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Chapter 1: Problem 40

[M] In Exercises \(37-40\) , determine if the columns of the matrix span \(\mathbb{R}^{4} .\) $$ \left[\begin{array}{rrrrr}{8} & {11} & {-6} & {-7} & {13} \\ {-7} & {-8} & {5} & {6} & {-9} \\ {11} & {7} & {-7} & {-9} & {-6} \\ {-3} & {4} & {1} & {8} & {7}\end{array}\right] $$

Short Answer

Expert verified
The columns of the matrix do not span \( \mathbb{R}^4 \).

Step by step solution

01

Understanding the Problem

The question asks if the columns of the given matrix span \( \mathbb{R}^{4} \). For the columns to span \( \mathbb{R}^{4} \), the matrix must have a pivot position in every row when in row-echelon form.
02

Forming the Augmented Matrix

The given matrix is \[\begin{bmatrix} 8 & 11 & -6 & -7 & 13 \-7 & -8 & 5 & 6 & -9 \11 & 7 & -7 & -9 & -6 \-3 & 4 & 1 & 8 & 7\end{bmatrix}.\]Since the matrix is already expressed in terms of its columns, we don't need to augment it further for this problem.
03

Perform Row Operations to Row-Reduce

We will perform row operations to get the matrix into row-echelon form. This involves: - Making the first entry of the first row a pivot and eliminating other entries in that column. - Repeating the process for subsequent rows and columns, aiming to get zeros below each pivot position.
04

Check for Pivot Positions

After row-reducing the matrix, if there are pivot positions in each of the four rows, this means the columns of the matrix span \( \mathbb{R}^{4} \). If any row doesn't have a pivot, then the columns do not span \( \mathbb{R}^{4} \).
05

Row Reduction Process

1. Start with the first column and make the first row's first element a pivot. Use row operations to clear the rest of the first column. 2. Move to the second row, second column, and create a new pivot. Use row operations to clear the other entries below this pivot.3. Continue this process for subsequent columns and rows.4. If you can form a pivot in all four rows (despite having only four columns), it suggests span. However, in this case, the columns of a 4x5 matrix cannot span \( \mathbb{R}^4 \) because it has more columns than pivot positions available.
06

Conclusion

Since we have 5 columns and only 4 rows, the matrix cannot have more than 4 pivots. Thus, the columns of this matrix do not span \( \mathbb{R}^{4} \) because we need exactly 4 pivot columns (one in each row) but have more columns than potential pivots.

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Key Concepts

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

Row-Reduction
Row-reduction is a key process in matrix theory that simplifies a matrix to make it easier to analyze. Through a series of elementary row operations, we transform a matrix into a simpler form, often the row-echelon form. These operations include:
  • Swapping two rows
  • Multiplying a row by a nonzero scalar
  • Adding a multiple of one row to another row
These operations are crucial not only for finding solutions to systems of equations but also for identifying key features of the matrix, such as pivot positions. In essence, row reduction helps reveal the structural properties of the matrix, making it easier to understand the span of its columns.
Pivot Positions
Pivot positions are specific locations in a matrix that provide critical information about the solution to a system of equations. A pivot position is where a leading 1 appears in a row after a matrix has been row-reduced to row-echelon form.
  • The position is in a column where a pivot (non-zero) is formed.
  • Pivots are used to determine the rank of the matrix, which is the number of pivots.
  • Pivot columns correspond to the basic variables in a system of linear equations.
In the context of determining whether columns span a vector space, like \( \mathbb{R}^{4} \), there must be a pivot position in every row, as this indicates each row has a leading term that can cover one of the dimensions of the space. Without this, the spanning set is incomplete.
Column Span
The column span of a matrix refers to the set of all possible linear combinations of its columns. If the columns of a matrix span a space, like \( \mathbb{R}^{4} \), it means every vector in that space can be expressed as a combination of the columns of the matrix.
  • To span \( \mathbb{R}^{4} \), a matrix needs at least four independent columns.
  • This is equivalent to having a pivot in each of the four rows of the matrix.
  • If there are not enough pivots, the matrix columns cannot span the space.
In practice, if a matrix has more columns than rows, the extra columns may be linearly dependent or not essential in spanning the space, reducing its effective span potential. Understanding the span is essential for determining how vectors can be linearly assembled from a matrix's columns.
Row-Echelon Form
Row-echelon form is an intermediate step in matrix transformation that greatly aids in solving and understanding a system of equations. A matrix is considered to be in row-echelon form when it satisfies the following rules:
  • All non-zero rows are above rows of all zero entries.
  • The leading entry of each non-zero row (called a pivot) is in a column to the right of the leading entry in the row above it.
  • All entries in a column below a pivot are zeros.
Achieving row-echelon form simplifies many operations, such as calculating determinants or solving systems of equations via back-substitution. In determining the span of a matrix, row-echelon form is used to easily identify pivot positions, which indicate which columns contribute to the span. Thus, transforming a matrix into row-echelon form is a fundamental step in matrix analysis.

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Most popular questions from this chapter

Determine by inspection whether the vectors are linearly independent. Justify each answer. \(\left[\begin{array}{r}{4} \\ {-2} \\\ {6}\end{array}\right],\left[\begin{array}{r}{6} \\ {-3} \\\ {9}\end{array}\right]\)

Construct a \(3 \times 3\) matrix, not in echelon form, whose columns span \(\mathbb{R}^{3} .\) Show that the matrix you construct has the desired property.

In Exercises 29鈥�32, find the elementary row operation that transforms the first matrix into the second, and then find the reverse row operation that transforms the second matrix into the first. $$ \left[\begin{array}{rrr}{0} & {-2} & {5} \\ {1} & {4} & {-7} \\ {3} & {-1} & {6}\end{array}\right],\left[\begin{array}{rrr}{1} & {4} & {-7} \\ {0} & {-2} & {5} \\ {3} & {-1} & {6}\end{array}\right] $$

Use the vector \(\mathbf{u}=\left(u_{1}, \ldots, u_{n}\right)\) to verify the following algebraic properties of \(\mathbb{R}^{n}\) . a. \(\mathbf{u}+(-\mathbf{u})=(-\mathbf{u})+\mathbf{u}=0\) b. \(c(d \mathbf{u})=(c d) \mathbf{u}\) for all scalars \(c\) and \(d\)

Let \(\mathbf{v}_{1}=\left[\begin{array}{r}{1} \\ {0} \\ {-1} \\\ {0}\end{array}\right], \mathbf{v}_{2}=\left[\begin{array}{r}{0} \\ {-1} \\\ {0} \\ {1}\end{array}\right], \mathbf{v}_{3}=\left[\begin{array}{r}{1} \\\ {0} \\ {0} \\ {-1}\end{array}\right]\) Does \(\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right\\}\) span \(\mathbb{R}^{4} ?\) Why or why not?
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