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Chapter 6: Problem 49

Find a basis for \(\operatorname{span}(1,1+x, 2 x)\) in \(\mathscr{P}_{1}\).

Short Answer

Expert verified
A basis is \(\{1, 1+x\}\).

Step by step solution

01

Understand the Problem

We need to find a basis for the span of the set \(\{1, 1+x, 2x\}\) within \(\mathscr{P}_{1}\), which is the space of polynomials with degree at most 1.
02

Identify the Elements in \(\mathscr{P}_{1}\)

Within \(\mathscr{P}_{1}\), any polynomial can be expressed in the form \(a_0 + a_1x\), which shows that the basis for \(\mathscr{P}_{1}\) is typically \(\{1, x\}\).
03

Write the Elements as Linear Combinations

Express the elements \(1\), \(1+x\), and \(2x\) as linear combinations of \(\{1, x\}\). This results in \(1 = 1\cdot 1 + 0\cdot x\), \(1+x = 1\cdot 1 + 1\cdot x\), and \(2x = 0\cdot 1 + 2\cdot x\).
04

Form the Matrix from Coefficients

Create a matrix using the coefficients from the linear combinations:\[\begin{bmatrix}1 & 1 & 0 \0 & 1 & 2\end{bmatrix}\]
05

Perform Row Reduction

Apply the row reduction process on the matrix to find the linearly independent columns:\[\begin{bmatrix}1 & 1 & 0 \0 & 1 & 2\end{bmatrix} \rightarrow \begin{bmatrix}1 & 0 & -2 \0 & 1 & 2\end{bmatrix}\]
06

Identify the Basis

The matrix in row-reduced form shows two pivot columns: the first and second. The vectors corresponding to these columns, \(1\) and \(1+x\), are linearly independent. Thus, \(\{1, 1+x\}\) is a basis for \(\operatorname{span}(1,1+x, 2x)\) in \(\mathscr{P}_{1}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Basis of a Vector Space
A basis of a vector space is a set of vectors that are both linearly independent and span the vector space. Think of it as the foundation of a house: every vector in the space can be built from this foundation using scalar multipliers.

Let's break it down:
  • **Spanning** means that any vector in the space can be formed as a linear combination of the basis vectors.
  • **Linear independence** ensures that no vector in the basis can be written as a combination of the others.
Together, these properties mean that a basis is the smallest set of vectors that fully describe the space. Finding a basis is crucial in linear algebra because it provides a way to uniquely describe every vector in a space.
Polynomial Vector Space
The polynomial vector space, particularly denoted as \(\mathscr{P}_{n}\), comprises all polynomials of degree at most \(n\). For example, \(\mathscr{P}_{1}\) includes every polynomial that looks like \(a_0 + a_1x\), with \(a_0\) and \(a_1\) being real numbers.

Here's an easy way to think about it:
  • Imagine a space where each point is a polynomial from 0 to degree 1.
  • The basis for \(\mathscr{P}_{1}\) is typically \(\{1, x\}\), because any polynomial like \(2 + 3x\) or \(-5 + 7x\) can be formed as a combination of these.
Knowing the basis helps you work with the space efficiently, allowing you to navigate through it and express any polynomial accurately and compactly.
Row Reduction
Row reduction is a process used in linear algebra to simplify matrices. It involves using elementary row operations to transform the matrix to row-echelon form, making it easier to identify pivot positions, which signal linearly independent vectors.

Here's how row reduction works:
  • Start by making the leading coefficient of the first row equal to 1 by scaling.
  • Eliminate all coefficients below this leading 1 to create a column with zero entries except the pivot.
  • Move to the next row and repeat until you have a staircase of leading ones.
Through this systematic transformation, you can easily determine which vectors in your matrix set are linearly independent, helping identify the basis for a given span within a vector space.
Linear Combination
A linear combination involves combining several vectors by multiplying each vector by a scalar (a constant) and then adding the results. It's a way to construct new vectors from existing ones.

In the exercise, finding a linear combination of \(\{1, 1+x, 2x\}\) determines which vectors can be expressed using others.
  • For example, to find if \(2x\) can be expressed as some combination of \(1\) and \(1+x\), you multiply each by some constant and sum them up.
  • If this combination equals \(2x\), then \(2x\) isn't necessary for our basis.
Thus, linear combinations are foundational in analyzing the structure of vector spaces and their bases. In general, if the linear combination equals a target vector, that target vector is not adding anything new to the space's span.

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Most popular questions from this chapter

Consider the subspace \(W\) of \(\mathscr{P}\), given by \(W=\operatorname{span}(\cos x, \sin x, x \cos x, x \sin x)\) (a) Find the matrix of \(D\) with respect to \(\mathcal{B}=\\{\cos x\) \(\sin x, x \cos x, x \sin x\\}\) (b) Compute the derivative of \(f(x)=\cos x+2 x \cos x\) indirectly, using Theorem \(6.26,\) and verify that it agrees with \(f^{\prime}(x)\) as computed directly.

Let \(T: V \rightarrow W\) be a linear transformation between two finite- dimensional vector spaces. (a) Prove that if \(\operatorname{dim} V<\operatorname{dim} W\), then \(T\) cannot be onto. (b) Prove that if \(\operatorname{dim} V>\operatorname{dim} W\), then \(T\) cannot be one-to-one.

(a) If a (9,4) linear code has generator matrix \(G\) and parity check matrix \(P\), what are the dimensions of \(G\) and \(P ?\) (b) Repeat part (a) for an ( \(n, k\) ) linear code

Let \(\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{n}\right\\}\) be a basis for a vector space \(V\) and let \(c_{1}, \ldots, c_{n}\) be nonzero scalars. Prove that \(\left\\{c_{1} \mathbf{v}_{1}, \ldots, c_{n} \mathbf{v}_{n}\right\\}\) is also a basis for \(V\).

\(\operatorname{let} A=\left[\begin{array}{rr}1 & 1 \\ -1 & 1\end{array}\right]\) and \(B=\left[\begin{array}{rr}1 & -1 \\ 1 & 0\end{array}\right] .\) Determine whether \(C\) is in span \((A, B)\) $$C=\left[\begin{array}{ll} 3 & -5 \\ 5 & -1 \end{array}\right]$$
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