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Chapter 7: Q15SE (page 395)

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

Short Answer

Expert verified

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 \({\rm{u}} = c{{\rm{a}}_1} + d{{\rm{a}}_2}\). It is verified that \(\mathop {\rm{x}}\limits^\^ \) is the minimum length solution of \[A{\rm{x}} = {\rm{b}}\].

Step by step solution

01

Find the pseudo-inverse of A

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 鈥淓nter.鈥�

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

02

Find a basis for Nul A

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 \({\rm{u}} = c{{\rm{a}}_1} + d{{\rm{a}}_2}\).

03

Compute \(\left\| {\mathop {\rm{x}}\limits^\^ } \right\|\) and \(\left\| {\mathop {\rm{x}}\limits^\^  + {\rm{u}}} \right\|\)

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 \[A{\rm{x}} = {\rm{b}}\].

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Most popular questions from this chapter

Question: In Exercises 1 and 2, convert the matrix of observations to mean deviation form, and construct the sample covariance matrix.

\(1.\,\,\left( {\begin{array}{*{20}{c}}{19}&{22}&6&3&2&{20}\\{12}&6&9&{15}&{13}&5\end{array}} \right)\)

Classify the quadratic forms in Exercises 9鈥�18. Then make a change of variable, \({\bf{x}} = P{\bf{y}}\), that transforms the quadratic form into one with no cross-product term. Write the new quadratic form. Construct \(P\) using the methods of Section 7.1.

11. \({\bf{2}}x_{\bf{1}}^{\bf{2}} - {\bf{4}}{x_{\bf{1}}}{x_{\bf{2}}} - x_{\bf{2}}^{\bf{2}}\)

Question: 14. Exercises 12鈥�14 concern an \(m \times n\) matrix \(A\) with a reduced singular value decomposition, \(A = {U_r}D{V_r}^T\), and the pseudoinverse \({A^ + } = {U_r}{D^{ - 1}}{V_r}^T\).

Given any \({\rm{b}}\) in \({\mathbb{R}^m}\), adapt Exercise 13 to show that \({A^ + }{\rm{b}}\) is the least-squares solution of minimum length. [Hint: Consider the equation \(A{\rm{x}} = {\rm{b}}\), where \(\mathop {\rm{b}}\limits^\^ \) is the orthogonal projection of \({\rm{b}}\) onto Col \(A\).

Determine which of the matrices in Exercises 1鈥�6 are symmetric.

\(\left( {\begin{array}{{}{}}1&2&2&1\\2&2&2&1\\2&2&1&2\end{array}} \right)\)

Orthogonally diagonalize the matrices in Exercises 13鈥�22, giving an orthogonal matrix\(P\)and a diagonal matrix\(D\). To save you time, the eigenvalues in Exercises 17鈥�22 are: (17)\( - {\bf{4}}\), 4, 7; (18)\( - {\bf{3}}\),\( - {\bf{6}}\), 9; (19)\( - {\bf{2}}\), 7; (20)\( - {\bf{3}}\), 15; (21) 1, 5, 9; (22) 3, 5.

14. \(\left( {\begin{aligned}{{}}{\,1}&{ - 5}\\{ - 5}&{\,\,1}\end{aligned}} \right)\)
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