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Chapter 6: Problem 7

\(\mathbf{}\) If \(\\{\mathbf{u}, \mathbf{v}, \mathbf{w}\\}\) is a basis of \(V,\) determin which of the following are bases. a. \(\\{\mathbf{u}+\mathbf{v}, \mathbf{u}+\mathbf{w}, \mathbf{v}+\mathbf{w}\\}\) b. \(\\{2 \mathbf{u}+\mathbf{v}+3 \mathbf{w}, 3 \mathbf{u}+\mathbf{v}-\mathbf{w}, \mathbf{u}-4 \mathbf{w}\\}\) c. \(\\{\mathbf{u}, \mathbf{u}+\mathbf{v}+\mathbf{w}\\}\) d. \(\\{\mathbf{u}, \mathbf{u}+\mathbf{w}, \mathbf{u}-\mathbf{w}, \mathbf{v}+\mathbf{w}\\}\)

Short Answer

Expert verified
Sets A and B are bases; sets C and D are not.

Step by step solution

01

Understanding the Exercise

We need to determine which sets form a basis. A basis of a vector space is a set of linearly independent vectors that span the space.
02

Criteria for Basis

A set of vectors forms a basis if it is linearly independent and spans the vector space. Since \(\{\mathbf{u}, \mathbf{v}, \mathbf{w}\}\) is a basis, any other basis must transform between them and remain independent.
03

Option A: Check Independence

Consider \(\{\mathbf{u}+\mathbf{v}, \mathbf{u}+\mathbf{w}, \mathbf{v}+\mathbf{w}\}\). Suppose \(a(\mathbf{u}+\mathbf{v}) + b(\mathbf{u}+\mathbf{w}) + c(\mathbf{v}+\mathbf{w}) = \mathbf{0}\). Simplify to find if only the trivial solution exists. If true, the set is independent.
04

Option A: Simplify & Test

Combine terms: \((a + b)\mathbf{u} + (a + c)\mathbf{v} + (b + c)\mathbf{w} = \mathbf{0}\). For independence: \(a + b = 0, a + c = 0, b + c = 0\). Solve to find \(a = b = c = 0\). Independent.
05

Option A: Conclusion

The set \(\{\mathbf{u}+\mathbf{v}, \mathbf{u}+\mathbf{w}, \mathbf{v}+\mathbf{w}\}\) is a basis as it is independent and spans \(V\).
06

Option B: Check Independence

Consider \(\{2 \mathbf{u}+\mathbf{v}+3 \mathbf{w}, 3 \mathbf{u}+\mathbf{v}-\mathbf{w}, \mathbf{u}-4 \mathbf{w}\}\). Assume \(a(2 \mathbf{u}+\mathbf{v}+3 \mathbf{w}) + b(3 \mathbf{u}+\mathbf{v}-\mathbf{w}) + c(\mathbf{u}-4 \mathbf{w}) = \mathbf{0}\).
07

Option B: Simplify & Test

Simplify: \((2a + 3b + c)\mathbf{u} + (a + b)\mathbf{v} + (3a - b - 4c)\mathbf{w} = \mathbf{0}\). Solve the system: \(2a + 3b + c = 0, a + b = 0, 3a - b - 4c = 0\). Find \(a = b = c = 0\). Independent.
08

Option B: Conclusion

The set \(\{2 \mathbf{u}+\mathbf{v}+3 \mathbf{w}, 3 \mathbf{u}+\mathbf{v}-\mathbf{w}, \mathbf{u}-4 \mathbf{w}\}\) is a basis as it is independent and spans \(V\).
09

Option C: Check Independence

Set \(\{\mathbf{u}, \mathbf{u}+\mathbf{v}+\mathbf{w}\}\). Assume \(a\mathbf{u} + b(\mathbf{u} + \mathbf{v} + \mathbf{w}) = \mathbf{0}\). Simplify to determine if the vectors are independent.
10

Option C: Simplify & Test

Expand: \((a + b)\mathbf{u} + b\mathbf{v} + b\mathbf{w} = \mathbf{0}\). Solve: \(a + b = 0, b = 0\). Which implies \(a = b = 0\). The equations do not cover the entire span, so not independent.
11

Option C: Conclusion

The set \(\{\mathbf{u}, \mathbf{u}+\mathbf{v}+\mathbf{w}\}\) is not a basis as it is not independent or spanning.
12

Option D: Check Independence

Set \(\{\mathbf{u}, \mathbf{u}+\mathbf{w}, \mathbf{u}-\mathbf{w}, \mathbf{v}+\mathbf{w}\}\). Since \(V\) is three-dimensional, having four vectors cannot be independent.
13

Final Conclusion

Sets A and B are bases since they are linearly independent and span the space. Sets C and D are not bases.

