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Chapter 4: Q10SE (page 191)

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?]

Short Answer

Expert verified

S must be a basis of V.

Step by step solution

01

Find a vector v in set V

Let \(S = \left\{ {{{\bf{v}}_1},\,{{\bf{v}}_2},.....,{{\bf{v}}_p}} \right\}\).

Assume that S does not span V.

Therefore, \(V \ne {\rm{Span}}\left\{ {{{\bf{v}}_1},{{\bf{v}}_2},....,{{\bf{v}}_p}} \right\}\).

Then, there must be a vector v in set V that is not in the \({\rm{Span}}\left\{ {{{\bf{v}}_1},{{\bf{v}}_2},....{{\bf{v}}_p}} \right\}\).

02

Check for linear independency

Let \(S'\) be a set formed by

\(\begin{aligned} S' &= S \cup \left\{ {\bf{v}} \right\}\\ &= \left\{ {{{\bf{v}}_1},{{\bf{v}}_2},.....,{{\bf{v}}_p},{\bf{v}}} \right\}\end{aligned}\).

The set \(S'\) is also a linearly independent set, but the number of elements in the set \(S'\) is more than \(S\)(contradiction).

So, S must be a basis of V.

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Most popular questions from this chapter

Let \(V\) and \(W\) be vector spaces, and let \(T:V \to W\) be a linear transformation. Given a subspace \(U\) of \(V\), let \(T\left( U \right)\) denote the set of all images of the form \(T\left( {\mathop{\rm x}\nolimits} \right)\), where x is in \(U\). Show that \(T\left( U \right)\) is a subspace of \(W\).

Question: Exercises 12-17 develop properties of rank that are sometimes needed in applications. Assume the matrix \(A\) is \(m \times n\).

14. Show that if \(Q\) is an invertible, then \({\mathop{\rm rank}\nolimits} AQ = {\mathop{\rm rank}\nolimits} A\). (Hint: Use Exercise 13 to study \({\mathop{\rm rank}\nolimits} {\left( {AQ} \right)^T}\).)

Use coordinate vector to test whether the following sets of poynomial span \({{\bf{P}}_{\bf{2}}}\). Justify your conclusions.

a. \({\bf{1}} - {\bf{3}}t + {\bf{5}}{t^{\bf{2}}}\), \( - {\bf{3}} + {\bf{5}}t - {\bf{7}}{t^{\bf{2}}}\), \( - {\bf{4}} + {\bf{5}}t - {\bf{6}}{t^{\bf{2}}}\), \({\bf{1}} - {t^{\bf{2}}}\)

b. \({\bf{5}}t + {t^{\bf{2}}}\), \({\bf{1}} - {\bf{8}}t - {\bf{2}}{t^{\bf{2}}}\), \( - {\bf{3}} + {\bf{4}}t + {\bf{2}}{t^{\bf{2}}}\), \({\bf{2}} - {\bf{3}}t\)

In Exercise 2, find the vector x determined by the given coordinate vector \({\left( x \right)_{\rm B}}\) and the given basis \({\rm B}\).

2. \({\rm B} = \left\{ {\left( {\begin{array}{*{20}{c}}{\bf{4}}\\{\bf{5}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\bf{6}}\\{\bf{7}}\end{array}} \right)} \right\},{\left( x \right)_{\rm B}} = \left( {\begin{array}{*{20}{c}}{\bf{8}}\\{ - {\bf{5}}}\end{array}} \right)\)

In Exercise 3, find the vector x determined by the given coordinate vector \({\left( x \right)_{\rm B}}\) and the given basis \({\rm B}\).

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