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In Exercises 11 and \(12,\) find the dimension of the subspace spanned by the given vectors. $$ \left[\begin{array}{l}{1} \\ {0} \\\ {2}\end{array}\right],\left[\begin{array}{l}{3} \\ {1} \\\ {1}\end{array}\right],\left[\begin{array}{r}{9} \\ {4} \\\ {-2}\end{array}\right],\left[\begin{array}{r}{-7} \\ {-3} \\\ {1}\end{array}\right] $$

Short Answer

Expert verified
The dimension of the subspace is 2.

Step by step solution

01

Set Up the Matrix

Arrange the vectors as columns in a matrix so that we can perform row operations to find the rank (dimension of the span).\[A = \begin{bmatrix} 1 & 3 & 9 & -7 \ 0 & 1 & 4 & -3 \ 2 & 1 & -2 & 1 \end{bmatrix}\]
02

Perform Row Operations

To find the rank of the matrix, perform elementary row operations to transform the matrix into row echelon form (REF) or reduced row echelon form (RREF). Start with the first row.1. To create zeros below the pivot in the first column, perform: - \(R_3 = R_3 - 2R_1\)After applying these operations, the matrix becomes:\[\begin{bmatrix} 1 & 3 & 9 & -7 \0 & 1 & 4 & -3 \0 & -5 & -20 & 15 \end{bmatrix}\]
03

Continue Row Reduction

Continue to simplify the second row to have leading 1’s:1. Apply \(R_3 = R_3 + 5R_2\) to create zero below the second pivot.The matrix will be:\[\begin{bmatrix} 1 & 3 & 9 & -7 \0 & 1 & 4 & -3 \0 & 0 & 0 & 0 \end{bmatrix}\]
04

Determine the Rank

The rank of the matrix is the number of non-zero rows in the echelon form. In this case, there are two non-zero rows. Therefore, the dimension of the subspace spanned by these vectors is 2.

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Key Concepts

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

Linear Combinations
Understanding linear combinations is fundamental in linear algebra and subspace analysis. A **linear combination** of vectors is created by multiplying each vector by a scalar and then adding the results together. In mathematical terms, if we have vectors \( \mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n \), a linear combination of these vectors can be expressed as: \[ c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + \cdots + c_n \mathbf{v}_n\] where \( c_1, c_2, \ldots, c_n \) are the scalars or coefficients.In our example, the goal is to determine the dimension of the subspace spanned by the given vectors. This involves examining all possible linear combinations of these vectors and how they construct a subspace in a larger vector space. Essentially, any vector in the subspace can be written as a linear combination of the given vectors.
  • The process usually involves solving systems of equations to see which combinations yield particular vectors.
  • Linear combinations not only help describe the span but also provide insight into how each vector influences the span.
Recognizing which vectors contribute to the dimension of the span leads us to the concepts of matrix rank and row operations.
Matrix Rank
Matrix rank is a crucial concept for determining the dimension of a vector space or subspace. The **rank** of a matrix is the maximum number of linearly independent row or column vectors in the matrix. It effectively tells us the dimension of the space spanned by the row or column vectors. In our exercise, we placed the given vectors as columns in a matrix and performed transformations to find its rank. Here's why matrix rank is essential:
  • It provides the number of non-zero rows in row-echelon form.
  • Rank reveals the number of independent vectors that span a subspace.
  • This property helps determine the dimension of the subspace.
The more dependent vectors you have (i.e., vectors that can be written as combinations of others), the fewer unique contributions to the span they make. In our solution, achieving a rank of 2 shows that two vectors are enough to describe the spanning subspace fully. The other vectors are linear combinations of these two.
Row Operations
Row operations are essential techniques in transforming matrices and finding their echelon forms. **Elementary row operations** include row swapping, multiplying a row by a non-zero scalar, and adding or subtracting rows.Why are row operations important?
  • They simplify matrices to forms (like reduced row echelon form) that are easier to analyze.
  • They preserve the row space and thus the rank of the matrix.
  • They help reveal the pivot positions, which indicate linear independence.
In our exercise, row operations reduce the given matrix to reveal its structure. For instance, applying \( R_3 = R_3 - 2R_1 \) and \( R_3 = R_3 + 5R_2 \) helped streamline the process by creating zeros below the pivot positions. By performing row operations, we were able to determine that the last row became zero, confirming that the solution's rank is 2. This outcome means only two vectors are linearly independent, which signifies the dimension of the subspace they span: showcasing how row operations illuminate the path to understanding matrix structure and vector spaces.

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Most popular questions from this chapter

Exercises 31 and 32 reveal an important connection between linear independence and linear transformations and provide practice using the definition of linear dependence. Let \(V\) and \(W\) be vector spaces, let \(T : V \rightarrow W\) be a linear transformation, and let \(\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{p}\right\\}\) be a subset of \(V .\) Show that if \(\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{p}\right\\}\) is linearly dependent in \(V,\) then the set of images, \(\left\\{T\left(\mathbf{v}_{1}\right), \ldots, T\left(\mathbf{v}_{p}\right)\right\\},\) is linearly dependent in \(W .\) This fact shows that if a linear transformation maps a set \(\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{p}\right\\}\) onto a linearly independent set \(\left\\{T\left(\mathbf{v}_{1}\right), \ldots, T\left(\mathbf{v}_{1}\right)\right\\},\) then the original set is linearly independent, too (because it cannot be linearly dependent).

Suppose \(\mathbb{R}^{4}=\operatorname{Span}\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{4}\right\\} .\) Explain why \(\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{4}\right\\}\) is a basis for \(\mathbb{R}^{4} .\)

Let \(\mathbf{v}_{1}=\left[\begin{array}{l}{1} \\ {0} \\\ {1}\end{array}\right], \mathbf{v}_{2}=\left[\begin{array}{l}{0} \\ {1} \\\ {1}\end{array}\right], \mathbf{v}_{3}=\left[\begin{array}{l}{0} \\ {1} \\\ {0}\end{array}\right],\) and let \(H\) be the set of vectors in \(\mathbb{R}^{3}\) whose second and third entries are equal. Then every vector in \(H\) has a unique expansion as a linear combination of \(\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3},\) because $$ \left[\begin{array}{l}{s} \\ {t} \\\ {t}\end{array}\right]=s\left[\begin{array}{l}{1} \\ {0} \\\ {1}\end{array}\right]+(t-s)\left[\begin{array}{l}{0} \\ {1} \\\ {1}\end{array}\right]+s\left[\begin{array}{l}{0} \\ {1} \\\ {0}\end{array}\right] $$ for any \(s\) and \(t .\) Is \(\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right\\}\) a basis for \(H ?\) Why or why not?

On any given day, a student is either healthy or ill. Of the students who are healthy today, 95\(\%\) will be healthy tomorrow. Of the students who are ill today, 55\(\%\) will still be ill tomorrow. a. What is the stochastic matrix for this situation? b. Suppose 20\(\%\) of the students are ill on Monday. What fraction or percentage of the students are likely to be ill on Tuesday? On Wednesday? c. If a student is well today, what is the probability that he or she will be well two days from now?

Exercises \(27-29\) concern an \(m \times n\) matrix \(A\) and what are often called the fundamental subspaces determined by \(A .\) Justify the following equalities: a. \(\operatorname{dim} \mathrm{Row} A+\operatorname{dim} \mathrm{Nul} A=n\) Number of columns of \(A\) b. \(\operatorname{dim} \operatorname{Col} A+\operatorname{dim} \mathrm{Nul} A^{T}=m\) Number of rows of \(A\)

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