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Chapter 1: Problem 4

Do the polynomials \(x^{3}-2 x^{2}+1,4 x^{2}-x+3\), and \(3 x-2\) generate \(\mathrm{P}_{3}(R)\) ? Justify your answer.

Short Answer

Expert verified
The given polynomials \(x^3-2x^2+1, 4x^2-x+3\), and \(3x-2\) do not generate \(\mathrm{P}_3(\mathbb{R})\) since they are not linearly independent, as their corresponding matrix has a rank of 3, which is less than the dimension of the vector space (4).

Step by step solution

01

Identify the given polynomials vector space

The vector space we are working with is \(\mathrm{P}_3(\mathbb{R})\), which contains all polynomial functions of degree less than or equal to 3 with real coefficients. In this vector space, polynomials can be added, subtracted, and multiplied by scalar values in accordance with the usual rules of polynomial arithmetic.
02

Write the given polynomials as vectors

We can treat the polynomials as vectors by representing their coefficients in the form of a 4-dimensional column vector. The three given polynomials can be written as: 1. \(p_1(x) = x^3-2x^2+1 \Rightarrow \begin{pmatrix} 1 \\ -2 \\ 0 \\ 1 \end{pmatrix}\) 2. \(p_2(x) = 4x^2-x+3 \Rightarrow \begin{pmatrix} 0 \\ 0 \\ 4 \\ -1 \end{pmatrix}\) 3. \(p_3(x) = 3x-2 \Rightarrow \begin{pmatrix} 0 \\ 0 \\ 0 \\ 3 \end{pmatrix}\)
03

Determine if the given vectors are linearly independent

Linear independence means that no vector in the set can be written as a linear combination of the other vectors. We can use Gaussian elimination to check for linear independence. To do this, we will arrange the vectors in a matrix and row reduce this matrix to its Reduced Row Echelon Form (RREF). If the RREF has a full rank, the vectors are linearly independent. Form the matrix \(A = \begin{pmatrix} 1 & 0 & 0 \\ -2 & 0 & 0 \\ 0 & 4 & 0 \\ 1 & -1 & 3 \end{pmatrix}\). The matrix A is already in its RREF, and the rank of A is 3, which is not full rank (4) for the vector space \(\mathrm{P}_3(\mathbb{R})\).
04

Conclude if the given polynomials generate \(\mathrm{P}_3(\mathbb{R})\)

The given polynomials, represented by their respective vectors, are not linearly independent since the rank of the corresponding matrix is less than the dimension of the vector space (4). Therefore, these polynomials do not form a basis and hence do not generate \(\mathrm{P}_3(\mathbb{R})\).

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Most popular questions from this chapter

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.)

The vectors \(u_{1}=(2,-3,1), u_{2}=(1,4,-2), u_{3}=(-8,12,-4), u_{4}=\) \((1,37,-17)\), and \(u_{5}=(-3,-5,8)\) generate \(\mathrm{R}^{3} .\) Find a subset of the set \(\left\\{u_{1}, u_{2}, u_{3}, u_{4}, u_{5}\right\\}\) that is a basis for \(\mathrm{R}^{3}\).

Show that if $$ M_{1}=\left(\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right), \quad M_{2}=\left(\begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right), \quad \text { and } \quad M_{3}=\left(\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right), $$ then the span of \(\left\\{M_{1}, M_{2}, M_{3}\right\\}\) is the set of all symmetric \(2 \times 2\) matrices.

Let \(W_{1}\) and \(W_{2}\) be subspaces of a vector space \(V\). Prove that \(V\) is the direct sum of \(W_{1}\) and \(W_{2}\) if and only if each vector in \(V\) can be uniquely written as \(x_{1}+x_{2}\), where \(x_{1} \in \mathrm{W}_{1}\) and $x_{2} \in \mathrm{W}_{2}$.

Let V be a vector space over a field of characteristic not equal to two. (a) Let u and v be distinct vectors in V. Prove that { u, v} is linearly independent if and only if { u + v, u- v} is linearly independent. (b) Let u, v, and w be distinct vectors in V. Prove that { u, v, w} is linearly independent if and only if { u + v, u + w, 'U + w} is linearly independent.
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