Question: Find the singular values of the matrices in Exercises 1-4.

2. \(\left( {\begin{array}{*{20}{c}}{ - 3}&0\\0&0\end{array}} \right)\)

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).

Consider the matrix as \(A = \left( {\begin{array}{*{20}{c}}{ - 3}&0\\0&0\end{array}} \right)\).

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

\(\begin{array}{c}{A^T}A = \left( {\begin{array}{*{20}{c}}{ - 3}&0\\0&0\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - 3}&0\\0&0\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{9 + 0}&{0 + 0}\\{0 + 0}&{0 + 0}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}9&0\\0&0\end{array}} \right)\end{array}\)

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

\(\begin{array}{c}\left( { - 9\lambda + {\lambda ^2}} \right) - 0 = 0\\{\lambda ^2} - 9\lambda = 0\\\lambda \left( {\lambda - 9} \right) = 0\end{array}\)

The eigenvalues of the matrix \({A^T}A\) are \({\lambda _1} = 9,{\lambda _2} = 0\).

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

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

\(\begin{array}{c}{\sigma _1} = \sqrt 9 \\ = 3\\{\sigma _2} = \sqrt 0 \\ = 0\end{array}\)

Thus, the singular values of the matrix \(A\) are \({\sigma _1} = 3,{\sigma _2} = 0\).