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Chapter 7: Q16SE (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).

16. \(A = \left( {\begin{array}{*{20}{c}}4&0&{ - 1}&{ - 2}&0\\{ - 5}&0&3&5&0\\{\,\,\,2}&{\,\,0}&{ - 1}&{ - 2}&0\\{\,\,\,6}&{\,\,0}&{ - 3}&{ - 6}&0\end{array}} \right)\)

Short Answer

Expert verified

The basis for NulA is \(\left\{ {{{\rm{a}}_1},{{\rm{a}}_2}} \right\} = \left\{ {\left( \begin{array}{l}0\\1\\0\\0\\0\end{array} \right)\left( \begin{array}{l}0\\0\\0\\0\\1\end{array} \right)} \right\}\), 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\(A\),\({U_r} = \left( {\begin{array}{*{20}{c}}{ - .337977}&{.936307}&{.095396}\\{.591763}&{.290230}&{ - .752053}\\{ - .231428}&{ - .062526}&{.206232}\\{ - .694283}&{ - .187578}&{ - .618696}\end{array}} \right){\rm{ }}\),\(D = \left( {\begin{array}{*{20}{c}}{12.9536}&0&0\\0&{1.44553}&0\\0&0&{.337763}\end{array}} \right){\rm{ }}\)and\({V_r} = \left( {\begin{array}{*{20}{c}}{ - .690099}&{.721920}&{.050939}\\0&0&0\\{.341800}&{.387156}&{ - .856320}\\{.637916}&{.573534}&{.513928}\\0&0&0\end{array}} \right){\rm{ }}\)

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\({A^ + } = \left( {\begin{array}{*{20}{c}}{.5}&0&{ - .05}&{ - .15}\\0&0&0&0\\0&2&{.5}&{1.5}\\{.5}&{ - 1}&{ - .35}&{ - 1.05}\\0&0&0&0\end{array}} \right)\).

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 \(\mathop {\rm{x}}\limits^\^ = \left( \begin{array}{l}2.3\\\,\,\,\,0\\\,5.0\\\, - .9\\\,\,\,\,\,0\end{array} \right)\).

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\(\left\{ {{{\rm{a}}_1},{{\rm{a}}_2}} \right\} = \left\{ {\left( \begin{array}{l}0\\1\\0\\0\\0\end{array} \right)\left( \begin{array}{l}0\\0\\0\\0\\1\end{array} \right)} \right\}\), such that any element of matrix NulA 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\(\mathop {\rm{x}}\limits^\^ = \left( \begin{array}{l}\,\,\,.7\\\,\,\,.7\\ - .8\\\,\,\,.8\\\,\,\,.6\end{array} \right)\), so find\(\left\| {\mathop {\rm{x}}\limits^\^ } \right\|\)as:

\(\begin{array}{c}\left\| {\mathop {\rm{x}}\limits^\^ } \right\| = \sqrt {2{{\left( {2.3} \right)}^2} + 2{{\left( 0 \right)}^2} + 2{{\left( {5.0} \right)}^2} + 2{{\left( { - .9} \right)}^2} + 2{{\left( 0 \right)}^2}} \\ = \sqrt {311/10} \end{array}\)

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

\(\left\| {\mathop {\rm{x}}\limits^\^ + {\rm{u}}} \right\| = \sqrt {{{\left( {311/10} \right)}^2} + 2{c^2} + 2{d^2}} \)

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: Mark Each statement True or False. Justify each answer. In each part, A represents an \(n \times n\) matrix.

