Question

We write vectors \(\mathbb{R}^{n}\) as rows. Find a basis and calculate the dimension of the following subspaces of \(\mathbb{R}^{4}\). a. \(\text {span}\\{(1,-1,2,0),(2,3,0,3),(1,9,-6,6)\\}\) b. \(\text {span}\\{(2,1,0,-1),(-1,1,1,1),(2,7,4,1)\\}\) c. \(\text {span}\\{(-1,2,1,0),(2,0,3,-1),(4,4,11,-3), (3,-2,2,-1)\\}\) d. span \(\\{(-2,0,3,1),(1,2,-1,0),(-2,8,5,3), (-1,2,2,1)\\}\)

Short answer

(a) Basis: \((1, -1, 2, 0), (0, 5, -4, 3)\); Dimension: 2. (b) Basis: \((2, 1, 0, -1), (1, 2, 1, 0), (0, 0, -1, -2)\); Dimension: 3. (c) Basis: \((-1, 2, 1, 0), (0, 4, 5, -1), (0, 0, 10, -2)\); Dimension: 3. (d) Basis: \((-2, 0, 3, 1), (0, 2, 1, 1), (0, 0, 4, 0), (0, 0, 0, 3)\); Dimension: 4.

Step-by-step solution

Setting Up the Problem

To find a basis and dimension for each subspace, we start by arranging each set of vectors as rows of a matrix. Our goal is to row reduce the matrix to its row echelon form (or reduced row echelon form), and identify the linearly independent rows which will constitute a basis. The number of these vectors will give us the dimension.

Solve Part (a)

Arrange the vectors \((1,-1,2,0), (2,3,0,3), (1,9,-6,6)\) as rows of a matrix. We get the matrix:\[\begin{bmatrix}1 & -1 & 2 & 0 \2 & 3 & 0 & 3 \1 & 9 & -6 & 6\end{bmatrix}\]Perform row operations to row reduce it:1. Subtract the first row from the third row:\[\begin{bmatrix}1 & -1 & 2 & 0 \2 & 3 & 0 & 3 \0 & 10 & -8 & 6\end{bmatrix}\]2. Subtract twice the first row from the second row:\[\begin{bmatrix}1 & -1 & 2 & 0 \0 & 5 & -4 & 3 \0 & 10 & -8 & 6\end{bmatrix}\]3. Subtract the second row from the third row:\[\begin{bmatrix}1 & -1 & 2 & 0 \0 & 5 & -4 & 3 \0 & 0 & 0 & 0\end{bmatrix}\]The non-zero rows form a basis for the subspace: \( (1, -1, 2, 0), (0, 5, -4, 3) \). The dimension is 2.

Solve Part (b)

Arrange the vectors \((2,1,0,-1), (-1,1,1,1), (2,7,4,1)\) as rows of a matrix. We have:\[\begin{bmatrix}2 & 1 & 0 & -1 \-1 & 1 & 1 & 1 \2 & 7 & 4 & 1\end{bmatrix}\]Perform row operations:1. Add the first row to the second: \[\begin{bmatrix}2 & 1 & 0 & -1 \1 & 2 & 1 & 0 \2 & 7 & 4 & 1\end{bmatrix}\]2. Subtract the first row from the third row:\[\begin{bmatrix}2 & 1 & 0 & -1 \1 & 2 & 1 & 0 \0 & 6 & 4 & 2\end{bmatrix}\]3. Subtract 2 times the second row from the third row:\[\begin{bmatrix}2 & 1 & 0 & -1 \1 & 2 & 1 & 0 \0 & 2 & 2 & 2\end{bmatrix}\]4. Subtract the third row from the second row:\[\begin{bmatrix}2 & 1 & 0 & -1 \1 & 2 & 1 & 0 \0 & 0 & -1 & -2\end{bmatrix}\]The non-zero rows forming a basis are \( (2, 1, 0, -1), (1, 2, 1, 0), (0, 0, -1, -2) \). The dimension is 3.

Solve Part (c)

Arrange the vectors \((-1,2,1,0), (2,0,3,-1), (4,4,11,-3), (3,-2,2,-1)\) as rows of a matrix:\[\begin{bmatrix}-1 & 2 & 1 & 0 \2 & 0 & 3 & -1 \4 & 4 & 11 & -3 \3 & -2 & 2 & -1\end{bmatrix}\]Perform row operations:1. Add twice the first row to the second:\[\begin{bmatrix}-1 & 2 & 1 & 0 \0 & 4 & 5 & -1 \4 & 4 & 11 & -3 \3 & -2 & 2 & -1\end{bmatrix}\]2. Subtract 4 times the first row from the third:\[\begin{bmatrix}-1 & 2 & 1 & 0 \0 & 4 & 5 & -1 \0 & -4 & 7 & -3 \3 & -2 & 2 & -1\end{bmatrix}\]3. Add 3 times the first row to the fourth:\[\begin{bmatrix}-1 & 2 & 1 & 0 \0 & 4 & 5 & -1 \0 & -4 & 7 & -3 \0 & 4 & 5 & -1\end{bmatrix}\]4. Add the second row to the fourth:\[\begin{bmatrix}-1 & 2 & 1 & 0 \0 & 4 & 5 & -1 \0 & -4 & 7 & -3 \0 & 0 & 10 & -2\end{bmatrix}\]The basis vectors are \( (-1, 2, 1, 0), (0, 4, 5, -1), (0, 0, 10, -2) \) and the dimension is 3.

