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Chapter 1: Problem 5

Show that the set \(\left\\{1, x, x^{2}, \ldots, x^{n}\right\\}\) is linearly independent in \(\mathrm{P}_{n}(F)\).

Short Answer

Expert verified
To show that the set \(S = \{1, x, x^2, \ldots, x^n\}\) is linearly independent in \(\mathrm{P}_n(F)\), we set up a linear combination of elements in the set equal to zero and treat it as a polynomial: \(p(x) = c_0 + c_1x + c_2x^2 + \cdots + c_nx^n = 0\). Observe that if \(p(x)\) has infinitely many zeros, then the coefficients must all be zero, making the solution trivial. Since F is a field with infinitely many elements, and the polynomial has infinitely many zeros, we conclude that the only solution is the trivial one, \(c_0 = c_1 = \cdots = c_n = 0\), proving that the set is linearly independent in \(\mathrm{P}_n(F)\).

Step by step solution

01

Set up a linear combination of the vectors (elements) in the set

Define a linear combination of the vectors (elements) in the set \(S\) using the scalar coefficients \(c_0, c_1, \ldots, c_n \in F\): $$ c_0(1) + c_1(x) + c_2(x^2) + \cdots + c_n(x^n) = 0 $$ Our goal is to show that the only solution for this equation is the trivial solution: \(c_0 = c_1 = \cdots = c_n = 0\).
02

Treat the linear combination as a polynomial

Observe that the left-hand side of the equation in step 1 represents a polynomial in variable x: $$ p(x) = c_0 + c_1x + c_2x^2 + \cdots + c_nx^n $$ Since this polynomial must be equal to the zero polynomial for all x, we can say that $$ p(x) = 0 \quad \forall x \in F $$
03

Prove that the coefficients are all zero

Note that p(x) is an nth-degree polynomial, which means it could have at most n non-zero roots. However, from step 2, we know that \(p(x) = 0\) for all \(x \in F\). Since F is a field, it must have infinitely many elements. Therefore, we have a contradiction because p(x) has infinitely many zeros. The only way to resolve this contradiction is if all coefficients \(c_i\) are zero: $$ c_0 = c_1 = c_2 = \cdots = c_n = 0 $$
04

Conclusion

Since the only solution for the linear combination of the vectors (elements) in the set \(S\) is the trivial solution (\(c_0 = c_1 = \cdots = c_n = 0\)), we can conclude that the set \(\{1, x, x^2, \ldots, x^n\}\) is linearly independent in \(\mathrm{P}_n(F)\).

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Most popular questions from this chapter

Determine whether the following sets are linearly dependent or linearly independent. (a) $\left\\{\left(\begin{array}{rr}1 & -3 \\ -2 & 4\end{array}\right),\left(\begin{array}{rr}-2 & 6 \\ 4 & -8\end{array}\right)\right\\}\( in \)\mathrm{M}_{2 \times 2}(R)$ (b) $\left\\{\left(\begin{array}{rr}1 & -2 \\ -1 & 4\end{array}\right),\left(\begin{array}{rr}-1 & 1 \\ 2 & -4\end{array}\right)\right\\}\( in \)\mathrm{M}_{2 \times 2}(R)$ (c) \(\left\\{x^{3}+2 x^{2},-x^{2}+3 x+1, x^{3}-x^{2}+2 x-1\right\\}\) in \(\mathrm{P}_{3}(R)\) (d) \(\left\\{x^{3}-x, 2 x^{2}+4,-2 x^{3}+3 x^{2}+2 x+6\right\\}\) in \(\mathrm{P}_{3}(R)\) (e) \(\\{(1,-1,2),(1,-2,1),(1,1,4)\\}\) in \(\mathbf{R}^{3}\) (f) \(\\{(1,-1,2),(2,0,1),(-1,2,-1)\\}\) in \(\mathbf{R}^{3}\) (g) $\left\\{\left(\begin{array}{rr}1 & 0 \\ -2 & 1\end{array}\right),\left(\begin{array}{rr}0 & -1 \\ 1 & 1\end{array}\right),\left(\begin{array}{rr}-1 & 2 \\ 1 & 0\end{array}\right),\left(\begin{array}{rr}2 & 1 \\ -4 & 4\end{array}\right)\right\\}\( in \)\mathrm{M}_{2 \times 2}(R)$ (h) $\left\\{\left(\begin{array}{rr}1 & 0 \\ -2 & 1\end{array}\right),\left(\begin{array}{rr}0 & -1 \\ 1 & 1\end{array}\right),\left(\begin{array}{rr}-1 & 2 \\ 1 & 0\end{array}\right),\left(\begin{array}{rr}2 & 1 \\ 2 & -2\end{array}\right)\right\\}\( in \)\mathrm{M}_{2 \times 2}(R)$ (i) \(\left\\{x^{4}-x^{3}+5 x^{2}-8 x+6,-x^{4}+x^{3}-5 x^{2}+5 x-3\right.\), \(\left.x^{4}+3 x^{2}-3 x+5,2 x^{4}+3 x^{3}+4 x^{2}-x+1, x^{3}-x+2\right\\}\) in \(\mathrm{P}_{4}(R)\) (j) \(\left\\{x^{4}-x^{3}+5 x^{2}-8 x+6,-x^{4}+x^{3}-5 x^{2}+5 x-3,\right.\), \(\left.x^{4}+3 x^{2}-3 x+5,2 x^{4}+x^{3}+4 x^{2}+8 x\right\\}\) in \(\mathrm{P}_{4}(R)\) \({ }^{3}\) The computations in Exercise \(2(\mathrm{~g}),(\mathrm{h}),(\mathrm{i})\), and \((\mathrm{j})\) are tedious unless technology is used.

Exercises 5 and 6 show why the definitions of matrix addition and scalar multiplication (as defined in Example 2) are the appropriate ones. Richard Gard ("Effects of Beaver on Trout in Sagehen Creek, California," J. Wildlife Management, 25, 221-242) reports the following number of trout having crossed beaver dams in Sagehen Creek. Upstream Crossings $$ \begin{array}{lccc} \hline & \text { Fall } & \text { Spring } & \text { Summer } \\ \hline \text { Brook trout } & 8 & 3 & 1 \\ \text { Rainbow trout } & 3 & 0 & 0 \\ \text { Brown trout } & 3 & 0 & 0 \\ \hline \end{array} $$ Upstream Crossings $$ \begin{array}{lccc} \hline & \text { Fall } & \text { Spring } & \text { Summer } \\ \hline \text { Brook trout } & 9 & 1 & 4 \\ \text { Rainbow trout } & 3 & 0 & 0 \\ \text { Brown trout } & 1 & 1 & 0 \\ \hline \end{array} $$Record the upstream and downstream crossings in two 3 x 3 matrices, and verify that the sum of these matrices gives the total number of crossings (both upstream and downstream) categorized by trout species and season

Let \(W\) be a subspace of a (not necessarily finite-dimensional) vector space \(V\). Prove that any basis for \(W\) is a subset of a basis for \(V\).

Prove that a subset \(W\) of a vector space \(V\) is a subspace of \(V\) if and only if \(0 \in \mathrm{W}\) and \(a x+y \in \mathrm{W}\) whenever \(a \in F\) and $x, y \in W$.

Show that if \(S_{1}\) and \(S_{2}\) are arbitrary subsets of a vector space V, then $\operatorname{span}\left(S_{1} \cup S_{2}\right)=\operatorname{span}\left(S_{1}\right)+\operatorname{span}\left(S_{2}\right)$. (The sum of two subsets is defined in the exercises of Section 1.3.)
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