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Let \({{\bf{v}}_1} = \left[ {\begin{array}{*{20}{c}}1\\0\\1\end{array}} \right]\), \({{\bf{v}}_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{1}}\\1\end{array}} \right]\), \({{\bf{v}}_{\bf{3}}} = \left[ {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{1}}\\{\bf{0}}\end{array}} \right]\), and let H be the set of vectors in \({\mathbb{R}^{\bf{3}}}\) whose second and third entries are equal. Then every vector in H has a unique expansion as a linear combination of \({{\bf{v}}_{\bf{1}}}\), \({{\bf{v}}_{\bf{2}}}\), \({{\bf{v}}_{\bf{3}}}\), because

\(\left[ {\begin{array}{*{20}{c}}s\\t\\t\end{array}} \right] = s\left[ {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{0}}\\{\bf{1}}\end{array}} \right] + \left( {t - s} \right)\left[ {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{1}}\\{\bf{1}}\end{array}} \right] + s\left[ {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{1}}\\{\bf{1}}\end{array}} \right]\)for any \(s\), and \(t\). Is \(\left\{ {{v_1},{v_2},{v_3}} \right\}\) a basis for H? Why or why not?

Short Answer

Expert verified

The set\(\left\{ {{v_1},{v_2},{v_3}} \right\}\)is not a basis for H.

Step by step solution

01

State the invertible matrix theorem

Recall the invertible matrix theorem, which states that if the square matrix is invertible, then the columns are linearly independent, and they form a basis for \({\mathbb{R}^n}\).

02

State the set of vectors in Span

The vectors\(\left\{ {{v_1},{v_2},{v_3}} \right\}\)must span H in order to provide a basis for H. So,\(H = {\rm{Span}}\left\{ {{v_1},{v_2},{v_3}} \right\}\).

The vectors in set H have identical second and third entries, and H is a subset of\(\left\{ {{v_1},{v_2},{v_3}} \right\}\).But there are some vectors that are not in H.

So,\(H\)is not in\({\rm{Span}}\left\{ {{v_1},{v_2},{v_3}} \right\}\).

Therefore,\(\left\{ {{v_1},{v_2},{v_3}} \right\}\)is not a basis for H.

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Most popular questions from this chapter

In statistical theory, a common requirement is that a matrix be of full rank. That is, the rank should be as large as possible. Explain why an m n matrix with more rows than columns has full rank if and only if its columns are linearly independent.

Let be a basis of\({\mathbb{R}^n}\). .Produce a description of an \(B = \left\{ {{{\bf{b}}_{\bf{1}}},....,{{\bf{b}}_n}\,} \right\}\)matrix A that implements the coordinate mapping \({\bf{x}} \mapsto {\left( {\bf{x}} \right)_B}\). Find it. (Hint: Multiplication by A should transform a vector x into its coordinate vector \({\left( {\bf{x}} \right)_B}\)). (See Exercise 21.)

Consider the polynomials , and \({p_{\bf{3}}}\left( t \right) = {\bf{2}}\) \({p_{\bf{1}}}\left( t \right) = {\bf{1}} + t,{p_{\bf{2}}}\left( t \right) = {\bf{1}} - t\)(for all t). By inspection, write a linear dependence relation among \({p_{\bf{1}}},{p_{\bf{2}}},\) and \({p_{\bf{3}}}\). Then find a basis for Span\(\left\{ {{p_{\bf{1}}},{p_{\bf{2}}},{p_{\bf{3}}}} \right\}\).

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

13. Show that if \(P\) is an invertible \(m \times m\) matrix, then rank\(PA\)=rank\(A\).(Hint: Apply Exercise12 to \(PA\) and \({P^{ - 1}}\left( {PA} \right)\).)

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

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