Question 9.6.3: Determine the singular value decomposition of the 5 × 3 matr...

We have

A A^{t}=\left[\begin{array}{lll}1 & 0 & 1 \\0 & 1 & 0 \\0 & 1 & 1 \\0 & 1 & 0 \\1 & 1 & 0\end{array}\right]\left[\begin{array}{lllll}1 & 0 & 0 & 0 & 1 \\0 & 1 & 1 & 1 & 1 \\1 & 0 & 1 & 0 & 0\end{array}\right]=\left[\begin{array}{lllll}2 & 0 & 1 & 0 & 1 \\0 & 1 & 1 & 1 & 1 \\1 & 1 & 2 & 1 & 1 \\0 & 1 & 1 & 1 & 1 \\1 & 1 & 1 & 1 & 2\end{array}\right],

which has the same nonzero eigenvalues as A^{t} A, \text { that is, } \lambda_{1}=5, \lambda_{2}=2, \text { and } \lambda_{3}=1 , and, additionally, \lambda_{4}=0, \text { and } \lambda_{5}=0 . Eigenvectors corresponding to these eigenvalues are, respectively,

x _{1}=(2,2,3,2,3)^{t}, \quad x _{2}=(2,-1,0,-1,0)^{t}, \quad x _{3}=(0,0,1,0,-1)^{t}, \quad x _{4}=(1,2,-1,0,-1)^{t},

and x _{5}=(0,1,0,-1,0)^{t}.

Both the sets \left\{ x _{1}, x _{2}, x _{3}, x _{4}\right\} \text { and }\left\{ x _{1}, x _{2}, x _{3}, x _{5}\right\} are orthogonal because they are eigenvectors associated with distinct eigenvalues of the symmetric matrix A A^{t} . However, x_{4} is not orthogonal to x_{5} . We will keep x_{4} as one of the eigenvectors used to form U and determine the fifth vector that will give an orthogonal set. For this we use the Gram Schmidt process as described in Theorem 9.8 on page 567. Using the notation in that theorem we have

v _{1}= x _{1}, v _{2}= x _{2}, v _{3}= x _{3}, v _{4}= x _{4},

and, because x_{5} is orthogonal to all but x_{4} ,

\begin{aligned}v _{5} &= x _{5}-\frac{ v _{4}^{t} x _{5}}{ v _{4}^{t} v _{4}} x _{4} \\&=(0,1,0,-1,0)^{t}-\frac{(1,2,-1,0,-1) \cdot(0,1,0,-1,0)^{t}}{\left\|(1,2,-1,0,-1)^{t}\right\|_{2}^{2}}(1,2,-1,0,-1) \\&=(0,1,0,-1,0)^{t}-\frac{2}{7}(1,2,-1,0,-1)^{t}=-\frac{1}{7}(2,-3,-2,7,-2)^{t}\end{aligned}.

It is easily verified that v _{5} \text { is orthogonal to } v _{4}= x _{4} . It is also orthogonal to the vectors in \left\{ v _{1}, v _{2}, v _{3}\right\} because it is a linear combination of x _{4} \text { and } x _{5} . Normalizing these vectors gives the columns of the matrix U in the factorization. Hence

U=\left[ u _{1}, u _{2}, u _{3}, u _{4}, u _{5}\right]=\left[\begin{array}{ccccc}\frac{\sqrt{30}}{15} & \frac{\sqrt{6}}{3} & 0 & \frac{\sqrt{7}}{7} & \frac{\sqrt{70}}{35} \\\frac{\sqrt{30}}{15} & -\frac{\sqrt{6}}{6} & 0 & \frac{2 \sqrt{7}}{7} & -\frac{3 \sqrt{70}}{70} \\\frac{\sqrt{30}}{10} & 0 & \frac{\sqrt{2}}{2} & -\frac{\sqrt{7}}{7} & -\frac{\sqrt{70}}{35} \\\frac{\sqrt{30}}{15} & -\frac{\sqrt{6}}{6} & 0 & 0 & \frac{\sqrt{70}}{10} \\\frac{\sqrt{30}}{10} & 0 & -\frac{\sqrt{2}}{2} & -\frac{\sqrt{7}}{7} & -\frac{\sqrt{70}}{35}\end{array}\right].

This is a different U from the one found in Example 2, but it gives a valid factorization A=U S V^{t}  using the same S and V as in that example.