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The first four Laguerre polynomials are \(1,1 - t,2 - 4t + {t^2}\), and \(6 - 18t + 9{t^2} - {t^5}\). Show that these polynomials form a basis of \({{\mathop{\rm P}\nolimits} _3}\).

Short Answer

Expert verified

It is proved that the first four Laguerre polynomials form a basis of \({{\mathop{\rm P}\nolimits} _3}\).

Step by step solution

01

Basis theorem

Theorem 12states that let\(V\)be a p-dimensional vector space;\(p \ge 1\), then anylinearly independent setof exactly \(p\) elements in \(V\) is automatically a basis for \(V\). Any set of exactly \(p\) elements that span \(V\) is automatically a basis for \(V\).

02

Show that the first four Laguerre polynomials form a basis of \({{\mathop{\rm P}\nolimits} _3}\)

The columns of the matrix are the coordinate vectors of the Laguerre polynomials corresponding to the standard basis\(\left\{ {1,t,{t^2},{t^3}} \right\}\)of\({{\mathop{\rm P}\nolimits} _3}\). Thus,

\(A = \left[ {\begin{array}{*{20}{c}}1&0&2&6\\0&{ - 1}&{ - 4}&{ - 18}\\0&0&1&9\\0&0&0&{ - 1}\end{array}} \right]\)

There are four pivot columns in the matrix; so its columns are linearly independent. The Laguerre polynomials themselves are linearly independent in\({{\mathop{\rm P}\nolimits} _3}\)because the coordinate vectors of Laguerre polynomials form a linearly independent set. The basis theorem asserts that the Laguerre polynomials form a basis for\({{\mathop{\rm P}\nolimits} _3}\)because there are four Hermite polynomials and\(\dim {{\mathop{\rm P}\nolimits} _3} = 4\).

Thus, it is proved that the first four Laguerre polynomials form a basis of \({{\mathop{\rm P}\nolimits} _3}\).

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Most popular questions from this chapter

In Exercises 21 and 22, mark each statement True or False. Justify

each answer.

22. a.A linearly independent set in a subspace H is a basis for H.

b. If a finite set S of nonzero vectors spans a vector space V, then some subset of S is a basis for V.

c. A basis is a linearly independent set that is as large as possible.

d. The standard method for producing a spanning set for Nul A, described in Section 4.2, sometimes fails to produce a basis for Nul A.

e. If B is an echelon form of a matrix A, then the pivot columns of B form a basis for Col A.

Question 3: On any given day, a student is either healthy or ill. Of the students who are healthy today, 95% will be healthy tomorrow. Of the students who are ill today, 55% will still be ill tomorrow.

a. What is the stochastic matrix for this situation?

b. Suppose 20% of the students are ill on Monday. What fraction or percentage of the students are likely to be ill on Tuesday? On Wednesday?

c. If a student is well today, what is the probability that he or she will be well two days from now?

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

Question:In Exercises 15–18, find a basis for the space spanned by the given vectors,\({{\bf{v}}_{\bf{1}}}, \ldots ,{{\bf{v}}_{\bf{5}}}\).

16. \(\left[ {\begin{array}{*{20}{c}}1\\{\bf{0}}\\{\bf{0}}\\{\bf{1}}\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}{ - {\bf{2}}}\\{\bf{1}}\\{ - {\bf{1}}}\\{\bf{1}}\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}{\bf{6}}\\{ - {\bf{1}}}\\{\bf{2}}\\{ - {\bf{1}}}\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}{\bf{5}}\\{ - {\bf{3}}}\\{\bf{3}}\\{ - {\bf{4}}}\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{3}}\\{ - {\bf{1}}}\\{\bf{1}}\end{array}} \right]\)

Suppose the solutions of a homogeneous system of five linear equations in six unknowns are all multiples of one nonzero solution. Will the system necessarily have a solution for every possible choice of constants on the right sides of the equations? Explain.

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