Exercises 23-26 concern a vector space V, a basis \(B = \left\{ {{{\bf{b}}_{\bf{1}}},....,{{\bf{b}}_n}\,} \right\}\)and the coordinate mapping \({\bf{x}} \mapsto {\left( {\bf{x}} \right)_B}\).

Given vectors, \({u_{\bf{1}}}\),….,\({u_p}\) and w in V, show that w is a linear combination of \({u_{\bf{1}}}\),….,\({u_p}\) if and only if \({\left( w \right)_B}\) is a linear combination of vectors \({\left( {{{\bf{u}}_{\bf{1}}}} \right)_B}\),….,\({\left( {{{\bf{u}}_p}} \right)_B}\).

Short Answer

Expert verified

\({\left( {\bf{w}} \right)_B}\) can be written as a linear combination of \({\left( {{{\bf{u}}_1}} \right)_B}\), \({\left( {{{\bf{u}}_2}} \right)_B}\),….,\({\left( {{{\bf{u}}_p}} \right)_B}\).

Step by step solution

01

Write w as a linear combination

Was a linear combinationof \({{\bf{u}}_1}\), \({{\bf{u}}_2}\),…., \({{\bf{u}}_p}\) can be written as:

\({\bf{w}} = {c_1}{{\bf{u}}_1} + {c_2}{{\bf{u}}_2} + .... + {c_p}{{\bf{u}}_p}\)

The first equation has a trivial solution if the second equation also has a trivial solution.

02

Check for \({\left( {\bf{w}} \right)_B}\)

If B is a basis for space V, then the coordinate mapping is one to one transformation. Therefore,

\(\begin{array}{c}{\left( {\bf{w}} \right)_B} = {\left( {{c_1}{{\bf{u}}_1} + {c_2}{{\bf{u}}_2} + .....{c_p}{{\bf{u}}_p}} \right)_B}\\ = {\left( {{c_1}{{\bf{u}}_1}} \right)_B} + {\left( {{c_2}{{\bf{u}}_2}} \right)_B} + .... + {\left( {{c_p}{{\bf{u}}_p}} \right)_B}\\ = {c_1}{\left( {{{\bf{u}}_1}} \right)_B} + {c_2}{\left( {{{\bf{u}}_2}} \right)_B} + .... + {c_p}{\left( {{{\bf{u}}_p}} \right)_B}\end{array}\)

Thus, the \({\left( {\bf{w}} \right)_B}\) can be written as a linear combination of \({\left( {{{\bf{u}}_1}} \right)_B}\), \({\left( {{{\bf{u}}_2}} \right)_B}\),….,\({\left( {{{\bf{u}}_p}} \right)_B}\).

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Most popular questions from this chapter

(calculus required) Define \(T:C\left( {0,1} \right) \to C\left( {0,1} \right)\) as follows: For f in \(C\left( {0,1} \right)\), let \(T\left( t \right)\) be the antiderivative \({\mathop{\rm F}\nolimits} \) of \({\mathop{\rm f}\nolimits} \) such that \({\mathop{\rm F}\nolimits} \left( 0 \right) = 0\). Show that \(T\) is a linear transformation, and describe the kernel of \(T\). (See the notation in Exercise 20 of Section 4.1.)

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.)

Question: Determine if the matrix pairs in Exercises 19-22 are controllable.

22. (M) \(A = \left( {\begin{array}{*{20}{c}}0&1&0&0\\0&0&1&0\\0&0&0&1\\{ - 1}&{ - 13}&{ - 12.2}&{ - 1.5}\end{array}} \right),B = \left( {\begin{array}{*{20}{c}}1\\0\\0\\{ - 1}\end{array}} \right)\).

(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 \({c_j}'{\mathop{\rm s}\nolimits} \).)

Find a basis for the set of vectors in\({\mathbb{R}^{\bf{3}}}\)in the plane\(x + {\bf{2}}y + z = {\bf{0}}\). (Hint:Think of the equation as a “system” of homogeneous equations.)

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