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Chapter 2: Problem 11

Show that \(\mathbb{R}^{3}=\operatorname{span}\left(\left[\begin{array}{l}1 \\\ 0 \\ 1\end{array}\right],\left[\begin{array}{l}1 \\ 1 \\\ 0\end{array}\right],\left[\begin{array}{l}0 \\ 1 \\\ 1\end{array}\right]\right)\)

Short Answer

Expert verified
The vectors span \( \mathbb{R}^3 \) since any vector \( \begin{bmatrix} x \\ y \\ z \end{bmatrix} \) can be expressed as a linear combination of them.

Step by step solution

01

Understand the Problem

We need to show that the vectors \( \mathbf{v}_1 = \begin{bmatrix} 1 \ 0 \ 1 \end{bmatrix}, \mathbf{v}_2 = \begin{bmatrix} 1 \ 1 \ 0 \end{bmatrix}, \mathbf{v}_3 = \begin{bmatrix} 0 \ 1 \ 1 \end{bmatrix} \) span \( \mathbb{R}^3 \). This means any vector in \( \mathbb{R}^3 \) can be expressed as a linear combination of these vectors.
02

Set Up the Linear Combination

Consider an arbitrary vector \( \mathbf{v} = \begin{bmatrix} x \ y \ z \end{bmatrix} \in \mathbb{R}^3 \). We need to find scalars \( a, b, c \) such that \( a \begin{bmatrix} 1 \ 0 \ 1 \end{bmatrix} + b \begin{bmatrix} 1 \ 1 \ 0 \end{bmatrix} + c \begin{bmatrix} 0 \ 1 \ 1 \end{bmatrix} = \begin{bmatrix} x \ y \ z \end{bmatrix} \).
03

Form the System of Equations

This demand results in the matrix equation: \[\begin{bmatrix} 1 & 1 & 0 \ 0 & 1 & 1 \ 1 & 0 & 1 \end{bmatrix} \begin{bmatrix} a \ b \ c \end{bmatrix} = \begin{bmatrix} x \ y \ z \end{bmatrix} \]This leads to the following system of equations:1. \( a + b = x \)2. \( b + c = y \)3. \( a + c = z \)
04

Solve the System of Equations

Solve these three equations to express \( a, b, \) and \( c \) in terms of \( x, y, \) and \( z \).- From equation (1), \( a = x - b \).- Substitute \( a = x - b \) into equation (3): \( x - b + c = z \) which simplifies to \( c = z - x + b \).- Substitute \( c = z - x + b \) into equation (2): \( b + z - x + b = y \), leading to \( 2b = y + x - z \).- Thus, \( b = \frac{y + x - z}{2} \).- Substitute \( b = \frac{y + x - z}{2} \) back into \( a = x - b \) and \( c = z - x + b \) to find: - \( a = x - \frac{y + x - z}{2} = \frac{x + z - y}{2} \) - \( c = z - x + \frac{y + x - z}{2} = \frac{y + z - x}{2} \)
05

Confirm the Solution

Check if \( \begin{bmatrix} a \ b \ c \end{bmatrix} \) as found holds true for all \( x, y, z \). We obtained expressions for \( a, b, \) and \( c \) that satisfy the original equations, thus any vector \( \begin{bmatrix} x \ y \ z \end{bmatrix} \) can indeed be expressed as a combination of the given vectors.

