Question

We often write vectors in \(\mathbb{R}^{n}\) as rows. In each case determine if \(\mathbf{x}\) lies in \(U=\) \(\operatorname{span}\\{\mathbf{y}, \mathbf{z}\\} .\) If \(\mathbf{x}\) is in \(U,\) write it as a linear combination of \(\mathbf{y}\) and \(\mathbf{z}\); if \(\mathbf{x}\) is not in \(U\), show why not. a. \(\mathbf{x}=(2,-1,0,1), \mathbf{y}=(1,0,0,1),\) and \(\mathbf{z}=(0,1,0,1)\) b. \(\mathbf{x}=(1,2,15,11), \mathbf{y}=(2,-1,0,2),\) and \(\quad \mathbf{z}=(1,-1,-3,1)\) c. \(\mathbf{x}=(8,3,-13,20), \mathbf{y}=(2,1,-3,5),\) and \(\mathbf{z}=(-1,0,2,-3)\) d. \(\mathbf{x}=(2,5,8,3), \mathbf{y}=(2,-1,0,5),\) and \(\mathbf{z}=(-1,2,2,-3)\)

Short answer

a. Yes; \( 2\mathbf{y} - \mathbf{z} \), b. No, c. No, d. Yes; \( 3\mathbf{y} + 4\mathbf{z} \).

Step-by-step solution

Define Linear Combination

A vector \( \mathbf{x} \) lies in the span of \( \{ \mathbf{y}, \mathbf{z} \} \) if there exist scalars \( a \) and \( b \) such that \( a\mathbf{y} + b\mathbf{z} = \mathbf{x} \). We need to solve this equation for each problem.

Step A: Set Up Equations

For (a), we want \( a(1,0,0,1) + b(0,1,0,1) = (2,-1,0,1) \). This gives us the system of equations: \[ a = 2 \]\[ b = -1 \]\[ 0 = 0 \]\[ a + b = 1 \]

Step A: Solve Equations

Solving the equations: 1. From \( a = 2 \), solve directly, thus, \( a = 2 \).2. From \( b = -1 \), solve directly, thus, \( b = -1 \).3. Check consistency with \( a + b = 1 \). Indeed, \( 2 - 1 = 1 \). So \( \mathbf{x} \) is in the span, expressed as \( 2\mathbf{y} - 1\mathbf{z} \).

Step B: Set Up Equations

For (b), set up \( a(2,-1,0,2) + b(1,-1,-3,1) = (1,2,15,11) \). This gives: \[ 2a + b = 1 \] \[ -a - b = 2 \] \[ 0a - 3b = 15 \] \[ 2a + b = 11 \]

Step B: Solve Equations

Notice \( 0a - 3b = 15 \) implies \( b = -5 \). Solving \( -a - (-5) = 2 \) gives \( a = -3 \). However, substitute into \( 2a + b = 11 \), gives \( 2(-3) - 5 = -11 eq 11 \). The system is inconsistent, so \( \mathbf{x} \) is not in the span.

Step C: Set Up Equations

(c) Setup \( a(2,1,-3,5) + b(-1,0,2,-3) = (8,3,-13,20) \) resulting in:\[ 2a - b = 8 \]\[ a = 3 \]\[ -3a + 2b = -13 \]\[ 5a - 3b = 20 \]

Step C: Solve Equations

Solving the system:1. \( a = 3 \) directly.2. Substitute \( a = 3 \) into \( 2a - b = 8 \); gives \( b = -2 \).3. Confirm with \( -3a + 2b = -13 \): \( -3(3) + 2(-2) = -13 \).4. Also, confirm \( 5a - 3b = 20 \): \( 5(3) - 3(-2) = 15 + 6 = 21 eq 20 \).Since the last check fails, \( \mathbf{x} \) is not in the span.

