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Consider the polynomials \({{\bf{p}}_{\bf{1}}}\left( t \right) = {\bf{1}} + t\), \({{\bf{p}}_{\bf{2}}}\left( t \right) = {\bf{1}} - t\), \({{\bf{p}}_{\bf{3}}}\left( t \right) = {\bf{4}}\), \({{\bf{p}}_{\bf{4}}}\left( t \right) = {\bf{1}} + {t^{\bf{2}}}\), and \({{\bf{p}}_{\bf{5}}}\left( t \right) = {\bf{1}} + {\bf{2}}t + {t^{\bf{2}}}\), and let H be the subspace of \({P_{\bf{5}}}\) spanned by the set \(S = \left\{ {{{\bf{p}}_{\bf{1}}},\,{{\bf{p}}_{\bf{2}}},\;{{\bf{p}}_{\bf{3}}},\,{{\bf{p}}_{\bf{4}}},\,{{\bf{p}}_{\bf{5}}}} \right\}\). Use the method described in the proof of the Spanning Set Theorem (Section 4.3) to produce a basis for H. (Explain how to select appropriate members of S.)

Short Answer

Expert verified

The basis of H is \(\left\{ {{{\bf{p}}_1},\;{{\bf{p}}_2},\,{{\bf{p}}_4}} \right\}\).

Step by step solution

01

Check for a combination of \({{\bf{p}}_{\bf{1}}}\) and \({{\bf{p}}_{\bf{2}}}\)

For a scalar \({c_1}\),

\(\begin{array}{c}{p_2} = {c_1}{p_1}\\1 - t = {c_1}\left( {1 + t} \right)\end{array}\).

The above equation is not true for any value of \({c_1}\). So, \({p_1}\) and \({p_2}\) are independent.

02

Check for a linear combination of \({{\bf{p}}_{\bf{3}}}\)

For scalars \({c_1}\) and \({c_2}\),

\(\begin{aligned}{c}{{\bf{p}}_3} &= {c_1}{{\bf{p}}_1} + {c_2}{{\bf{p}}_2}\\4 &= {c_1}\left( {1 + t} \right) + {c_2}\left( {1 - t} \right)\\ &= {c_1} + {c_2} + \left( {{c_1} - {c_2}} \right)t\end{aligned}\).

For the above equation, \({c_1} = 2\) and \({c_2} = 2\).

So, \({{\bf{p}}_3}\) can be obtained from \({{\bf{p}}_1}\) and \({{\bf{p}}_2}\).

03

Check for a linear combination of \({{\bf{p}}_{\bf{4}}}\)

For the scalars \({c_1}\), \({c_2}\), and \({c_3}\),

\(\begin{aligned} {{\bf{p}}_4} &= {c_1}{{\bf{p}}_1} + {c_2}{{\bf{p}}_2} + {c_3}{{\bf{p}}_3}\\t + {t^2} &= {c_1}\left( {1 + t} \right) + {c_2}\left( {1 - t} \right) + {c_3}\left( 4 \right)\\ &= {c_1} + {c_2} + 4{c_3} + \left( {{c_1} - {c_2}} \right)t\end{aligned}\).

As \({t^2}\) is not present in RHS, \({p_4}\) is an independent work.

04

Check for a linear combination of \({{\bf{p}}_{\bf{5}}}\)

For the scalars \({c_1}\), \({c_2}\), \({c_3}\), and \({c_4}\),

\(\begin{aligned}{c}{{\bf{p}}_5} &= {c_1}{{\bf{p}}_1} + {c_2}{{\bf{p}}_2} + {c_3}{{\bf{p}}_3} + {c_4}{{\bf{p}}_4}\\1 + 2t + {t^2} &= {c_1}\left( {1 + t} \right) + {c_2}\left( {1 - t} \right) + {c_3}\left( 4 \right) + {c_4}\left( {t + {t^2}} \right)\\ &= {c_1} + {c_2} + 4{c_3} + \left( {{c_1} - {c_2} + {c_4}} \right){t^2}\end{aligned}\).

For the above equation, \({c_1} = 1\), \({c_2} = 0\), \({c_3} = 0\), and \({c_4} = 1\).

So, \({p_5}\) is a dependent work.

The basis is formed by the independent vectors, so the basis of H is \(\left\{ {{{\bf{p}}_1},\;{{\bf{p}}_2},\,{{\bf{p}}_4}} \right\}\).

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Most popular questions from this chapter

[M] Let \(A = \left[ {\begin{array}{*{20}{c}}7&{ - 9}&{ - 4}&5&3&{ - 3}&{ - 7}\\{ - 4}&6&7&{ - 2}&{ - 6}&{ - 5}&5\\5&{ - 7}&{ - 6}&5&{ - 6}&2&8\\{ - 3}&5&8&{ - 1}&{ - 7}&{ - 4}&8\\6&{ - 8}&{ - 5}&4&4&9&3\end{array}} \right]\).

  1. Construct matrices \(C\) and \(N\) whose columns are bases for \({\mathop{\rm Col}\nolimits} A\) and \({\mathop{\rm Nul}\nolimits} A\), respectively, and construct a matrix \(R\) whose rows form a basis for Row\(A\).
  2. Construct a matrix \(M\) whose columns form a basis for \({\mathop{\rm Nul}\nolimits} {A^T}\), form the matrices \(S = \left[ {\begin{array}{*{20}{c}}{{R^T}}&N\end{array}} \right]\) and \(T = \left[ {\begin{array}{*{20}{c}}C&M\end{array}} \right]\), and explain why \(S\) and \(T\) should be square. Verify that both \(S\) and \(T\) are invertible.

In Exercise 17, Ais an \(m \times n\] matrix. Mark each statement True or False. Justify each answer.

17. a. The row space of A is the same as the column space of \({A^T}\].

b. If B is any echelon form of A, and if B has three nonzero rows, then the first three rows of A form a basis for Row A.

c. The dimensions of the row space and the column space of A are the same, even if Ais not square.

d. The sum of the dimensions of the row space and the null space of A equals the number of rows in A.

e. On a computer, row operations can change the apparent rank of a matrix.

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

If the null space of an \({\bf{8}} \times {\bf{5}}\) matrix A is 2-dimensional, what is the dimension of the row space of A?

In Exercise 6, find the coordinate vector of x relative to the given basis \({\rm B} = \left\{ {{b_{\bf{1}}},...,{b_n}} \right\}\).

6. \({b_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{2}}}\end{array}} \right),{b_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{5}}\\{ - {\bf{6}}}\end{array}} \right),x = \left( {\begin{array}{*{20}{c}}{\bf{4}}\\{\bf{0}}\end{array}} \right)\)

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