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Key Concepts

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

Vector Space
Understanding vector spaces is key when delving into linear algebra. A vector space is a collection of vectors that can be added together and multiplied by scalars, complying with specific rules (axioms). These spaces allow us to explore the concepts of linear transformations and the positioning of points_
  • Vectors in a vector space must adhere strictly to closure under addition and scalar multiplication.
  • There must exist a zero vector in the space, serving as the additive identity.
  • Vectors must also observe distributive, associative, and commutative properties for operations.
Understanding these fundamentals is vital, especially as we aim to solve problems concerning bases and spans, such as the one at hand.
Linear Independence
A fundamental aspect of vector spaces is determining whether a set of vectors is linearly independent. Linear independence means no vector in the set can be written as a combination of the others. For example, if you have vectors \(\mathbf{u}, \mathbf{v}, \,\text{and}\, \mathbf{w}\), they are linearly independent if only the trivial combination \(a\mathbf{u} + b\mathbf{v} + c\mathbf{w} = \mathbf{0}\) holds true for \(a = b = c = 0\)._
  • Checking linear independence often involves setting up a linear combination equal to zero.
  • If the only solution is the trivial one, the vectors are independent.
  • This concept is crucial in determining whether a set of vectors forms a basis.
Basis of Vector Space
The basis of a vector space is a set of linearly independent vectors that fully span the entire space. Every vector in the space can be expressed as a linear combination of the basis vectors. In more simple terms, the basis provides the minimal 'scaffolding' for building every possible vector in the space._
  • A set must be linearly independent to qualify as a basis.
  • The set must also be able to span the entire vector space.
  • If the dimension of the space is \(n\), the basis must consist of exactly \(n\) vectors.
Being able to identify or construct a basis is a powerful tool in linear algebra, assisting in understanding the structure of the vector space in question.
Span of Vectors
The span of vectors is the set of all possible linear combinations of a given set of vectors. Essentially, it represents all the places you can reach using those vectors, scaled and added in numerous ways. If a collection of vectors spans a vector space, it means that vector space can be constituted from these vectors._
  • To determine a span, consider any linear combination of the vectors.
  • If you can express every vector in the space as such a combination, then these vectors span the space.
  • The concept of span helps determine whether a set of vectors can serve as a complete basis for the space.
Grasping the span of vectors concept paves the way to understanding more complex structures and behaviors within the realm of linear algebra.

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Most popular questions from this chapter

Which of the following are subspaces of \(\mathbf{M}_{22}\) ? Support your answer. a. \(U=\left\\{\left[\begin{array}{ll}a & b \\ 0 & c\end{array}\right] \mid a, b,\right.\) and \(c\) in \(\left.\mathbb{R}\right\\}\) b. \(U=\left\\{\left[\begin{array}{ll}a & b \\ c & d\end{array}\right] \mid a+b=c+d ; a, b, c, d\right.\) in \(\left.\mathbb{R}\right\\}\) c. \(U=\left\\{A \mid A \in \mathbf{M}_{22}, A=A^{T}\right\\}\) d. \(U=\left\\{A \mid A \in \mathbf{M}_{22}, A B=0\right\\}, B\) a fixed \(2 \times 2\) matrix e. \(U=\left\\{A \mid A \in \mathbf{M}_{22}, A^{2}=A\right\\}\) f. \(U=\left\\{A \mid A \in \mathbf{M}_{22}, A\right.\) is not invertible \(\\}\) g. \(U=\left\\{A \mid A \in \mathbf{M}_{22}, B A C=C A B\right\\}, B\) and \(C\) fixed \(2 \times 2\) matrices

Use Taylor's theorem to derive the binomial theorem: $$(1+x)^{n}=\left(\begin{array}{l}n \\\0\end{array}\right)+\left(\begin{array}{l}n \\\1\end{array}\right) x+\left(\begin{array}{l}n \\\2\end{array}\right) x^{2}+\cdots+\left(\begin{array}{l}n \\\n\end{array}\right) x^{n}$$\ Here the binomial coefficients \(\left(\begin{array}{c}n \\\ r\end{array}\right)\) are defined by $$\left(\begin{array}{l}n \\\r\end{array}\right)=\frac{n !}{r !(n-r) !}$$ where \(n !=n(n-1) \cdots 2 \cdot 1\) if \(n \geq 1\) and \(0 !=1\).

Which of the following subsets of \(V\) are independent? a. \(V=\mathbf{P}_{2} ;\left\\{x^{2}+1, x+1, x\right\\}\) b. \(V=\mathbf{P}_{2} ;\left\\{x^{2}-x+3,2 x^{2}+x+5, x^{2}+5 x+1\right\\}\) c. \(V=\mathbf{M}_{22} ;\left\\{\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right],\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right],\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\right\\}\) d. \(V=\mathbf{M}_{22}\) \(\left\\{\left[\begin{array}{rr}-1 & 0 \\ 0 & -1\end{array}\right],\left[\begin{array}{rr}1 & -1 \\ -1 & 1\end{array}\right],\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right],\left[\begin{array}{rr}0 & -1 \\ -1 & 0\end{array}\right]\right\\}\) e. \(V=\mathbf{F}[1,2] ;\left\\{\frac{1}{x}, \frac{1}{x^{2}}, \frac{1}{x^{3}}\right\\}\) f. \(V=\mathbf{F}[0,1] ;\left\\{\frac{1}{x^{2}+x-6}, \frac{1}{x^{2}-5 x+6}, \frac{1}{x^{2}-9}\right\\}\)

Let \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\) denote vectors in a vector space \(V\). Show that: $$\begin{array}{l}\text { a. } \operatorname{span}\\{\mathbf{u}, \mathbf{v}, \mathbf{w}\\}=\operatorname{span}\\{\mathbf{u}+\mathbf{v}, \mathbf{u}+\mathbf{w}, \mathbf{v}+\mathbf{w}\\} \\\\\text { b. } \operatorname{span}\\{\mathbf{u}, \mathbf{v}, \mathbf{w}\\}=\operatorname{span}\\{\mathbf{u}-\mathbf{v}, \mathbf{u}+\mathbf{w}, \mathbf{w}\\} \end{array}$$

If \(X\) and \(Y\) are two sets of vectors in a vector space \(V\), and if \(X \subseteq Y\), show that \(\operatorname{span} X \subseteq \operatorname{span} Y\).
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