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Chapter 4: Q35E (page 191)

(M) Let \(H = {\bf{Span}}\left\{ {{{\bf{v}}_{\bf{1}}},\;{{\bf{v}}_{\bf{2}}}} \right\}\) and \(B = \left\{ {{{\bf{v}}_{\bf{1}}},\,{{\bf{v}}_{\bf{2}}}} \right\}\). Show that x is in the H and find the B coordinate vector of x, for

\({{\bf{v}}_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{{\bf{11}}}\\{ - {\bf{5}}}\\{{\bf{10}}}\\{\bf{7}}\end{array}} \right)\), \({{\bf{v}}_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{{\bf{14}}}\\{ - {\bf{8}}}\\{{\bf{13}}}\\{{\bf{10}}}\end{array}} \right)\), \({\bf{x}} = \left( {\begin{array}{*{20}{c}}{{\bf{19}}}\\{ - {\bf{13}}}\\{{\bf{18}}}\\{{\bf{15}}}\end{array}} \right)\)

Short Answer

Expert verified

\({\left( {\bf{x}} \right)_B} = \left( {\begin{array}{*{20}{c}}{ - \frac{5}{3}}\\{\frac{8}{3}}\end{array}} \right)\)

Step by step solution

01

Write the augmented matrix \(\left( {\begin{array}{*{20}{c}}{{{\bf{v}}_1}}&{{{\bf{v}}_2}}&{\bf{x}}\end{array}} \right)\)

\(\left( {\begin{array}{*{20}{c}}{{{\bf{v}}_1}}&{{{\bf{v}}_2}}&{\bf{x}}\end{array}} \right)\)

The augmented matrixcan be written as:

\(\left( {\begin{array}{*{20}{c}}{{{\bf{v}}_1}}&{{{\bf{v}}_2}}&{\bf{x}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{11}&{14}&{19}\\{ - 5}&{ - 8}&{ - 13}\\{10}&{13}&{18}\\7&{10}&{15}\end{array}} \right)\)

02

Write the augmented matrix in the echelon form

Consider matrix \(A = \left( {\begin{array}{*{20}{c}}{11}&{14}&{19}\\{ - 5}&{ - 8}&{ - 13}\\{10}&{13}&{18}\\7&{10}&{15}\end{array}} \right)\).

Use code in the MATLAB to obtain the row-reduced echelon form as shown below:

\(\begin{array}{l} > > {\rm{ A }} = {\rm{ }}\left( {{\rm{ }}\begin{array}{*{20}{c}}{11}&{14}&{19;\;\;\begin{array}{*{20}{c}}{ - 5}&{ - 8}&{ - 13;\;\;\begin{array}{*{20}{c}}{10}&{13}&{18;\;\;\begin{array}{*{20}{c}}7&{10}&{15}\end{array}}\end{array}}\end{array}}\end{array}} \right);\\ > > {\rm{ U}} = {\rm{rref}}\left( {\rm{A}} \right)\end{array}\)

\(\left( {\begin{array}{*{20}{c}}{11}&{14}&{19}\\{ - 5}&{ - 8}&{ - 13}\\{10}&{13}&{18}\\7&{10}&{15}\end{array}} \right) \sim \left( {\begin{array}{*{20}{c}}1&0&{ - \frac{5}{3}}\\0&1&{\frac{8}{3}}\\0&0&0\\0&0&0\end{array}} \right)\)

03

Write the system of equations for x

For\({\bf{x}}\), using the augmented matrix, you get:

\({\bf{x}} = {c_1}{{\bf{x}}_1} + {c_2}{{\bf{x}}_2}\)

So, the values of\({c_1}\)and\({c_2}\)are:

\({c_1} = - \frac{5}{3}\)and\({c_2} = \frac{8}{3}\)

Therefore, the B-coordinate for vector \({\bf{x}}\) is \({\left( {\bf{x}} \right)_B} = \left( {\begin{array}{*{20}{c}}{ - \frac{5}{3}}\\{\frac{8}{3}}\end{array}} \right)\).

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Most popular questions from this chapter

Suppose the solutions of a homogeneous system of five linear equations in six unknowns are all multiples of one nonzero solution. Will the system necessarily have a solution for every possible choice of constants on the right sides of the equations? Explain.

Let be a linear transformation from a vector space \(V\) \(T:V \to W\)in to vector space \(W\). Prove that the range of T is a subspace of . (Hint: Typical elements of the range have the form \(T\left( {\mathop{\rm x}\nolimits} \right)\) and \(T\left( {\mathop{\rm w}\nolimits} \right)\) for some \({\mathop{\rm x}\nolimits} ,\,{\mathop{\rm w}\nolimits} \)in \(V\).)\(W\)

(M) Let \(H = {\mathop{\rm Span}\nolimits} \left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2}} \right\}\) and \(K = {\mathop{\rm Span}\nolimits} \left\{ {{{\mathop{\rm v}\nolimits} _3},{{\mathop{\rm v}\nolimits} _4}} \right\}\), where

\({{\mathop{\rm v}\nolimits} _1} = \left( {\begin{array}{*{20}{c}}5\\3\\8\end{array}} \right),{{\mathop{\rm v}\nolimits} _2} = \left( {\begin{array}{*{20}{c}}1\\3\\4\end{array}} \right),{{\mathop{\rm v}\nolimits} _3} = \left( {\begin{array}{*{20}{c}}2\\{ - 1}\\5\end{array}} \right),{{\mathop{\rm v}\nolimits} _4} = \left( {\begin{array}{*{20}{c}}0\\{ - 12}\\{ - 28}\end{array}} \right)\)

Then \(H\) and \(K\) are subspaces of \({\mathbb{R}^3}\). In fact, \(H\) and \(K\) are planes in \({\mathbb{R}^3}\) through the origin, and they intersect in a line through 0. Find a nonzero vector w that generates that line. (Hint: w can be written as \({c_1}{{\mathop{\rm v}\nolimits} _1} + {c_2}{{\mathop{\rm v}\nolimits} _2}\) and also as \({c_3}{{\mathop{\rm v}\nolimits} _3} + {c_4}{{\mathop{\rm v}\nolimits} _4}\). To build w, solve the equation \({c_1}{{\mathop{\rm v}\nolimits} _1} + {c_2}{{\mathop{\rm v}\nolimits} _2} = {c_3}{{\mathop{\rm v}\nolimits} _3} + {c_4}{{\mathop{\rm v}\nolimits} _4}\) for the unknown \({c_j}'{\mathop{\rm s}\nolimits} \).)

Let S be a maximal linearly independent subset of a vector space V. In other words, S has the property that if a vector not in S is adjoined to S, the new set will no longer be linearly independent. Prove that S must be a basis of V. [Hint: What if S were linearly independent but not a basis of V?]

If A is a \({\bf{6}} \times {\bf{4}}\) matrix, what is the smallest possible dimension of Null A?

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