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Suppose \({\mathbb{R}^4} = Span\left\{ {{v_1},...,{v_4}} \right\}\). Explain why \(\left\{ {{v_1},...,{v_4}} \right\}\) is a basis for \({\mathbb{R}^4}\).

Short Answer

Expert verified

The set \(\left\{ {{v_1},...,{v_4}} \right\}\) is a basis for \({\mathbb{R}^4}\).

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 the columns form a basis for \({\mathbb{R}^n}\).

02

State the basis theorem

It is given that\(\left\{ {{v_1},...,{v_4}} \right\}\)is the set of vectors. It forms a square matrix of the order\(4 \times 4\)as shown below:

\(A = \left[ {\begin{array}{*{20}{c}}{{{\bf{v}}_1}}&{{{\bf{v}}_2}}&{{{\bf{v}}_3}}&{{{\bf{v}}_4}}\end{array}} \right]\)

As\({\mathbb{R}^4} = {\rm{Span}}\left\{ {{v_1},...,{v_4}} \right\}\), the columns of A span\({\mathbb{R}^4}\). Also, the columns are linearly independent and form a basis.

Thus, \(\left\{ {{v_1},...,{v_4}} \right\}\) is a basis for \({\mathbb{R}^4}\).

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Most popular questions from this chapter

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 \({P^{ - 1}}\left( {PA} \right)\).)

A homogeneous system of twelve linear equations in eight unknowns has two fixed solutions that are not multiples of each other, and all other solutions are linear combinations of these two solutions. Can the set of all solutions be described with fewer than twelve homogeneous linear equations? If so, how many? Discuss.

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 \({\left( {\bf{x}} \right)_B}\)).

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

19. \(A = \left( {\begin{array}{*{20}{c}}{.9}&1&0\\0&{ - .9}&0\\0&0&{.5}\end{array}} \right),B = \left( {\begin{array}{*{20}{c}}0\\1\\1\end{array}} \right)\).

In Exercise 7, find the coordinate vector \({\left( x \right)_{\rm B}}\) of x relative to the given basis \({\rm B} = \left\{ {{b_{\bf{1}}},...,{b_n}} \right\}\).

7. \({b_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{1}}}\\{ - {\bf{3}}}\end{array}} \right),{b_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{ - {\bf{3}}}\\{\bf{4}}\\{\bf{9}}\end{array}} \right),{b_{\bf{3}}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{ - {\bf{2}}}\\{\bf{4}}\end{array}} \right),x = \left( {\begin{array}{*{20}{c}}{\bf{8}}\\{ - {\bf{9}}}\\{\bf{6}}\end{array}} \right)\)

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