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In Exercises 27-30, use coordinate vectors to test the linear independence of the sets of polynomials. Explain your work.

Short Answer

Expert verified

\({\bf{1}} + {\bf{2}}{t^{\bf{3}}}\),\({\bf{2}} + t - {\bf{3}}{t^{\bf{2}}}\),\(t + {\bf{2}}{t^2} - {t^{\bf{3}}}\)

Step by step solution

01

Write the polynomials in the standard vector form

The vectorsof the given polynomials can be written as follows:

\(1 + 2{t^3} \equiv \left( {\begin{array}{*{20}{c}}1\\0\\0\\2\end{array}} \right)\),\(\left( {2 + t - 3{t^2}} \right) \equiv \left( {\begin{array}{*{20}{c}}2\\1\\{ - 3}\\0\end{array}} \right)\),\(\left( { - t + 2{t^2} - {t^3}} \right) \equiv \left( {\begin{array}{*{20}{c}}0\\{ - 1}\\2\\{ - 1}\end{array}} \right)\)

02

Form the matrix using the vectors

The matrix formed by using the vectors of the polynomials is:

\(A = \left( {\begin{array}{*{20}{c}}1&2&0\\0&1&{ - 1}\\0&{ - 3}&2\\2&0&{ - 1}\end{array}} \right)\)

03

Write the matrix in the echelon form

\(\left( {\begin{array}{*{20}{c}}1&2&0\\0&1&{ - 1}\\0&{ - 3}&2\\2&0&{ - 1}\end{array}} \right) \sim \left( {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\\0&0&0\end{array}} \right)\)

As the matrix has a pivot in each column, its columns are linearly independent.

So, the polynomials are linearly independent.

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

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?]

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

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

16. If \(A\) is an \(m \times n\) matrix of rank\(r\), then a rank factorization of \(A\) is an equation of the form \(A = CR\), where \(C\) is an \(m \times r\) matrix of rank\(r\) and \(R\) is an \(r \times n\) matrix of rank \(r\). Such a factorization always exists (Exercise 38 in Section 4.6). Given any two \(m \times n\) matrices \(A\) and \(B\), use rank factorizations of \(A\) and \(B\) to prove that rank\(\left( {A + B} \right) \le {\mathop{\rm rank}\nolimits} A + {\mathop{\rm rank}\nolimits} B\).

(Hint: Write \(A + B\) as the product of two partitioned matrices.)

(M) Determine whether w is in the column space of \(A\), the null space of \(A\), or both, where

\({\mathop{\rm w}\nolimits} = \left( {\begin{array}{*{20}{c}}1\\1\\{ - 1}\\{ - 3}\end{array}} \right),A = \left( {\begin{array}{*{20}{c}}7&6&{ - 4}&1\\{ - 5}&{ - 1}&0&{ - 2}\\9&{ - 11}&7&{ - 3}\\{19}&{ - 9}&7&1\end{array}} \right)\)

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