Question 9.6.2: Determine the singular value decomposition of the 5 × 3 matr...
Determine the singular value decomposition of the 5 × 3 matrix
A=\left[\begin{array}{lll}1 & 0 & 1 \\0 & 1 & 0 \\0 & 1 & 1 \\0 & 1 & 0 \\1 & 1 & 0\end{array}\right].
We found in Example 1 that A has the singular values s_{1}=\sqrt{5}, s_{2}=\sqrt{2} , and s_{3}=1 , so
S=\left[\begin{array}{ccc}\sqrt{5} & 0 & 0 \\0 & \sqrt{2} & 0 \\0 & 0 & 1 \\0 & 0 & 0 \\0 & 0 & 0\end{array}\right].
Eigenvectors of A^{t} A corresponding to s_{1}=\sqrt{5}, s_{2}=\sqrt{2} \text { and } s_{3}=1 , are, respectively, (1,2,1)^{t},(1,-1,1)^{t}, \text { and }(-1,0,1)^{t} (see Exercise 5). Normalizing these vectors and using the values for the columns of V gives
V=\left[\begin{array}{ccc} \frac{\sqrt{6}}{6} & \frac{\sqrt{3}}{3} & -\frac{\sqrt{2}}{2} \\ \frac{\sqrt{6}}{3} & -\frac{\sqrt{3}}{3} & 0 \\ \frac{\sqrt{6}}{6} & \frac{\sqrt{3}}{3} & \frac{\sqrt{2}}{2} \end{array}\right] \quad \text { and } \quad V^{t}=\left[\begin{array}{ccc} \frac{\sqrt{6}}{6} & \frac{\sqrt{6}}{3} & \frac{\sqrt{6}}{6} \\ \frac{\sqrt{3}}{3} & -\frac{\sqrt{3}}{3} & \frac{\sqrt{3}}{3} \\ -\frac{\sqrt{2}}{2} & 0 & \frac{\sqrt{2}}{2} \end{array}\right]
The first 3 columns of U are therefore
\begin{aligned}& u _{1}=\frac{1}{\sqrt{5}} \cdot A\left(\frac{\sqrt{6}}{6}, \frac{\sqrt{6}}{3}, \frac{\sqrt{6}}{6}\right)^{t}=\left(\frac{\sqrt{30}}{15}, \frac{\sqrt{30}}{15}, \frac{\sqrt{30}}{10}, \frac{\sqrt{30}}{15}, \frac{\sqrt{30}}{10}\right)^{t} \\& u _{2}=\frac{1}{\sqrt{2}} \cdot A\left(\frac{\sqrt{3}}{3},-\frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3}\right)^{t}=\left(\frac{\sqrt{6}}{3},-\frac{\sqrt{6}}{6}, 0,-\frac{\sqrt{6}}{6}, 0\right)^{t} \\& u _{3}=1 \cdot A\left(-\frac{\sqrt{2}}{2}, 0, \frac{\sqrt{2}}{2}\right)^{t}=\left(0,0, \frac{\sqrt{2}}{2}, 0,-\frac{\sqrt{2}}{2}\right)^{t}\end{aligned}
To determine the two remaining columns of U we first need two vectors x _{4} \text { and } x _{5} so that \left\{ u _{1}, u _{2}, u _{3}, x _{4}, x _{5}\right\} is a linearly independent set. Then we apply the Gram Schmidt process to obtain u _{4} \text { and } u _{5} \text { so that }\left\{ u _{1}, u _{2}, u _{3}, u _{4}, u _{5}\right\} is an orthogonal set. Two vectors that satisfy are
(1,1,-1,1,-1)^{t} \quad \text { and }(0,1,0,-1,0)^{t} .
Normalizing the vectors u _{i} , for i = 1, 2, 3, 4, and 5 produces the matrix U and the singular value decomposition as
A=U S V^{t}=\left[\begin{array}{ccccc}\frac{\sqrt{30}}{15} & \frac{\sqrt{6}}{3} & 0 & \frac{\sqrt{5}}{5} & 0 \\\frac{\sqrt{30}}{15} & -\frac{\sqrt{6}}{6} & 0 & \frac{\sqrt{5}}{5} & \frac{\sqrt{2}}{2} \\\frac{\sqrt{30}}{10} & 0 & \frac{\sqrt{2}}{2} & -\frac{\sqrt{5}}{5} & 0 \\\frac{\sqrt{30}}{15} & -\frac{\sqrt{6}}{3} & 0 & \frac{\sqrt{5}}{5} & -\frac{\sqrt{2}}{2} \\\frac{\sqrt{30}}{10} & 0 & -\frac{\sqrt{2}}{2} & -\frac{\sqrt{5}}{5} & 0\end{array}\right]\left[\begin{array}{ccc}\sqrt{5} & 0 & 0 \\0 & \sqrt{2} & 0 \\0 & 0 & 1 \\0 & 0 & 0 \\0 & 0 & 0\end{array}\right]
\times\left[\begin{array}{ccc}\frac{\sqrt{6}}{6} & \frac{\sqrt{6}}{3} & \frac{\sqrt{6}}{6} \\\frac{\sqrt{3}}{3} & -\frac{\sqrt{3}}{3} & \frac{\sqrt{3}}{3} \\-\frac{\sqrt{2}}{2} & 0 & \frac{\sqrt{2}}{2}\end{array}\right].