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Chapter 4: Q37E (page 191)

(M) Determine whether w is in the column space of \(A\), the null space of \(A\), or both, where

\({\mathop{\rm w}\nolimits} = \left( {\begin{array}{*{20}{c}}1\\1\\{ - 1}\\{ - 3}\end{array}} \right),A = \left( {\begin{array}{*{20}{c}}7&6&{ - 4}&1\\{ - 5}&{ - 1}&0&{ - 2}\\9&{ - 11}&7&{ - 3}\\{19}&{ - 9}&7&1\end{array}} \right)\)

Short Answer

Expert verified

\({\mathop{\rm w}\nolimits} \)is inCol A, and w is not in Nul A.

Step by step solution

01

Write an augmented matrix

Consider the augmented matrix \(\left( {\begin{array}{*{20}{c}}A&{\mathop{\rm w}\nolimits} \end{array}} \right)\) shown below:

\(\left( {\begin{array}{*{20}{c}}A&w\end{array}} \right) = \left( {\begin{array}{*{20}{c}}7&6&{ - 4}&1&1\\{ - 5}&{ - 1}&0&{ - 2}&1\\9&{ - 11}&7&{ - 3}&{ - 1}\\{19}&{ - 9}&7&1&{ - 3}\end{array}} \right)\)

02

Convert the matrix into row-reduced echelon form

Consider the matrix \(A = \left( {\begin{array}{*{20}{c}}7&6&{ - 4}&1&1\\{ - 5}&{ - 1}&0&{ - 2}&1\\9&{ - 11}&7&{ - 3}&{ - 1}\\{19}&{ - 9}&7&1&{ - 3}\end{array}} \right)\).

Use the following code in MATLAB to obtain the row-reduced echelon form.

\(\begin{array}{l} > > {\mathop{\rm A}\nolimits} = \left( {7\,\,\,6\,\,\, - 4\,\,\,1\,\,\,1;\,\, - 5\,\,\, - 1\,\,\,0\,\,\, - 2\,\,\,1;\,\,9\,\,\, - 11\,\,\,7\,\,\, - 3\,\,\, - 1;\,19\,\,\, - 9\,\,\,7\,\,\,1\,\,\, - 3} \right)\\ > > {\mathop{\rm U}\nolimits} = {\mathop{\rm rref}\nolimits} \left( {\mathop{\rm A}\nolimits} \right)\end{array}\)

\(\left( {\begin{array}{*{20}{c}}7&6&{ - 4}&1&1\\{ - 5}&{ - 1}&0&{ - 2}&1\\9&{ - 11}&7&{ - 3}&{ - 1}\\{19}&{ - 9}&7&1&{ - 3}\end{array}} \right) \sim \left( {\begin{array}{*{20}{c}}1&0&0&{\frac{{ - 1}}{{95}}}&{\frac{1}{{95}}}\\0&1&0&{\frac{{39}}{{19}}}&{\frac{{ - 20}}{{19}}}\\0&0&1&{\frac{{267}}{{95}}}&{\frac{{ - 172}}{{95}}}\\0&0&0&0&0\end{array}} \right)\)

03

Determine whether w is in the column space of A

A typical vector v in Col A has the property that the equation \(A{\mathop{\rm x}\nolimits} = {\mathop{\rm v}\nolimits} \) is consistent.

The system of equations of matrix A is consistent.

Thus, \({\mathop{\rm w}\nolimits} \) is inCol A.

04

Determine whether w is in the null space of A

A typical vector v in Nul A has the property that \(A{\mathop{\rm v}\nolimits} = 0\).

Use the code in MATLAB to compute the matrix \({\mathop{\rm Aw}\nolimits} \) as shown below:

\(\begin{array}{l} > > {\mathop{\rm A}\nolimits} = \left( {7\,\,\,6\,\,\, - 4\,\,\,1;\,\, - 5\,\,\, - 1\,\,\,0\,\,\, - 2;\,\,9\,\,\, - 11\,\,\,7\,\,\, - 3;\,19\,\,\, - 9\,\,\,7\,\,\,1} \right)\\ > > {\mathop{\rm w}\nolimits} = \left( {1;\,\,1;\,\, - 1;\,\, - 3} \right)\\ > > {\mathop{\rm U}\nolimits} = {\mathop{\rm A}\nolimits} * w\end{array}\)

\(\begin{array}{c}{\mathop{\rm A}\nolimits} {\mathop{\rm w}\nolimits} = \left( {\begin{array}{*{20}{c}}7&6&{ - 4}&1\\{ - 5}&{ - 1}&0&{ - 2}\\9&{ - 11}&7&{ - 3}\\{19}&{ - 9}&7&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}1\\1\\{ - 1}\\{ - 3}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{14}\\0\\0\\0\end{array}} \right)\end{array}\)

Thus, w is not in Nul A.

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Most popular questions from this chapter

Let \(H\) be an \(n\)-dimensional subspace of an \(n\)-dimensional vector space \(V\). Show that \(H = V\).

Justify the following equalities:

a.\({\rm{dim Row }}A{\rm{ + dim Nul }}A = n{\rm{ }}\)

b.\({\rm{dim Col }}A{\rm{ + dim Nul }}{A^T} = m\)

In Exercises 29 and 30, V is a nonzero finite-dimensional vector space, and the vectors listed belong to V. Mark each statement True or False. Justify each answer. (These questions are some difficult than those 19 and 20.)

30.

a. If there exists a linearly dependent set \(\left\{ {{{\mathop{\rm v}\nolimits} _1},...,{{\mathop{\rm v}\nolimits} _p}} \right\}\) in \(V\), then \(\dim V \le p\).

b. If every set of \(p\) elements in \(V\) fails to span \(V\), then \(\dim V > p\).

c. If \(p \ge 2\), and \(\dim V = p\), then every set of \(p - 1\) nonzero vectors is linearly independent.

In Exercise 4, find the vector x determined by the given coordinate vector \({\left( x \right)_{\rm B}}\) and the given basis \({\rm B}\).

4. \({\rm B} = \left\{ {\left( {\begin{array}{*{20}{c}}{ - {\bf{1}}}\\{\bf{2}}\\{\bf{0}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\bf{3}}\\{ - {\bf{5}}}\\{\bf{2}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\bf{4}}\\{ - {\bf{7}}}\\{\bf{3}}\end{array}} \right)} \right\},{\left( x \right)_{\rm B}} = \left( {\begin{array}{*{20}{c}}{ - {\bf{4}}}\\{\bf{8}}\\{ - {\bf{7}}}\end{array}} \right)\)

Consider the following two systems of equations:

\(\begin{array}{c}5{x_1} + {x_2} - 3{x_3} = 0\\ - 9{x_1} + 2{x_2} + 5{x_3} = 1\\4{x_1} + {x_2} - 6{x_3} = 9\end{array}\) \(\begin{array}{c}5{x_1} + {x_2} - 3{x_3} = 0\\ - 9{x_1} + 2{x_2} + 5{x_3} = 5\\4{x_1} + {x_2} - 6{x_3} = 45\end{array}\)

It can be shown that the first system of a solution. Use this fact and the theory from this section to explain why the second system must also have a solution. (Make no row operations.)
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