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We often write vectors in \(\mathbb{R}^{n}\) as rows. If \(a \neq 0\) is a scalar, show that \(\operatorname{span}\\{a \mathbf{x}\\}=\operatorname{span}\\{\mathbf{x}\\}\) for every vector \(\mathbf{x}\) in \(\mathbb{R}^{n}\)

Short Answer

Expert verified
The span of \( a \mathbf{x} \) equals the span of \( \mathbf{x} \) since scalar multiples do not change the span.

Step by step solution

01

Understanding Vector Span

To start, let's recall what the span of a vector is. The span of a vector \( \mathbf{x} \) in \( \mathbb{R}^{n} \) is the set of all possible scalar multiples of \( \mathbf{x} \). In mathematical terms, it's defined as \( \operatorname{span}\{\mathbf{x}\} = \{ \lambda \mathbf{x} : \lambda \in \mathbb{R} \} \). Therefore, any scalar multiple \( a \mathbf{x} \) of the vector \( \mathbf{x} \) would also belong to this span.
02

Explore Span of Scalar Multiple

Consider the scalar multiple \( a \mathbf{x} \) of the vector \( \mathbf{x} \). The span of \( a \mathbf{x} \) is \( \operatorname{span}\{a \mathbf{x}\} = \{ \lambda (a \mathbf{x}) : \lambda \in \mathbb{R} \} \). This simplifies to \( \{ (\lambda a) \mathbf{x} : \lambda \in \mathbb{R} \} \). Here, \( \lambda a \) is just another scalar, which we'll denote by \( \mu \), giving us \( \{ \mu \mathbf{x} : \mu \in \mathbb{R} \} \).
03

Equivalence of Spans

Now, notice that the set \( \{ \mu \mathbf{x} : \mu \in \mathbb{R} \} \) is precisely the same as \( \operatorname{span}\{\mathbf{x}\} = \{ \lambda \mathbf{x} : \lambda \in \mathbb{R} \} \). Since \( \mu \) can take any real value as \( \lambda a \) varies over all real numbers, \( \operatorname{span}\{a \mathbf{x}\} = \operatorname{span}\{\mathbf{x}\} \). This completes the proof.

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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 concept in linear algebra. It is a collection of objects called vectors, where you can perform operations like vector addition and scalar multiplication. These operations must follow certain rules to preserve the structure of the vector space. For a set of vectors to be considered a vector space, the following conditions must be satisfied:
  • Closure under addition: Adding any two vectors from the set must produce another vector in the same set.
  • Closure under scalar multiplication: Multiplying a vector by any scalar (a real number) must also produce a vector in the same set.
  • Associativity and Communtativity: The operations of vector addition and scalar multiplication are associative and commutative.
These properties ensure that linear combinations of vectors, which are integral in defining concepts like vector span, still belong to the vector space. Understanding these operations and rules is essential for analyzing more complex topics like transformations and vector dynamics. Linear combinations and other transformations maintain the properties of vector spaces.
Scalars
In linear algebra, a scalar is simply a real number that you use to scale a vector. Scalars are crucial because they allow you to stretch or shrink vectors, which helps form linear combinations.
  • Scalar Multiplication: When you multiply a vector by a scalar, you change the vector's length but not its direction, unless the scalar is negative, in which case the vector also reverses direction.
  • Identity Scalar: Multiplying by 1 leaves a vector unchanged, reinforcing the notion that 1 is the identity for multiplication in this context.
  • Zero Scalar: Multiplying any vector by 0 gives the zero vector, which is crucial for understanding vector span and dependencies.
Scalars effectively aid in rescaling the basis vectors of a vector space, which is fundamental in understanding more advanced topics like eigenvectors and matrix factorization. Neighboring concepts like vector span rely heavily on the functional use of scalars to express combinations and dependencies.
Vector Span
The span of a vector in a vector space refers to all possible vectors you can reach by scaling and adding that vector. This concept is important because it provides a way to describe all the possibilities one can achieve from a vector through linear combinations.
  • Definition: The span of a vector \( \mathbf{x} \) is the set of all possible scalar multiples of \( \mathbf{x} \). Mathematically, it’s expressed as \( \text{span}\{\mathbf{x}\} = \{ \lambda \mathbf{x} : \lambda \in \mathbb{R} \} \).
  • Properties: If you take any scalar multiple of a vector \( a\mathbf{x} \), it's clear that this also falls within the vector’s span because \( a\lambda \) is just a different real number when accumulated from \( \lambda \times a \).
  • Equivalence of Spans: As proven in your exercise, \( \text{span}\{a \mathbf{x}\} = \text{span}\{\mathbf{x}\} \), because all linear combinations produced by multiplying with any real number scalar remain part of the same set.
Understanding vector span is pivotal in disciplines like machine learning and computer graphics, where combinations of features or directions create larger spaces of possible outcomes or transformations.

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Most popular questions from this chapter

If \(A\) is \(2 \times 2\) and diagonalizable, show that \(C(A)=\\{X \mid X A=A X\\}\) has dimension 2 or 4. [Hint: If \(P^{-1} A P=D,\) show that \(X\) is in \(C(A)\) if and only if \(P^{-1} X P\) is in \(C(D) .\)

a. Show that \(\mathbf{x} \cdot \mathbf{y}=\frac{1}{4}\left[\|\mathbf{x}+\mathbf{y}\|^{2}-\|\mathbf{x}-\mathbf{y}\|^{2}\right]\) for all \(\mathbf{x}\), \(\mathbf{y}\) in \(\mathbb{R}^{n}\) b. Show that \(\|\mathbf{x}\|^{2}+\|\mathbf{y}\|^{2}=\frac{1}{2}\left[\|\mathbf{x}+\mathbf{y}\|^{2}+\|\mathbf{x}-\mathbf{y}\|^{2}\right]\) for all \(\mathbf{x}, \mathbf{y}\) in \(\mathbb{R}^{n}\)

In each case find a basis of the null space of \(A\). Then compute rank \(A\) and verify (1) of Theorem 5.4.2. a. \(A=\left[\begin{array}{rrr}3 & 1 & 1 \\ 2 & 0 & 1 \\ 4 & 2 & 1 \\ 1 & -1 & 1\end{array}\right]\) b. \(A=\left[\begin{array}{rrrrr}3 & 5 & 5 & 2 & 0 \\ 1 & 0 & 2 & 2 & 1 \\ 1 & 1 & 1 & -2 & -2 \\ -2 & 0 & -4 & -4 & -2\end{array}\right]\)

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 \(\mathbf{z}=(-1,2,2,-3)\)

If \(A\) is \(m \times n\) show that $$ \operatorname{col}(A)=\left\\{A \mathbf{x} \mid \mathbf{x} \text { in } \mathbb{R}^{n}\right\\} $$

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