Solve Part (d)

Arrange the vectors \((-2,0,3,1), (1,2,-1,0), (-2,8,5,3), (-1,2,2,1)\) as rows of a matrix:\[\begin{bmatrix}-2 & 0 & 3 & 1 \1 & 2 & -1 & 0 \-2 & 8 & 5 & 3 \-1 & 2 & 2 & 1\end{bmatrix}\]Perform row operations:1. Add twice the first row to the second:\[\begin{bmatrix}-2 & 0 & 3 & 1 \0 & 2 & 1 & 1 \-2 & 8 & 5 & 3 \-1 & 2 & 2 & 1\end{bmatrix}\]2. Add the first row to the fourth row:\[\begin{bmatrix}-2 & 0 & 3 & 1 \0 & 2 & 1 & 1 \-2 & 8 & 5 & 3 \-3 & 2 & 5 & 2\end{bmatrix}\]3. Add the first row to the third row:\[\begin{bmatrix}-2 & 0 & 3 & 1 \0 & 2 & 1 & 1 \0 & 8 & 8 & 4 \-3 & 2 & 5 & 2\end{bmatrix}\]4. Subtract 4 times the second row from the third row:\[\begin{bmatrix}-2 & 0 & 3 & 1 \0 & 2 & 1 & 1 \0 & 0 & 4 & 0 \-3 & 2 & 5 & 2\end{bmatrix}\]5. Add 1.5 times the second row to the fourth row:\[\begin{bmatrix}-2 & 0 & 3 & 1 \0 & 2 & 1 & 1 \0 & 0 & 4 & 0 \0 & 0 & 0 & 3\end{bmatrix}\]The basis consists of the vectors \((-2, 0, 3, 1), (0, 2, 1, 1), (0, 0, 4, 0), (0, 0, 0, 3)\) and the dimension is 4.

Key concepts

Vector Spaces

In the world of linear algebra, a vector space is a collection of objects called vectors, which can be added together and multiplied by scalars (numbers) to produce another vector. Vector spaces are central to mathematical concepts, and each operates under specific rules and properties that help us manipulate vectors effectively. For example, when working in the vector space \(\mathbb{R}^4\), this implies that any vector we work with has four components or entries, each belonging to the real numbers \(\mathbb{R}\). This framework forms the basis of countless operations and calculations in computer science, physics, and engineering.

• A vector space must satisfy two key operations: vector addition and scalar multiplication.
• Each operation must follow specific rules, such as commutativity (vectors can be added in any order) and associativity (grouped in any way).
• The zero vector must be included in the space, acting as an identity element for addition.

These fundamental properties provide structure, allowing us to determine other key aspects of vector spaces, such as bases and dimensions. In practical terms, understanding vector spaces helps us analyze systems of linear equations and transforms many areas of linearly-related operations.

Row Reduction

Row reduction is a method used in linear algebra to simplify matrices, which are rectangular arrays of numbers, into a form that's easier to work with. This process involves using elementary row operations to achieve row echelon form (REF) or reduced row echelon form (RREF), making it simpler to identify solutions within linear systems. By performing a series of strategic operations, row reduction helps reveal essential qualities of the matrix structure, such as its rank and potential to be a basis for a vector space.

• Elementary row operations include row swapping, multiplying a row by a non-zero scalar, and adding or subtracting a multiple of one row from another.
• While REF can signify a matrix form where every leading entry (first non-zero number in a row) is below and to the right of the leading entries in the rows above, RREF is more restrictive, requiring leading entries to be the only non-zero numbers in their columns.
• Row reduction is a systematic process, and it's a pivotal technique for solving systems of equations, determining invertibility of matrices, and more importantly, finding basis vectors.

Understanding how to row reduce effectively aids in solving many types of vector-related problems efficiently and accurately.

Basis and Dimension

The concept of a basis is fundamental in vector spaces as it provides a way to represent all vectors in that space using a minimal set of vectors. A basis for a space \(V\) is a set of linearly independent vectors that span \(V\). This implies that any vector in the space can be expressed as a linear combination of basis vectors. The number of vectors in the basis is called the dimension of the vector space, and it essentially tells us the 'size' or 'degree of freedom' within the space.

• A basis must consist of vectors that are linearly independent, meaning no vector in the set can be written as a combination of the others.
• The dimension of the vector space gives crucial information about the vector space, such as the number of parameters needed to describe any point within it.
• Finding a basis often requires row-reducing a matrix to identify the pivot columns, which correspond to the basis vectors.

For example, in the solution of our original exercise, we row-reduced matrices to find the basis vectors corresponding to non-zero rows, and established the dimension of subspaces from the number of such basis vectors. This process not only aids in understanding the fundamental structure of vector spaces but also opens doors to various applications across different fields.