \(A = U\sum {V^T}\).

Determine the eigenvalues and eigenvector of \({A^T}A\)

Consider the matrix as \(A = \left( {\begin{array}{*{20}{c}}{ - 2}&0\\0&0\end{array}} \right)\).

Compute the matrix \({A^T}A\) as shown below:

\(\begin{array}{c}{A^T}A = \left( {\begin{array}{*{20}{c}}{ - 2}&0\\0&0\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - 2}&0\\0&0\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{4 + 0}&{0 + 0}\\{0 + 0}&{0 + 0}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}4&0\\0&0\end{array}} \right)\end{array}\)

The characteristic equation of \({A^T}A\) to obtain the eigenvalues of the matrix is shown below:

\(\begin{array}{c}{\lambda ^2} - 4\lambda = 0\\\lambda \left( {\lambda - 4} \right) = 0\end{array}\)

Thus, the eigenvalues of the matrix \({A^T}A\) are \({\lambda _1} = 0,{\lambda _2} = 4\).

It is observed that the eigenvalue of \({A^T}A\) is in decreasing order.

The associated unit eigenvectors of \({\lambda _2} = 4\) is \(\left( {\begin{array}{*{20}{c}}1\\0\end{array}} \right)\) and \({\lambda _1} = 0\) is \(\left( {\begin{array}{*{20}{c}}0\\1\end{array}} \right)\).

Construct   \(V\) and determine the singular values of the matrix

The square roots of the eigenvaluesof \({A^T}A\), represented by \({\sigma _1}, \ldots ,{\sigma _n}\) are known as the singular values of \(A\) and they have been arranged in decreasing order. In other words, \({\sigma _i} = \sqrt {{\lambda _i}} \) for \(1 \le i \le n\). The lengths of the vectors \(A{{\bf{v}}_1}, \ldots ,A{{\bf{v}}_n}\) are called the singular values of A from equation (2).

Use the eigenvectors to construct \(V\) as shown below:

\(\begin{array}{c}V = \left( {\begin{array}{*{20}{c}}{{{\bf{v}}_1}}&{{{\bf{v}}_2}}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)\end{array}\)

Obtain the singular values of the matrices as shown below:

\(\begin{array}{c}{\sigma _1} = \sqrt 4 \\ = 2\\{\sigma _2} = \sqrt 0 \\ = 0\end{array}\)

Construct the matrix \(\sum \) as shown below:

\(\sum = \left( {\begin{array}{*{20}{c}}2&0\\0&0\end{array}} \right)\)

Construct the matrix U

Compute \({{\bf{u}}_1}\) as shown below:

\(\begin{array}{c}{{\bf{u}}_1} = \frac{1}{{{\sigma _1}}}A{{\bf{v}}_1}\\ = \frac{1}{2}\left( {\begin{array}{*{20}{c}}{ - 2}&0\\0&0\end{array}} \right)\left( {\begin{array}{*{20}{c}}1\\0\end{array}} \right)\\ = \frac{1}{2}\left( {\begin{array}{*{20}{c}}{ - 2 + 0}\\{0 + 0}\end{array}} \right)\\ = \frac{1}{2}\left( {\begin{array}{*{20}{c}}{ - 2}\\0\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - 1}\\0\end{array}} \right)\end{array}\)

Since \(A{{\bf{v}}_2} = 0\) the only column for \(U\) that has been discovered is \({{\bf{u}}_1}\).

Extend \(\left\{ {{{\bf{u}}_1}} \right\}\) to an orthonormal basis in \({\mathbb{R}^2}\) to obtain other columns of \(U\). Therefore, \({{\bf{u}}_2} = \left( {\begin{array}{*{20}{c}}0\\1\end{array}} \right)\).

Use \({{\bf{u}}_1}\) and \({{\bf{u}}_2}\) to construct the matrix \(U\) as shown below:

\(U = \left( {\begin{array}{*{20}{c}}{ - 1}&0\\0&1\end{array}} \right)\)

Determine the singular value decomposition of the matrix

Obtain the singular value decomposition of \(A\) as shown below:

\(\begin{array}{c}A = U\sum {V^T}\\ = \left( {\begin{array}{*{20}{c}}{ - 1}&0\\0&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}2&0\\0&0\end{array}} \right)\left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)\end{array}\)

Thus, the singular value decomposition of \(A\) is \(A = \left( {\begin{array}{*{20}{c}}{ - 1}&0\\0&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}2&0\\0&0\end{array}} \right)\left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)\).