Question: In Exercises 15 and 16, construct the pseudo-inverse of \(A\). Begin by using a matrix program to produce the SVD of \(A\), or, if that is not available, begin with an orthogonal diagonalization of \({A^T}A\). Use the pseudo-inverse to solve \(A{\rm{x}} = {\rm{b}}\), for \({\rm{b}} = \left( {6, - 1, - 4,6} \right)\) and let \(\mathop {\rm{x}}\limits^\^ \)be the solution. Make a calculation to verify that \(\mathop {\rm{x}}\limits^\^ \) is in Row \(A\). Find a nonzero vector \({\rm{u}}\) in Nul\(A\), and verify that \(\left\| {\mathop {\rm{x}}\limits^\^ } \right\| < \left\| {\mathop {\rm{x}}\limits^\^ + {\rm{u}}} \right\|\), which must be true by Exercise 13(c).

15. \(A = \left[ {\begin{array}{*{20}{c}}{ - 3}&{ - 3}&{ - 6}&6&{\,\,1}\\{ - 1}&{ - 1}&{ - 1}&1&{ - 2}\\{\,\,\,0}&{\,\,0}&{ - 1}&1&{ - 1}\\{\,\,\,0}&{\,\,0}&{ - 1}&1&{ - 1}\end{array}} \right]\)

The reduced SVD of the matrix\(A\)is\(A = {U_r}D{V_r}^T\), so that its pseudo inverse is\({A^ + } = {V_r}{D^{ - 1}}{U_r}^T\).

For the given matrix\(A\), \[{U_r} = \left[ {\begin{array}{*{20}{c}}{.966641}&{.253758}&{ - .034804}\\{185205}&{.786338}&{.589382}\\{.125107}&{.398296}&{.570709}\\{.125107}&{.398296}&{.570709}\end{array}} \right]{\rm{ }}\], \[D = \left[ {\begin{array}{*{20}{c}}{9.84443}&0&0\\0&{2.62466}&0\\0&0&{1.09467}\end{array}} \right]{\rm{ }}\]and \[{V_r} = \left[ {\begin{array}{*{20}{c}}{ - .313388}&{.009549}&{.633795}\\{ - .313388}&{.009549}&{.633795}\\{ - .633380}&{.023005}&{ - .313529}\\{ - .633380}&{.023005}&{ - .313529}\\{.035148}&{.999379}&{.002322}\end{array}} \right]{\rm{ }}\]

Use the MATLAB for calculating the pseudo inverse of matrix A by using the following steps:

1. Enter the matrix \({U_r}{\rm{ }}\), \(D\) and \({V_r}^T\)into the product program,
2. Run the program.
3. Press “Enter.”

The product is obtained as \({A^ + } = \left[ {\begin{array}{*{20}{c}}{ - .05}&{ - .35}&{.325}&{.325}\\{ - .05}&{ - .35}&{.325}&{.325}\\{ - .05}&{.15}&{ - .175}&{ - .175}\\{.05}&{ - .15}&{.175}&{.175}\\{.10}&{ - .30}&{ - .150}&{ - .150}\end{array}} \right]\).

Also, find the solution of the system \(A{\rm{x}} = {\rm{b}}\] as \[\mathop {\rm{x}}\limits^\^ = {A^ + }{\rm{b}}\)by using the same MATLAB program.

The solution is obtained as \(\mathop {\rm{x}}\limits^\^ = \left[ \begin{array}{l}\,\,\,.7\\\,\,\,.7\\ - .8\\\,\,\,.8\\\,\,\,.6\end{array} \right]\).

On running the row reduction program for the system\({A^T}{\rm{z}} = \mathop {\rm{x}}\limits^\^ \), it is obtained that its rank is equal to the number of free variables. So, the system has a solution such that\(\mathop {\rm{x}}\limits^\^ \)is in\({\rm{Col}}\,{A^T} = {\rm{Row}}\,A\).

Moreover, the basis for NulA is \(\left\{ {{{\rm{a}}_1},{{\rm{a}}_2}} \right\} = \left\{ {\left[ \begin{array}{l}0\\0\\1\\1\\0\end{array} \right]\left[ \begin{array}{l} - 1\\\,\,\,1\\\,\,\,0\\\,\,\,0\\\,\,\,0\end{array} \right]} \right\}\), such that any element of matrix Nul A is calculated as \({\rm{u}} = c{{\rm{a}}_1} + d{{\rm{a}}_2}\).

As \(\mathop {\rm{x}}\limits^\^ = \left[ \begin{array}{l}\,\,\,.7\\\,\,\,.7\\ - .8\\\,\,\,.8\\\,\,\,.6\end{array} \right]\), find \(\left\| {\mathop {\rm{x}}\limits^\^ } \right\|\) as:

\(\begin{array}{c}\left\| {\mathop {\rm{x}}\limits^\^ } \right\| = \sqrt {2{{\left( {0.7} \right)}^2} + 2{{\left( {0.7} \right)}^2} + 2{{\left( { - 0.8} \right)}^2} + 2{{\left( {0.8} \right)}^2} + 2{{\left( {0.6} \right)}^2}} \\ = \sqrt {131/50} \end{array}\)

So,\(\left\| {\mathop {\rm{x}}\limits^\^ + {\rm{u}}} \right\|\)is calculated as:

\(\left\| {\mathop {\rm{x}}\limits^\^ + {\rm{u}}} \right\| = \sqrt {{{\left( {131/50} \right)}^2} + 2{c^2} + 2{d^2}} \)

Thus, for nonzero matrix \({\rm{u}}\), \(\left\| {\mathop {\rm{x}}\limits^\^ } \right\| < \left\| {\mathop {\rm{x}}\limits^\^ + {\rm{u}}} \right\|\). This implies that \(\mathop {\rm{x}}\limits^\^ \) is the minimum length solution of \[A{\rm{x}} = {\rm{b}}\].