Question

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

4. \(\left( {\begin{array}{*{20}{c}}3&0\\8&3\end{array}} \right)\)

Short answer

The singular values of the matrix \(A\) are \({\sigma _1} = 9,{\sigma _2} = 1\).

Step-by-step solution

Definition of Singular values

The square roots of the eigenvaluesof \({A^T}A\), represented by \({\sigma _1}, \ldots ,{\sigma _n}\) are known as the singular valuesof \(A\) 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 ofA from equation (2).

Determine the singular values of the matrix

Consider the matrix as \(A = \left( {\begin{array}{*{20}{c}}3&0\\8&3\end{array}} \right)\).

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

\(\begin{array}{c}{A^T}A = \left( {\begin{array}{*{20}{c}}3&8\\0&3\end{array}} \right)\left( {\begin{array}{*{20}{c}}3&0\\8&3\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{9 + 64}&{0 + 24}\\{0 + 24}&{0 + 9}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{73}&{24}\\{24}&9\end{array}} \right)\end{array}\)

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

\(\begin{array}{c}\left( {657 - 82\lambda + {\lambda ^2}} \right) - 576 = 0\\{\lambda ^2} - 82\lambda + 81 = 0\\\left( {\lambda - 1} \right)\left( {\lambda - 81} \right) = 0\end{array}\)

Thus, the eigenvalues of the matrix \({A^T}A\) are \({\lambda _1} = 81,{\lambda _2} = 1\).

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

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

\(\begin{array}{c}{\sigma _1} = \sqrt {81} \\ = 9\\{\sigma _2} = \sqrt 1 \\ = 1\end{array}\)

Thus, the singular values of the matrix \(A\) are \({\sigma _1} = 9,{\sigma _2} = 1\).