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Chapter 4: Q8SE (page 191)

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

Short Answer

Expert verified

H is a subspace of V, and B is also in V. As B is a linearly independent set in V, B must also be a basis for V.

Step by step solution

01

Find the dimension of H and V

The bases of H and V have exactly n vectors because H is an n-dimensional subspace of V.

For \(n = 0\),

\(H = V = \left\{ 0 \right\}\).

02

Check for \(H = V\)

For subspace H is n-dimensional subspace, there is a B for H. Bmust have n elements and be linearly independent.

Since H is a subspace of V and B is also in V, B is a linearly independent set in V. So, B must also be a basis for V. Hence, Hand V are the same.

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Most popular questions from this chapter

If a\({\bf{6}} \times {\bf{3}}\)matrix A has a rank 3, find dim Nul A, dim Row A, and rank\({A^T}\).

Let \(B = \left\{ {\left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{4}}}\end{array}} \right),\,\left( {\begin{array}{*{20}{c}}{ - {\bf{2}}}\\{\bf{9}}\end{array}} \right)\,} \right\}\). Since the coordinate mapping determined by B is a linear transformation from \({\mathbb{R}^{\bf{2}}}\) into \({\mathbb{R}^{\bf{2}}}\), this mapping must be implemented by some \({\bf{2}} \times {\bf{2}}\) matrix A. Find it. (Hint: Multiplication by A should transform a vector x into its coordinate vector \({\left( {\bf{x}} \right)_B}\)).

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

Explain what is wrong with the following discussion: Let \({\bf{f}}\left( t \right) = {\bf{3}} + t\) and \({\bf{g}}\left( t \right) = {\bf{3}}t + {t^{\bf{2}}}\), and note that \({\bf{g}}\left( t \right) = t{\bf{f}}\left( t \right)\). Then, \(\left\{ {{\bf{f}},{\bf{g}}} \right\}\) is linearly dependent because g is a multiple of f.

Question 11: Let \(S\) be a finite minimal spanning set of a vector space \(V\). That is, \(S\) has the property that if a vector is removed from \(S\), then the new set will no longer span \(V\). Prove that \(S\) must be a basis for \(V\).
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