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We often write vectors in \(\mathbb{R}^{n}\) as rows. Is \(\mathbb{R}^{2}\) a subspace of \(\mathbb{R}^{3}\) ? Defend your answer.

Short Answer

Expert verified
No, \( \mathbb{R}^2 \) is not a subspace of \( \mathbb{R}^3 \) because they differ in dimensions.

Step by step solution

01

Understanding Vector Spaces

Vectors in \( \mathbb{R}^n \) have \( n \) components. \( \mathbb{R}^2 \) refers to vectors with two components, such as \([x, y]\). \( \mathbb{R}^3 \) refers to vectors with three components, such as \([x, y, z]\). To be a subspace, we need to check if \( \mathbb{R}^2 \) fits within \( \mathbb{R}^3 \) in a certain specific form.
02

Definition of a Subspace

A subspace of a vector space is a subset that is also a vector space under the same operations. Specifically, for \( \mathbb{R}^2 \) to be a subspace of \( \mathbb{R}^3 \), vectors from \( \mathbb{R}^2 \) must fit into \( \mathbb{R}^3 \) while fulfilling conditions related to vector addition, scalar multiplication, and must include the zero vector from \( \mathbb{R}^3 \).
03

Checking Direct Inclusion

A vector from \( \mathbb{R}^2 \), being of form \([x, y]\), cannot directly fit into \( \mathbb{R}^3 \) without modification because \( \mathbb{R}^3 \) requires three components. To be a subspace, we'd need a clear embedding for these vectors into \( \mathbb{R}^3 \).
04

Embedding Criteria

One common embedding of \( \mathbb{R}^2 \) into \( \mathbb{R}^3 \) is taking vectors \([x, y]\) and mapping them to \([x, y, 0]\), which fits them into \( \mathbb{R}^3 \). However, this is only a subset and not \( \mathbb{R}^2 \) as a whole, since it depends on the specific vector form \([x, y, 0]\) rather than the entirety of \( \mathbb{R}^3 \).
05

Conclusion

\( \mathbb{R}^2 \) as a whole cannot be a subspace of \( \mathbb{R}^3 \) because it cannot fulfill the necessary criteria (particularly, the closure under vector operations in \( \mathbb{R}^3 \)) without a specific embedding that reduces its dimensionality into \( \mathbb{R}^3 \). Thus, only specific subsets of \( \mathbb{R}^3 \) related to \( \mathbb{R}^2 \)'s structure or form can be subspaces.

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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 collection of vectors that can be added together and multiplied by scalars. This means any linear combination of vectors from the space will also belong to that space.
Let's take a closer look at the characteristics of vector spaces:
  • They must include the zero vector, the additive identity, meaning adding it to any vector doesn't change the vector.
  • They follow closure under addition, where adding any two vectors from the space results in another vector within the same space.
  • They possess closure under scalar multiplication, which means multiplying a vector by a scalar retains a vector in the same space.
In our exercise, we're evaluating if \( \mathbb{R}^{2} \), which involves 2-component vectors like \([x,y]\), can be a subspace of \( \mathbb{R}^{3} \), the space of 3-component vectors like \([x,y,z]\). A direct inclusion isn't possible as the number of dimensions must be compatible without transformation.
Embedding
Embedding is a process that allows us to "fit" one mathematical structure within another. When we embed \( \mathbb{R}^{2} \) into \( \mathbb{R}^{3} \), we need a method to incorporate 2-component vectors while respecting vector space rules.
A common approach is to map a vector \([x, y]\) from \( \mathbb{R}^{2} \) to \([x, y, 0]\) in \( \mathbb{R}^{3} \). This maintains the vector properties in the higher-dimensional space, but please note:
  • This kind of embedding considers the zero extension in the dimension we are increasing. Specifically, the third component is set to zero, providing a flat plane in \( \mathbb{R}^{3} \).
  • Such an embedding forms a valid subspace because it maintains vector addition and scalar multiplication within the projected plane.
However, this embedding demonstrates that while certain parts of \( \mathbb{R}^{3} \) behave like \( \mathbb{R}^{2} \), \( \mathbb{R}^{2} \) in its entirety cannot be a natural subspace of \( \mathbb{R}^{3} \).
Dimension
Dimension in vector spaces refers to the number of coordinates needed to specify any vector.
In \( \mathbb{R}^{2} \), vectors need two numbers for specification, typically shown as \([x, y]\). For \( \mathbb{R}^{3} \), three numbers are necessary, shown as \([x, y, z]\).
  • Each dimension in a vector space adds a degree of freedom. It represents an additional direction one can move in within that space.
  • For a subspace like \([x, y, 0]\) within \( \mathbb{R}^{3} \), the dimension effectively remains 2, since the zero holds the dimension fixed in the additional direction.
Thus, a subspace doesn't inherit the parent’s full dimensionality unless all dimensions are actively utilized. Because \( \mathbb{R}^{2} \) lacks a component to align with \( \mathbb{R}^{3} \)'s third dimension, it can't fully join the higher dimensional space without imposing dimensional restrictions.