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Key Concepts

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

Span
Span is a fundamental concept in understanding vector spaces. It refers to the set of all linear combinations possible from a given set of vectors. For example, when we say that vectors span \(\mathbb{R}^3\), it means that any vector in this three-dimensional space can be formed by summing multiples (or scalars) of these vectors.
  • Imagine you have vectors \(\mathbf{v}_1\), \(\mathbf{v}_2\), and \(\mathbf{v}_3\). If the span of these vectors equals \(\mathbb{R}^3\), these vectors are capable of covering or 'filling' the entire space when combined in various ways.
  • In our example, we demonstrate that any vector in \(\mathbb{R}^3\) can be represented as a combination of our specific vectors \(\begin{bmatrix}1 & 0 & 1 \end{bmatrix}, \begin{bmatrix}1 & 1 & 0 \end{bmatrix}, \begin{bmatrix}0 & 1 & 1 \end{bmatrix}\).
By confirming the span, we ensure these vectors are sufficiently 'strong' or versatile to describe the entire space, showing that they are independent and provide a complete description of that vector space.
Linear Combination
The concept of a linear combination is crucial for grasping how vectors interact within a vector space. A linear combination involves adding multiple vectors together, each multiplied by a scalar, to produce a new vector.
  • Given three vectors \(\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\), any vector \(\mathbf{v}\) in \(\mathbb{R}^3\) can be expressed as \(a\mathbf{v}_1 + b\mathbf{v}_2 + c\mathbf{v}_3\), where \(a, b,\text{ and } c\) are scalars.
  • In our exercise, we set an arbitrary vector \(\begin{bmatrix} x & y & z \end{bmatrix}\) and write it as a linear combination of \(\mathbf{v}_1, \mathbf{v}_2,\text{ and } \mathbf{v}_3\).
This approach demonstrates the ability to construct any vector in the space by adjusting the scalar multipliers, thus proving the sufficiency and comprehensiveness of the chosen spanning vectors.
System of Equations
To determine the coefficients for expressing a vector as a linear combination, we resort to solving a system of equations. This method provides the specific scalar multipliers needed.
  • We start by setting a linear equation for each dimension of the vector. For our example, we have three equations representing each component of the vector: \(a + b = x\), \(b + c = y\), and \(a + c = z\).
  • These equations result from comparing each component (\(x, y, z\)) of the target vector with the respective components of the linear combination of \(\mathbf{v}_1, \mathbf{v}_2, \text{ and } \mathbf{v}_3\).
Solving this system allows us to find values for \(a, b, \text{ and } c\), confirming that our chosen vectors can indeed be combined to represent any vector in \(\mathbb{R}^3\). This approach verifies the effectiveness of our spanning vectors and their linear combinations.

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Most popular questions from this chapter

describe the span of the given vectors (a) geometrically and (b) algebraically $$\left[\begin{array}{r} 1 \\ 0 \\ -1 \end{array}\right],\left[\begin{array}{r} -1 \\ 1 \\ 0 \end{array}\right],\left[\begin{array}{r} 0 \\ -1 \\ 1 \end{array}\right]$$

describe the span of the given vectors (a) geometrically and (b) algebraically \(\left[\begin{array}{l}0 \\ 0\end{array}\right],\left[\begin{array}{l}3 \\\ 4\end{array}\right]\)

(a) If the columns of an \(n \times n\) matrix \(A\) are linearly independent as vectors in \(\mathbb{R}^{n}\), what is the rank of \(A\) ? Explain. (b) If the rows of an \(n \times n\) matrix \(A\) are linearly independent as vectors in \(\mathbb{R}^{n}\), what is the rank of \(A\) ? Explain.

(a) Prove that if \(\mathbf{u}_{1}, \ldots, \mathbf{u}_{m}\) are vectors in \(\mathbb{R}^{n}, S=\) \(\left\\{\mathbf{u}_{1}, \mathbf{u}_{2}, \ldots, \mathbf{u}_{k}\right\\},\) and \(T=\left\\{\mathbf{u}_{1}, \ldots, \mathbf{u}_{k}, \mathbf{u}_{k+1}, \ldots,\right.\) \(\left.\mathbf{u}_{m}\right\\},\) then \(\operatorname{span}(S) \subseteq \operatorname{span}(T) .\) (Hint: Rephrase this question in terms of linear combinations. (b) Deduce that if \(\mathbb{R}^{n}=\operatorname{span}(S),\) then \(\mathbb{R}^{n}=\operatorname{span}(T)\) also.

Determine if the vector b is in the span of the columns of the matrix \(A\) $$A=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right], \mathbf{b}=\left[\begin{array}{l} 5 \\ 6 \end{array}\right]$$
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