Question: In parts (a)-(d), let U be the matrix formed by normalizing each column of matrix A in Exercise 35.

1. Compute \({U^T}U\) and \(U{U^T}\). How do they differ?
2. Generate a random vector y in\({\mathbb{R}^{\bf{8}}}\), and compute\({\bf{p}} = U{U^T}{\bf{y}}\)and\({\bf{z}} = {\bf{y}} - {\bf{p}}\). Explain why p is in col A. Verfiy that z is orthogonal to p.
3. Verfiy that z is orthogonal to each column of U.
4. Notice that \({\bf{y}} = {\bf{p}} + {\bf{z}}\), with p in Col A. Explain why z is in \({\left( {{\bf{Col}}A} \right)^ \bot }\). (The significance of this decomposition of y will be explained in the next section.)

Use the following MATLAB command to obtain the transpose of A.

\( > > A = \left( \begin{array}{l}\begin{array}{*{20}{c}}{ - 6}&{ - 3}&6&1\end{array};\,\begin{array}{*{20}{c}}{ - 1}&2&1&{ - 6}\end{array};\,\begin{array}{*{20}{c}}3&6&3&{ - 2;\,\begin{array}{*{20}{c}}6&{ - 3}&6&{ - 1}\end{array}}\end{array};\\\begin{array}{*{20}{c}}2&{ - 1}&2&{3;\,\begin{array}{*{20}{c}}{ - 3}&6&3&{2;\,\,\begin{array}{*{20}{c}}{ - 2}&{ - 1}&2&{ - 3;\,\,\begin{array}{*{20}{c}}1&2&1&6\end{array}}\end{array}}\end{array}}\end{array}\end{array} \right);\)

Find the matrix \(U\).

\( > > U = \frac{A}{{10}};\)

The matrix Uis:

\(U = \left( {\begin{array}{*{20}{c}}{ - 0.6000}&{ - 0.3000}&{0.6000}&{0.1000}\\{ - 0.1000}&{0.2000}&{0.1000}&{ - 0.6000}\\{0.3000}&{0.6000}&{0.3000}&{ - 0.2000}\\{0.6000}&{ - 0.3000}&{0.6000}&{ - 0.1000}\\{0.2000}&{ - 0.1000}&{0.2000}&{0.3000}\\{ - 0.3000}&{0.6000}&{0.3000}&{0.2000}\\{ - 0.2000}&{ - 0.1000}&{0.2000}&{ - 0.3000}\\{0.1000}&{0.2000}&{0.1000}&{0.6000}\end{array}} \right)\)

Compute the product \({U^T}U\) and \(U{U^T}\).

\(\begin{array}{c} > > P\_1 = U'*U\\ > > P\_2 = U*U'\end{array}\)

The product \({U^T}U\) and \(U{U^T}\) is shown below:

\({U^T}U = \left( {\begin{array}{*{20}{c}}{1.0000}&0&0&0\\0&{1.0000}&0&0\\0&0&{1.0000}&0\\0&0&0&{1.0000}\end{array}} \right)\)

And,

\(U{U^T} = \left( {\begin{array}{*{20}{c}}{0.8200}&0&{ - 0.2000}&{0.0800}\\0&{0.4200}&{0.2400}&0\\{ - 0.2000}&{0.2400}&{0.5800}&{0.2000}\\{0.0800}&0&{0.2000}&{0.8200}\\{0.0600}&{ - 0.2000}&0&{0.2400}\\{0.2000}&{0.0600}&{0.3200}&{ - 0.2000}\\{0.2400}&{0.2000}&0&{0.0600}\\0&{ - 0.3200}&{0.0600}&0\end{array}} \right)\)

The matrix \({U^T}U\) is of order \(4 \times 4\) , and the matrix \(U{U^T}\) is of order \(8 \times 8\).

Consider the following column vector \(y = {\left( {\begin{array}{*{20}{c}}1&1&1&1&1&1&1&1\end{array}} \right)^T}\).

Use the following command in the MATLAB to obtain the random vector:

\(\begin{array}{l} > > {\bf{y}} = {\rm{ones}}\left( {1,8} \right)\\ > > {\rm{p}} = {\rm{U}}*{\rm{U}}*{\rm{y}}\\ > > {\rm{z}} = {\rm{y}} - {\rm{p}}\end{array}\)

So, the vector z is shown below:

\(z = \left( {\begin{array}{*{20}{c}}{ - 0.2000}\\{0.6000}\\{ - 0.2000}\\{ - 0.2000}\\{0.6000}\\{ - 0.2000}\\{0.6000}\\{0.6000}\end{array}} \right)\)

As \({\bf{p}} = U{U^T}{\bf{y}}\), therefore, pis the column space of U.

Find the product of z and p.

\( > > {\rm{z}}*{\bf{p}}\)

The product \({\rm{z}}*{\rm{p}}\) gives the result as zero. Therefore z and p are orthogonal to each other.

Multiply each column of U with zby using the following MATLAB commands:

\(\begin{array}{l} > > {\rm{z}}*{\rm{U}}\left( {;1} \right)\\ > > {\rm{z}}*{\rm{U}}\left( {;2} \right)\\ > > {\rm{z}}*{\rm{U}}\left( {;3} \right)\\ > > {\rm{z}}*{\rm{U}}\left( {;4} \right)\end{array}\)

The result of all the products is zero. Therefore z is orthogonal to each column of U.

As it is proved in step 3, that zis orthogonal to each column of A. Therefore z must be orthogonal to every vector in the column space of A.

Thus, z is in the column space of A.