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We often write vectors in \(\mathbb{R}^{n}\) as rows. If \(\mathbf{x} \neq 0\) in \(\mathbb{R}^{n},\) determine all subspaces of \(\operatorname{span}\\{\mathbf{x}\\}\).

Short Answer

Expert verified
The subspaces are \( \operatorname{span}\{\mathbf{x}\} \) and the trivial subspace \( \{\mathbf{0}\} \).

Step by step solution

01

Understand the Problem

We need to find all subspaces of the span of a vector \( \mathbf{x} \) in \( \mathbb{R}^{n} \). A span is the set of all scalar multiples of a vector. So, \( \operatorname{span}\{\mathbf{x}\} \) is essentially the line through the origin and the vector \( \mathbf{x} \).
02

Identify Properties of a Span

In \( \mathbb{R}^{n} \), the span of a single non-zero vector \( \mathbf{x} \) is a 1-dimensional subspace. The span \( \operatorname{span}\{\mathbf{x}\} \) contains vectors of the form \( c\mathbf{x} \) where \( c \) is a scalar.
03

Consider Subspaces of a Line

For any non-zero vector \( \mathbf{x} \), the only possible subspaces of \( \operatorname{span}\{\mathbf{x}\} \) are itself and the trivial subspace containing the zero vector \( \mathbf{0} \).
04

Conclusion

Therefore, the subspaces of \( \operatorname{span}\{\mathbf{x}\} \) are \( \operatorname{span}\{\mathbf{x}\} \) itself and \( \{\mathbf{0}\} \), the trivial subspace. This is because there are no smaller non-trivial subspaces within a 1-dimensional subspace.

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Key Concepts

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

Span
The concept of "span" is fundamental in linear algebra. When we talk about the span of a vector, we're referring to all the possible vectors that can be created by scaling that vector by any real number. Let's picture this in \( \mathbb{R}^n \), the "n-dimensional real coordinate space."

The span of a vector \( \mathbf{x} \) is written as \( \operatorname{span}\{\mathbf{x}\} \), and it symbolizes the collection of all vectors that look like \( c\mathbf{x} \) where \( c \) is any real number. This means you can make \( \operatorname{span}\{\mathbf{x}\} \) by simply multiplying \( \mathbf{x} \) by any real number you like:
  • When \( c = 0 \), you get the zero vector.
  • For other values of \( c \), you get different points along the line formed by \( \mathbf{x} \).

This set forms a line that passes through both the origin and the point represented by the vector \( \mathbf{x} \). It's a 1-dimensional subspace, meaning it's a flat, straight line in space without any thickness or height. The idea of span is crucial because it helps define how vectors in a set create a space in the vector world.
Vector Spaces
Vector spaces are a key structure in linear algebra. A vector space is essentially a collection of vectors along with two operations: addition and scalar multiplication, that adhere to specific rules. Imagine a vector space as a giant playground where vectors can frolic, obeying certain guidelines:
  • Closure under addition: If you add two vectors in the vector space, you'll get another vector that's also inside the vector space.
  • Closure under scalar multiplication: Multiply a vector by a real number (a scalar), and you'll end up with another vector still in the vector space.
Other rules include the existence of a zero vector (the vector that when added to any other vector doesn’t change it) and something called distributive and associative properties.These properties mean you can add vectors together and scale them by numbers without ever "breaking" the vector space. In the vector space \( \mathbb{R}^{n} \), this is just how lines, planes, or any "flatter" surfaces are built using vectors. Each of these flatter spaces is called a subspace, which must logically obey these vector space rules.
Trivial Subspace
When we discuss the trivial subspace, we're focusing on the simplest subspace possible within any vector space — the one that consists only of the zero vector \( \mathbf{0} \).

Though it seems very plain, the trivial subspace is essential. It's essentially the "identity element" under addition for all vectors. For example:
  • Adding \( \mathbf{0} \) to any vector doesn't change the vector.
Despite being just a single point, this subspace plays a role in the definition and structure of all other subspaces. It shows up naturally in any scenario where vectors are being combined or manipulated because a vector space requires the presence of \( \mathbf{0} \) as part of its structure.

Understanding trivial subspaces helps cement the comprehension of the backdrop against which more complex vector combinations are formed, emphasizing the inherent simplicity and balance within the algebraic structures.

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Most popular questions from this chapter

a. Can a \(3 \times 4\) matrix have independent columns? Independent rows? Explain. b. If \(A\) is \(4 \times 3\) and rank \(A=2,\) can \(A\) have independent columns? Independent rows? Explain. c. If \(A\) is an \(m \times n\) matrix and rank \(A=m\), show that \(m \leq n\) d. Can a nonsquare matrix have its rows independent and its columns independent? Explain. e. Can the null space of a \(3 \times 6\) matrix have dimension 2? Explain. f. Suppose that \(A\) is \(5 \times 4\) and \(\operatorname{null}(A)=\mathbb{R} \mathbf{x}\) for some column \(\mathbf{x} \neq \mathbf{0} .\) Can \(\operatorname{dim}(\operatorname{im} A)=2 ?\)

If \(A\) is \(n \times n\), show that every eigenvalue of \(A^{T} A\) is nonnegative. [Hint: Compute \(\|A \mathbf{x}\|^{2}\) where \(\mathbf{x}\) is an eigenvector.]

In each case, find all \((a, b, c, d)\) in \(\mathbb{R}^{4}\) such that the given set is orthogonal. a. \(\\{(1,2,1,0),(1,-1,1,3),(2,-1,0,-1),\) \((a, b, c, d)\\}\) b. \(\\{(1,0,-1,1),(2,1,1,-1),(1,-3,1,0),\) \((a, b, c, d)\\}\)

Let \(A\) be any \(m \times n\) matrix and write \(K=\left\\{\mathbf{x} \mid A^{T} A \mathbf{x}=\mathbf{0}\right\\} .\) Let \(\mathbf{b}\) be an \(m\) -column. Show that if \(\mathbf{z}\) is an \(n\) -column such that \(\|\mathbf{b}-A \mathbf{z}\|\) is minimal, then all such vectors have the form \(\mathbf{z}+\mathbf{x}\) for some \(\mathbf{x} \in K\). [Hint: \(\|\mathbf{b}-A \mathbf{y}\|\) is minimal if and only if \(\left.A^{T} A \mathbf{y}=A^{T} \mathbf{b} .\right]\)

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. \(A=\left[\begin{array}{rrrrr}3 & 5 & 5 & 2 & 0 \\ 1 & 0 & 2 & 2 & 1 \\ 1 & 1 & 1 & -2 & -2 \\ -2 & 0 & -4 & -4 & -2\end{array}\right]\)

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