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Chapter 1: Problem 22

Construct a \(3 \times 3\) matrix \(A,\) with nonzero entries, and a vector \(\mathbf{b}\) in \(\mathbb{R}^{3}\) such that \(\mathbf{b}\) is not in the set spanned by the columns of \(A .\)

Short Answer

Expert verified
Matrix: A = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]; Vector: b = [1, 0, 0]. b is not in the span of A's columns.

Step by step solution

01

Define Matrix A

We need a 3x3 matrix with non-zero entries. Consider the matrix \[A = \begin{bmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 \end{bmatrix}\] is a simple choice where each entry is non-zero.
02

Check Columns for Span

The columns of matrix \(A\) are \(\mathbf{v}_1 = \begin{bmatrix} 1 \ 4 \ 7 \end{bmatrix}\), \(\mathbf{v}_2 = \begin{bmatrix} 2 \ 5 \ 8 \end{bmatrix}\), and \(\mathbf{v}_3 = \begin{bmatrix} 3 \ 6 \ 9 \end{bmatrix}\). These columns determine the vector space spanned by them.
03

Determine Linear Dependence

To check if the columns of \(A\) are linearly dependent, calculate the determinant of \(A\). Using the determinant formula for a 3x3 matrix, \[\text{det}(A) = 1(5 \cdot 9 - 6 \cdot 8) - 2(4 \cdot 9 - 6 \cdot 7) + 3(4 \cdot 8 - 5 \cdot 7).\]Calculating the determinant results in 0, showing columns are linearly dependent.
04

Identify Vector b

Choose a vector \(\mathbf{b}\) in \(\mathbb{R}^3\) that does not lie in the span of the dependent columns of \(A\). One option is \[\mathbf{b} = \begin{bmatrix} 1 \ 0 \ 0 \end{bmatrix},\]which cannot be expressed as a linear combination of the columns since det(A)=0 limits the span to 2D.
05

Verify b is Outside the Span

Attempt to express \(\mathbf{b}\) as a combination of \(\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\). The equation \(c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + c_3 \mathbf{v}_3 = \mathbf{b}\) results in contradictions because of linear dependence, confirming \(\mathbf{b}\) is not in the span.

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Key Concepts

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

Span of Vectors
When we talk about the span of vectors, we're essentially discussing all the possible vectors you can create using linear combinations of a set of vectors. Imagine vectors as arrows in space, and the span is all the places you can reach by layering and scaling these arrows.

The span of vectors is particularly intriguing because it forms a subspace. If you have a set of vectors, say columns from a matrix, their span will form a plane or a line, depending on the number of vectors and their linear independence.
  • For example, the span of three vectors that are linearly independent in \(\mathbb{R}^3\) would cover the whole 3-dimensional space.
  • If they are linearly dependent, their span might only form a plane or even a line, which is not the entire space.
In the original exercise, you were asked to find a vector outside of this span. This means that the vector cannot be formed by any combination of the given set of vectors. This usually happens when the vectors you're trying to span are linearly dependent.
Determinant of a Matrix
The determinant is a special number that can be calculated from a square matrix. It's a valuable tool in linear algebra that provides a lot of information about the matrix.

One main characteristic is its use in determining whether a set of vectors is linearly dependent. If the determinant of a matrix is zero, it implies that the columns of the matrix are linearly dependent. So, the volume described by those vectors collapses to a lower dimension.
  • For instance, a zero determinant for a 3x3 matrix means the vectors lie on a 2D plane.
  • In contrast, a non-zero determinant indicates the vectors span a space of full dimension, such as filling the entire 3D space.
In the context of the original problem, calculating the determinant of matrix \(A\) was crucial. Since \(\text{det}(A) = 0\), it confirmed the columns are linearly dependent, and their span is limited to a two-dimensional plane within \(\mathbb{R}^3\).
Linear Combination
A linear combination involves summing multiple vectors, each multiplied by a scalar. Imagine you're cooking a meal 鈥� each vector is like an ingredient, and the scalars are the amounts you use of each one. By adjusting these quantities, you create different flavors, which in our vector world creates new vectors.

In linear algebra, if a vector \(\mathbf{b}\) can be written as \(c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + c_3 \mathbf{v}_3\), it's part of the span of \(\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\).
  • If you can't find any such scalars to express \(\mathbf{b}\), it lies outside of the span created by those vectors.
  • This was the case in our exercise, where the zero determinant of matrix \(A\) restricted possible linear combinations to a plane, making it impossible to express some vectors, like \(\begin{bmatrix} 1 \ 0 \ 0 \end{bmatrix}\), in \(\mathbb{R}^3\).
Therefore, understanding linear combinations is crucial for grasping how vectors interact in spans and spaces.

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Most popular questions from this chapter

Determine by inspection whether the vectors are linearly independent. Justify each answer. \(\left[\begin{array}{r}{3} \\ {5} \\\ {-1}\end{array}\right],\left[\begin{array}{l}{0} \\ {0} \\\ {0}\end{array}\right],\left[\begin{array}{r}{-6} \\ {5} \\\ {4}\end{array}\right]\)

[M] In Exercises \(37-40\) , determine if the columns of the matrix span \(\mathbb{R}^{4} .\) $$ \left[\begin{array}{rrrr}{5} & {-7} & {-4} & {9} \\ {6} & {-8} & {-7} & {5} \\\ {4} & {-4} & {-9} & {-9} \\ {-9} & {11} & {16} & {7}\end{array}\right] $$

How many pivot columns must a \(5 \times 7\) matrix have if its columns span \(\mathbb{R}^{5}\) ? Why?

Describe the possible echelon forms of the matrix. Use the notation of Example 1 in Section 1.2. \(A\) is a \(4 \times 3\) matrix, \(A=\left[\begin{array}{lll}{a_{1}} & {a_{2}} & {a_{3}}\end{array}\right],\) such that \(\left\\{a_{1}, a_{2}\right\\}\) is linearly independent and \(a_{3}\) is not in \(\operatorname{Span}\left\\{\mathbf{a}_{1}, a_{2}\right\\}\)

Balance the chemical equations in Exercises \(5-10\) using the vector equation approach discussed in this section. The following reaction between potassium permanganate \(\left(\mathrm{KMnO}_{4}\right)\) and manganese sulfate in water produces manganese dioxide, potassium sulfate, and sulfuric acid: \(\mathrm{KMnO}_{4}+\mathrm{MnSO}_{4}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{MnO}_{2}+\mathrm{K}_{2} \mathrm{SO}_{4}+\mathrm{H}_{2} \mathrm{SO}_{4}\) [For each compound, construct a vector that lists the numbers of atoms of potassium (K), manganese, oxygen, sulfur, and hydrogen.]
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