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Chapter 4: Problem 15

Show that the vector space \(V=\mathbf{P}(t)\) of real polynomials cannot be spanned by a finite number of polynomials.

Short Answer

Expert verified
Assuming a finite set of polynomials \(S=\{p_1(t), p_2(t), ..., p_n(t)\}\) can span V, let m be the maximum degree among polynomials in S. Create a new polynomial \(q(t) = t^{m+1}\) with degree m + 1, which cannot be a linear combination of the polynomials in S. This contradiction shows that V cannot be spanned by a finite number of polynomials.

Step by step solution

01

Define necessary concepts

To approach this problem, we first need to define what it means for a set of polynomials to span a vector space. A set of polynomials S spans a vector space V if every element in V can be expressed as a linear combination of the polynomials in S.
02

Assume a finite set of polynomials can span V

For the sake of contradiction, let's assume that there exists a finite set of polynomials \(S=\{p_1(t), p_2(t), ..., p_n(t)\}\) that can span the vector space V of real polynomials.
03

Consider the degrees of the polynomials in S

Notice that each polynomial in S has a degree associated with it. Let the maximum degree of the polynomials in S be m, i.e., \(\text{degree}(p_i(t)) \leq m\) for all \(i = 1, 2, ..., n\).
04

Construct a new polynomial with a higher degree than m

Now let's create a new polynomial \(q(t) = t^{m+1}\). We can observe that the degree of \(q(t)\) is m + 1, which is greater than the maximum degree m of the polynomials in S.
05

Realize that q(t) cannot be a linear combination of the polynomials in S

Since the degree of any linear combination of the polynomials in S cannot be higher than m, it's impossible to construct the polynomial \(q(t)\) as a linear combination of the polynomials in S (otherwise, it would contradict the maximum degree we assumed for the polynomials in S). In other words, there is no way to write \(q(t)\) as a linear combination of the polynomials \(p_1(t), p_2(t), ..., p_n(t)\), because the degree of any linear combination would be at most m, and we need a polynomial of degree m + 1 to represent \(q(t)\).
06

Conclude that V cannot be spanned by a finite set of polynomials

Since we've shown that the polynomial \(q(t)\) cannot be created as a linear combination of any finite set of polynomials with maximum degree m, this contradicts our initial assumption that there exists a finite set of polynomials that can span V. The vector space V of real polynomials cannot be spanned by a finite number of polynomials.

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Most popular questions from this chapter

Determine whether the following polynomials \(u, v, w\) in \(\mathbf{P}(t)\) are linearly dependent or independent: (a) \(u=t^{3}-4 t^{2}+3 t+3, \quad v=t^{3}+2 t^{2}+4 t-1, \quad w=2 t^{3}-t^{2}-3 t+5\); (b) \(u=t^{3}-5 t^{2}-2 t+3, v=t^{3}-4 t^{2}-3 t+4, w=2 t^{3}-17 t^{2}-7 t+9\).

For \(M \in M_{n \times n}(C)\), let \(\bar{M}\) be the matrix such that \((\bar{M})_{i j}=\overline{M_{i j}}\) for all \(i, j\), where $\overline{M_{i j}}\( is the complex conjugate of \)M_{i j}$. (a) Prove that \(\operatorname{det}(\bar{M})=\overline{\operatorname{det}(M)}\). (b) A matrix \(Q \in \mathrm{M}_{n \times n}(C)\) is called unitary if $Q Q^{*}=I\(, where \)Q^{*}=\overline{Q^{t}}\(. Prove that if \)Q$ is a unitary matrix, then \(|\operatorname{det}(Q)|=1\).

Find the dimension and a basis of the subspace \(W\) of \(\mathbf{P}_{3}(t)\) spanned by \\[u=t^{3}+2 t^{2}-3 t+4, \quad v=2 t^{3}+5 t^{2}-4 t+7, \quad w=t^{3}+4 t^{2}+t+2\\]

Determine which of the following subspaces of \(\mathbf{R}^{4}\) are identical: \\[\begin{array}{c} U_{1}=\operatorname{span}[(1,2,1,4),(2,4,1,5),(3,6,2,9)], \quad U_{2}=\operatorname{span}[(1,2,1,2),(2,4,1,3)] \\ U_{3}=\operatorname{span}[(1,2,3,10),(2,4,3,11)] \end{array}\\]

Let \(S\) and \(T\) be arbitrary nonempty subsets (not necessarily subspaces) of a vector space \(V\) and let \(k\) be a scalar. The sum \(S+T\) and the scalar product \(k S\) are defined by \\[S+T=(u+v: u \in S, v \in T\\}, \quad k S=\\{k u: u \in S\\}\\] \([\mathrm{We} \text { also write } w+S \text { for }\\{w\\}+S .]\) Let \\[S=\\{(1,2),(2,3)\\}, \quad T=\\{(1,4),(1,5),(2,5)\\}, \quad w=(1,1), \quad k=3\\] Find: \((\mathrm{a}) S+T,(\mathrm{b}) w+S,(\mathrm{c}) k S,(\mathrm{d}) k T,(\mathrm{c}) k S+k T,(\mathrm{f}) k(S+T)\).
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