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Chapter 4: Problem 11

In Exercises \(7-14\) , cither use an appropriate theorem to show that the given set, \(W,\) is a vector space, or find a specific example to the contrary. $$ \left\\{\left[\begin{array}{c}{b-2 d} \\ {5+d} \\ {b+3 d} \\\ {d}\end{array}\right] : b, d \text { real }\right\\} $$

Short Answer

Expert verified
The set \( W \) is not a vector space because it does not contain the zero vector.

Step by step solution

01

Understand the components of the vector set

The given set \( W \) comprises vectors of the form \[ \begin{bmatrix} b - 2d \ 5 + d \ b + 3d \ d \end{bmatrix} \] where \( b \) and \( d \) are real numbers.
02

Define the zero vector for the vector space

For a set of vectors to be a vector space, it must contain the zero vector. The zero vector in this context is the vector where all components are zero: \( \begin{bmatrix} 0 \ 0 \ 0 \ 0 \end{bmatrix} \).
03

Check if the zero vector is in the set

Set each component equal to zero: \( b - 2d = 0 \), \( 5 + d = 0 \), \( b + 3d = 0 \), \( d = 0 \). From \( d = 0 \), we substitute into \( 5 + d = 0 \), resulting in a contradiction since 5 cannot equal zero. Thus, the zero vector is not in the set.
04

Conclusion based on the zero vector check

Since the zero vector is not in the set, \( W \) cannot be a vector space. The existence of a zero vector is crucial for a set to be considered a vector space.

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Key Concepts

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

Zero Vector
In the world of vector spaces, the zero vector plays a pivotal role. It is the vector that serves as the identity element for vector addition. In simpler terms, when you add the zero vector to any other vector, you end up with the same vector you started with.

For example, if you have a vector \[\begin{bmatrix} x_1 \ x_2 \end{bmatrix}\] and you add it to the zero vector \[\begin{bmatrix} 0 \ 0 \end{bmatrix}\], you will still have \[\begin{bmatrix} x_1 \ x_2 \end{bmatrix}\].
  • The zero vector is essential for verifying whether a set is a vector space.
  • A set of vectors cannot qualify as a vector space if it lacks the zero vector.
This concept was tested in the exercise, and it was determined that the given set did not contain the zero vector. As a result, we concluded that the set isn鈥檛 a vector space.
Real Numbers
Real numbers are an integral part of defining components in vectors. They consist of all the numbers we typically think of, including whole numbers, fractions, decimals, and irrational numbers like \( \pi \) and \( \sqrt{2} \). These numbers are used as coefficients and scalar elements in vector spaces.

Each component of a vector is generally a real number, fulfilling two important roles in vector spaces:
  • Providing specific, scalable values for the coordinates of vectors.
  • Acting as scalars in scalar multiplication, allowing us to stretch or shrink vectors.
In the solution for the given exercise, we considered variables \( b \) and \( d \) as real numbers, using them to determine if the set meets the criteria for a vector space.
Vector Set
A vector set is the collection of vectors that we can analyze to determine if it forms a vector space. These vectors can originate from various combinations of coefficients and variables. When establishing a vector space, we look for certain characteristics within the vector set:

Here is what a vector set should show:
  • Closure under addition: Adding any two vectors from the set should yield another vector within the set.
  • Closure under scalar multiplication: Multiplying any vector in the set by a real number should still result in a vector within the set.
  • Inclusion of the zero vector: As previously discussed, the set must include the zero vector.
The vector set in the exercise failed to meet one of these crucial criteria, indicating that it does not form a vector space.
Properties of Vector Spaces
Vector spaces come with several properties, making them indispensable structures in mathematics and physics. To qualify as a vector space, a set of vectors needs to satisfy several well-defined properties:

  • Existence of the zero vector, which acts as an additive identity.
  • Associative and commutative properties of vector addition.
  • Existence of additive inverses: For each vector, there should be another vector that can be added to make the zero vector.
  • Distributive properties involving vectors and scalars over addition.
  • Compatibility of scalar multiplication with field multiplication.
  • Containment: All linear combinations of the vectors in the set must be present within the set.
In the given exercise, not all these properties were fulfilled, notably the absence of the zero vector, leading to the conclusion that the set \( W \) cannot be classified as a vector space.

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Most popular questions from this chapter

Exercises 31 and 32 reveal an important connection between linear independence and linear transformations and provide practice using the definition of linear dependence. Let \(V\) and \(W\) be vector spaces, let \(T : V \rightarrow W\) be a linear transformation, and let \(\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{p}\right\\}\) be a subset of \(V .\) Suppose that \(T\) is a one-to-one transformation, so that an equation \(T(\mathbf{u})=T(\mathbf{v})\) always implies \(\mathbf{u}=\mathbf{v} .\) Show that if the set of images \(\left\\{T\left(\mathbf{v}_{1}\right), \ldots, T\left(\mathbf{v}_{p}\right)\right\\}\) is linearly dependent, then \(\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{p}\right\\}\) is linearly dependent. This fact shows that a one-to-one linear transformation maps a linearly independent set onto a linearly independent set (because in this case the set of images cannot be linearly dependent).

Consider the polynomials \(\mathbf{p}_{1}(t)=1+t^{2}\) and \(\mathbf{p}_{2}(t)=1-\) \(t^{2} .\) Is \(\left\\{\mathbf{p}_{1}, \mathbf{p}_{2}\right\\}\) a linearly independent set in \(\mathbb{P}_{3} ?\) Why or why not?

In Exercises \(15-18,\) find a basis for the space spanned by the given vectors, \(\mathbf{v}_{1}, \ldots, \mathbf{v}_{5}\) $$ \left[\begin{array}{l}{1} \\ {0} \\ {0} \\\ {1}\end{array}\right],\left[\begin{array}{r}{-2} \\ {1} \\ {-1} \\\ {1}\end{array}\right],\left[\begin{array}{r}{6} \\ {-1} \\ {2} \\\ {-1}\end{array}\right],\left[\begin{array}{r}{5} \\ {-3} \\ {3} \\\ {-4}\end{array}\right],\left[\begin{array}{r}{0} \\ {3} \\ {-1} \\\ {1}\end{array}\right] $$

\(\mathcal{B}\) and \(\mathcal{C}\) are bases for a vector space \(V\) Mark each statement True or False. Justify each answer. a. The columns of the change-of-coordinates matrix \(c \leftarrow \mathcal{B}\) are \(\mathcal{B}\) -coordinate vectors of the vectors in \(\mathcal{C}\) . b. If \(V=\mathbb{R}^{n}\) and \(\mathcal{C}\) is the standard basis for \(V,\) then \(c \leftarrow \mathcal{B}\) is the same as the change-of-coordinates matrix \(P_{\mathcal{B}}\) introduced in Section \(4.4 .\)

In Exercises \(25-28\) , show that the given signal is a solution of the difference equation. Then find the general solution of that difference equation. $$ y_{k}=2 k-4 ; y_{k+2}+\frac{3}{2} y_{k+1}-y_{k}=1+3 k $$
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