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Chapter 3: Problem 14

Prove that if \(\\{\mathbf{v}, \mathbf{w}\\}\) is a linearly independent subset of a vector space, then \(\\{\mathbf{v}+\mathbf{w}, \mathbf{w}\\}\) is also linearly independent.

Short Answer

Expert verified
The proof reveals that, as per the definition of linear independence, the only solution to the assumed equation in Step 1 is when the scalars are both zero. Thus the set \(\{\mathbf{v} + \mathbf{w}, \mathbf{w}\}\) is confirmed to be linearly independent as claimed.

Step by step solution

01

Assume a linear combination

To begin with, assume a linear combination of the second set of vectors is equal to the zero vector. This can be expressed as \(a(\mathbf{v}+\mathbf{w}) + b\mathbf{w} = 0\), where \(a\) and \(b\) are scalars.
02

Simplify expression

Next, simplify the above equation by distributing the scalar \(a\). This gives the equation \(a\mathbf{v} + a\mathbf{w} + b\mathbf{w} = 0\). This can be further simplified to \(a\mathbf{v} + (a+b)\mathbf{w} = 0\).
03

Apply linear independence of the first set

The original set \(\{\mathbf{v}, \mathbf{w}\}\) is linearly independent, which by definition implies that the only way \(c\mathbf{v} + d\mathbf{w} = 0\) is if both \(c = 0\) and \(d = 0\). Comparing with the equation obtained in step 2, it can be concluded that \(a = 0\) and \(a+b = 0\).
04

Derive the values of the scalars

From above, since \(a = 0\) and \(a+b = 0\), the only solution is \(a = 0\) and \(b = 0\).
05

Conclude the proof

The only solution to the equation assumed in Step 1 is \(a = 0\) and \(b = 0\), thus confirming that the set \(\{\mathbf{v} + \mathbf{w}, \mathbf{w}\}\) is linearly independent. The original assumption is thereby proven.

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Key Concepts

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

Vector Spaces
A vector space is a fundamental mathematical concept used in linear algebra. It's essentially a collection of objects known as vectors, which can be added together or multiplied by scalars. These operations must satisfy specific rules or axioms. A simple way to think about vector spaces is like a playground where vectors can move around freely, but always following some rules!
  • **Additive Identity**: There's a zero vector in the space, which acts like adding zero in normal arithmetic. This means adding the zero vector to any vector in the space does not change the original vector.
  • **Additive Inverse**: For every vector, there is another vector in the same space that's like subtracting it. Together, they "cancel out" to zero.
  • **Associativity and Commutativity**: These properties ensure that vectors can be grouped in any way during addition, and switching their order won鈥檛 change the result.
These properties make vector spaces predictable and reliable, which is why they form the backbone of linear algebra and its applications. Understanding them is essential when examining concepts like linear combinations and independence.
Linear Combinations
A linear combination refers to an expression made by multiplying vectors by scalars and then adding them together. Consider it like making a vector smoothie by blending together different vector "ingredients," each weighted by a scalar.For example, if you have vectors \(\mathbf{v}\) and \(\mathbf{w}\), a linear combination could look like \(a\mathbf{v} + b\mathbf{w}\), where \(a\) and \(b\) are scalars.
  • Linear combinations allow you to explore all possible vectors within a vector space.
  • They are used to determine things like span, which is the set of all possible vectors you can get through different linear combinations.
  • This concept is particularly critical when determining linear independence; if there's a way to create the zero vector (nothing!) without all scalars being zero, the vectors aren't independent.
Linear combinations help us understand how vectors relate to each other within a vector space, making them a core concept in fields like systems of equations and transformations.
Scalars
Scalars are the key players when it comes to manipulating vectors in a vector space. They're just plain numbers, which are used to "scale" vectors. In algebra, when we multiply a vector by a scalar, we're essentially changing its length while keeping it pointing in the same direction. This is like adjusting the volume of a song; the tune remains unchanged, but its intensity varies.
  • Scalars can be positive, negative, or zero:
    • An increase makes the vector longer if positive.
    • A negative scalar reverses the direction of the vector.
    • Zero "flattens" the vector to the zero vector.
  • They're the "number parts" of a linear combination, helping define the final shape and size of the resulting vector.
  • Scalars apply crucially when proving linear independence. If modifying scalars in a combination leads you back to the zero vector without using the trivial (all scalars zero) combination, linear independence is compromised.
Understanding scalars is crucial as they empower us to flexibly scale vectors and analyze vector spaces with precision.

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Most popular questions from this chapter

Suppose that \(v\) is a linear combination of \(w_{1}\) and \(w_{2}\) and that \(w_{1}\) and \(w_{2}\) are linear combinations of \(x_{1}\) and \(x_{2}\). Show that \(v\) is a linear combination of \(x_{1}\) and \(\mathrm{x}_{2}\).

Determine whether \(\left\\{\left[\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right] \cdot\left[\begin{array}{ll}1 & 0 \\ 1 & 0\end{array}\right],\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right],\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]\right\\}\) is a basis for \(M(2,2) .\)

Give a geometric argument to show that it is impossible for a set with two elements to span \(\mathbb{R}^{3}\).

a. Describe a basis for \(\operatorname{M}(4,3)\). b. What is \(\operatorname{dim} \mathbb{M}(4,3)\) ? c. What is \(\operatorname{dim} \mathbb{M}(m, n)\) ?

Suppose \(S\) is a subspace of a vector space \(V\). Prove that if \(\mathbf{v}_{1}, \ldots, \mathbf{v}_{n} \in S\), then \(\operatorname{span}\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{n}\right\\} \subseteq S\)
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