Step D: Set Up Equations

For (d), setup equation: \( a(2,-1,0,5) + b(-1,2,2,-3) = (2,5,8,3) \), resulting in:\[ 2a - b = 2 \]\[ -a + 2b = 5 \]\[ 0 + 2b = 8 \]\[ 5a - 3b = 3 \]

Step D: Solve Equations

1. From \( 2b = 8 \), \( b = 4 \).2. Substitute \( b = 4 \) in \( 2a - 4 = 2 \); \( 2a = 6 \) so \( a = 3 \).3. Confirm \( -a + 2b = 5 \): \( -3 + 8 = 5 \).4. Check \( 5a - 3b = 3 \): \( 15 - 12 = 3 \).All checks valid, \( \mathbf{x} \) is in the span as \( 3\mathbf{y} + 4\mathbf{z} \).

Key concepts

Vectors

In linear algebra, vectors are fundamental building blocks. A vector is a quantity that has both magnitude and direction. It can be represented in various dimensional spaces, like \(\mathbb{R}^2\) or \(\mathbb{R}^3\), but in this exercise, we deal with vectors in \(\mathbb{R}^4\). Vectors are often written as ordered lists of numbers, for instance, \(\mathbf{x} = (x_1, x_2, x_3, x_4)\).

• Vectors can be added together, meaning their corresponding components are added.
• They can also be multiplied by scalars, meaning each component of the vector is multiplied by the scalar.

Understanding vectors is crucial because they are used to represent physical quantities like force and velocity. In terms of the exercise, vectors such as \(\mathbf{x}\), \(\mathbf{y}\), and \(\mathbf{z}\) serve as specific examples from which we explore other linear algebra concepts like span and linear combinations. Knowing how to work with vectors helps make sense of more complex operations in linear algebra.

Linear Combination

A key concept related to vectors is the idea of a linear combination. A linear combination involves taking some vectors and combining them using specific weights or scalars, which are usually represented by \(a\), \(b\), etc. For instance, for vectors \(\mathbf{y}\) and \(\mathbf{z}\), a linear combination can be expressed as \(a\mathbf{y} + b\mathbf{z}\).

• The scalars \(a\) and \(b\) determine the contribution of each vector to the combination.
• Through solving equations, it's determined whether certain vectors like \(\mathbf{x}\) can be expressed as a linear combination of others.

In the given exercises, finding such combinations involves setting up equations based on vector components and solving them to see if there are solutions for the scalars. Linear combinations are foundational for other operations involving vectors, especially in determining the span.

Span of Vectors

The span of vectors refers to the collection of all possible vectors that can be formed through linear combinations of a set of vectors. This defines a subspace of the vector space. For example, the span of \(\{\mathbf{y}, \mathbf{z}\}\) is all vectors that can be written in the form \(a\mathbf{y} + b\mathbf{z}\).

• The set of vectors forming the span can vary from a line to a plane or higher dimensional space, depending on the number of vectors and their arrangement.
• In the exercise, checking whether \(\mathbf{x}\) lies in the span involves seeing if there exist scalars that satisfy the equation for a linear combination of \(\mathbf{y}\) and \(\mathbf{z}\).

Mastering the concept of span is vital for understanding vector spaces and subspaces in linear algebra, as it determines the potential of vectors to generate certain spaces within larger vector settings.

Consistency of Equations

Consistency of equations in linear algebra refers to whether a system of equations has at least one solution. In the exercises, determining if a vector lies in the span of others involves solving a system of linear equations. These are equations formed from the components of the vectors involved.

• A consistent system has solutions that satisfy all equations, showing that \(\mathbf{x}\) can be represented as a linear combination of \(\mathbf{y}\) and \(\mathbf{z}\).
• An inconsistent system has no solutions that satisfy all equations, indicating that \(\mathbf{x}\) is not in the span.

Solving these systems involves using methods like substitution or elimination. It's crucial to ensure all equations in the system align; otherwise, inconsistencies will show \(\mathbf{x}\) doesn't fit as a linear combination. Grasping consistency is important for analyzing whether certain linear relationships exist between vectors.