  1. If A is orthogonally diagonizable, then A is symmetric.
  2. If A is an orthogonal matrix, then A is symmetric.
  3. If A is an orthogonal matrix, then \(\left\| {A{\bf{x}}} \right\| = \left\| {\bf{x}} \right\|\) for all x in \({\mathbb{R}^n}\).
  4. The principal axes of a quadratic from \({{\bf{x}}^T}A{\bf{x}}\) can be the columns of any matrix P that diagonalizes A.
  5. If P is an \(n \times n\) matrix with orthogonal columns, then \({P^T} = {P^{ - {\bf{1}}}}\).
  6. If every coefficient in a quadratic form is positive, then the quadratic form is positive definite.
  7. If \({{\bf{x}}^T}A{\bf{x}} > {\bf{0}}\) for some x, then the quadratic form \({{\bf{x}}^T}A{\bf{x}}\) is positive definite.
  8. By a suitable change of variable, any quadratic form can be changed into one with no cross-product term.
  9. The largest value of a quadratic form \({{\bf{x}}^T}A{\bf{x}}\), for \(\left\| {\bf{x}} \right\| = {\bf{1}}\) is the largest entery on the diagonal A.
  10. The maximum value of a positive definite quadratic form \({{\bf{x}}^T}A{\bf{x}}\) is the greatest eigenvalue of A.
  11. A positive definite quadratic form can be changed into a negative definite form by a suitable change of variable \({\bf{x}} = P{\bf{u}}\), for some orthogonal matrix P.
  12. An indefinite quadratic form is one whose eigenvalues are not definite.
  13. If P is an \(n \times n\) orthogonal matrix, then the change of variable \({\bf{x}} = P{\bf{u}}\) transforms \({{\bf{x}}^T}A{\bf{x}}\) into a quadratic form whose matrix is \({P^{ - {\bf{1}}}}AP\).
  14. If U is \(m \times n\) with orthogonal columns, then \(U{U^T}{\bf{x}}\) is the orthogonal projection of x onto ColU.
  15. If B is \(m \times n\) and x is a unit vector in \({\mathbb{R}^n}\), then \(\left\| {B{\bf{x}}} \right\| \le {\sigma _{\bf{1}}}\), where \({\sigma _{\bf{1}}}\) is the first singular value of B.
  16. A singular value decomposition of an \(m \times n\) matrix B can be written as \(B = P\Sigma Q\), where P is an \(m \times n\) orthogonal matrix and \(\Sigma \) is an \(m \times n\) diagonal matrix.
  17. If A is \(n \times n\), then A and \({A^T}A\) have the same singular values.

Determine which of the matrices in Exercises 7–12 are orthogonal. If orthogonal, find the inverse.

7. \(\left( {\begin{aligned}{{}{}}{.6}&{\,\,\,.8}\\{.8}&{ - .6}\end{aligned}} \right)\)

Question: Let \({x_1}\,,{x_2}\) denote the variables for the two-dimensional data in Exercise 1. Find a new variable \({y_1}\) of the form \({y_1} = {c_1}{x_1} + {c_2}{x_2}\), with\(c_1^2 + c_2^2 = 1\), such that \({y_1}\) has maximum possible variance over the given data. How much of the variance in the data is explained by \({y_1}\)?

25.Let \({\bf{T:}}{\mathbb{R}^{\bf{n}}} \to {\mathbb{R}^{\bf{m}}}\) be a linear transformation. Describe how to find a basis \(B\) for \({\mathbb{R}^n}\) and a basis \(C\) for \({\mathbb{R}^m}\) such that the matrix for \(T\) relative to \(B\) and \(C\) is an \(m \times n\) “diagonal” matrix.

Question: 2. Let \(\left\{ {{{\bf{u}}_1},{{\bf{u}}_2},....,{{\bf{u}}_n}} \right\}\) be an orthonormal basis for \({\mathbb{R}_n}\) , and let \({\lambda _1},....{\lambda _n}\) be any real scalars. Define

\(A = {\lambda _1}{{\bf{u}}_1}{\bf{u}}_1^T + ..... + {\lambda _n}{\bf{u}}_n^T\)

a. Show that A is symmetric.

b. Show that \({\lambda _1},....{\lambda _n}\) are the eigenvalues of A
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