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Most popular questions from this chapter

Show that the only matrix similar to a scalar matrix \(A=r I, r\) in \(\mathbb{R},\) is \(A\) itself.

Let \(A\) be \(n \times n\) with \(n\) distinct real eigenvalues. If \(A C=C A,\) show that \(C\) is diagonalizable.

In each case either show that the statement is true or give an example showing that it is false. a. Every independent set in \(\mathbb{R}^{n}\) is orthogonal. b. If \(\\{\mathbf{x}, \mathbf{y}\\}\) is an orthogonal set in \(\mathbb{R}^{n},\) then \(\\{\mathbf{x}, \mathbf{x}+\mathbf{y}\\}\) is also orthogonal. c. If \(\\{\mathbf{x}, \mathbf{y}\\}\) and \(\\{\mathbf{z}, \mathbf{w}\\}\) are both orthogonal in \(\mathbb{R}^{n}\), then \(\\{\mathbf{x}, \mathbf{y}, \mathbf{z}, \mathbf{w}\\}\) is also orthogonal. d. If \(\left\\{\mathbf{x}_{1}, \mathbf{x}_{2}\right\\}\) and \(\left\\{\mathbf{y}_{1}, \mathbf{y}_{2}, \mathbf{y}_{3}\right\\}\) are both or- thogonal and \(\mathbf{x}_{i} \cdot \mathbf{y}_{j}=0\) for all \(i\) and \(j,\) then \(\left\\{\mathbf{x}_{1}, \mathbf{x}_{2}, \mathbf{y}_{1}, \mathbf{y}_{2}, \mathbf{y}_{3}\right\\}\) is orthogonal. e. If \(\left\\{\mathbf{x}_{1}, \mathbf{x}_{2}, \ldots, \mathbf{x}_{n}\right\\}\) is orthogonal in \(\mathbb{R}^{n}\), then \(\mathbb{R}^{n}=\operatorname{span}\left\\{\mathbf{x}_{1}, \mathbf{x}_{2}, \ldots, \mathbf{x}_{n}\right\\}\) f. If \(\mathbf{x} \neq \mathbf{0}\) in \(\mathbb{R}^{n},\) then \(\\{\mathbf{x}\\}\) is an orthogonal set.

We often write vectors in \(\mathbb{R}^{n}\) as rows. If \(a_{1}, a_{2}, \ldots, a_{k}\) are nonzero scalars, show that \(\operatorname{span}\left\\{a_{1} \mathbf{x}_{1}, a_{2} \mathbf{x}_{2}, \ldots, a_{k} \mathbf{x}_{k}\right\\}=\) \(\operatorname{span}\left\\{\mathbf{x}_{1}, \mathbf{x}_{2}, \ldots, \mathbf{x}_{k}\right\\}\) for any vectors \(\mathbf{x}_{i}\) in \(\mathbb{R}^{n}\)

A matrix obtained from \(A\) by deleting rows and columns is called a submatrix of \(A .\) If \(A\) has an invertible \(k \times k\) submatrix, show that rank \(A \geq k\). [Hint: Show that row and column operations carry \(A \rightarrow\left[\begin{array}{rr}I_{k} & P \\ 0 & Q\end{array}\right]\) in block form.] Remark: It can be shown that rank \(A\) is the largest integer \(r\) such that \(A\) has an invertible \(r \times r\) submatrix.

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