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

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

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 \(A = U\sum {V^T}\).

Let \(A = \left( {\begin{aligned}{{}}{ - 3}&0\\0&{ - 2}\end{aligned}} \right)\). The transpose of A is:

\({A^T} = \left( {\begin{aligned}{{}}{ - 3}&0\\0&{ - 2}\end{aligned}} \right)\)

Find the product \({A^T}A\).

\(\begin{aligned}{}{A^T}A = \left( {\begin{aligned}{{}}{ - 3}&0\\0&{ - 2}\end{aligned}} \right)\left( {\begin{aligned}{{}}{ - 3}&0\\0&{ - 2}\end{aligned}} \right)\\ = \left( {\begin{aligned}{{}}9&0\\0&4\end{aligned}} \right)\end{aligned}\)

So, the eigenvalues of the matrix are 9 and 4. The eigenvectors can be calculated as,

\(\begin{aligned}{}\left( {\begin{aligned}{{}}9&0\\0&4\end{aligned}} \right)\left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\end{aligned}} \right) = \left( {\begin{aligned}{{}}0\\0\end{aligned}} \right)\\9{x_1} = 0\\4{x_2} = 0\end{aligned}\)

So, the eigenvectors of the matrix are\(\left( {\begin{aligned}{{}}1\\0\end{aligned}} \right)\) and \(\left( {\begin{aligned}{{}}0\\1\end{aligned}} \right)\).

The singular value is the square root of eigenvalues of \({A^T}A\) , that are, 3 and 2. Thus the matrix \(\Sigma \) is:

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

Consider the following equations:

\(\begin{array}{c}\Sigma U = AV\\\left( {\begin{array}{*{20}{c}}3&0\\0&2\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{y_1}}&{{y_3}}\\{{y_2}}&{{y_4}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{ - 3}&0\\0&{ - 2}\end{array}} \right)\left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)\\\left( {\begin{array}{*{20}{c}}{3{y_1}}&0\\0&{2{y_4}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{ - 3}&0\\0&{ - 2}\end{array}} \right)\end{array}\)

On comparing two sides of the equation \({y_1} = - 1\), \({y_2} = 0\), \({y_3} = 0\) and \({y_4} = - 1\)

The matrix U can be written as,

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

Then by using \(A = U\sum {V^T}\).

\(A = \left( {\begin{array}{*{20}{c}}{ - 1}&0\\0&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}3&0\\0&2\end{array}} \right)\left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)\)

Hence, the SVD of A can be written as, \(A = \left( {\begin{array}{*{20}{c}}{ - 1}&0\\0&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}3&0\\0&2\end{array}} \right)\left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)\).