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Chapter 7: Problem 29

Describe the nullspace of the given matrix, both parametrically and geometrically in \(\mathbf{R}^{3}\). \(\left(\begin{array}{rrr}1 & -1 & -1 \\ 2 & -2 & -2 \\ -4 & 4 & 4\end{array}\right)\)

Short Answer

Expert verified
The nullspace is a plane spanned by \((1, 1, 0)\) and \((1, 0, 1)\) in \(\mathbf{R}^3\).

Step by step solution

01

Write the matrix equation

We want to find the nullspace of the matrix \(A\), meaning all vectors \(\mathbf{x}\) such that \(A\mathbf{x} = \mathbf{0}\). Given \(A = \begin{pmatrix} 1 & -1 & -1 \ 2 & -2 & -2 \ -4 & 4 & 4 \end{pmatrix}\), the equation is \(A\begin{pmatrix} x_1 \ x_2 \ x_3 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \end{pmatrix}\).
02

Form the system of linear equations

The matrix equation \(A\mathbf{x} = \mathbf{0}\) translates to the following system of linear equations: 1. \(x_1 - x_2 - x_3 = 0\)2. \(2x_1 - 2x_2 - 2x_3 = 0\)3. \(-4x_1 + 4x_2 + 4x_3 = 0\).
03

Simplify the system of equations

Note that the second equation is just twice the first equation and the third equation is \(-4 \) times the first equation. Thus, these are linearly dependent. We can reduce the system to just one equation:\(x_1 - x_2 - x_3 = 0\).
04

Express as parametric vector form

We solve \(x_1 - x_2 - x_3 = 0\) for one variable in terms of the others. Let's express \(x_1\) in terms of free variables \(t\) and \(s\) where \(x_2 = t\) and \(x_3 = s\).\(x_1 = x_2 + x_3 = t + s\).Thus, the solution vector \(\begin{pmatrix} x_1 \ x_2 \ x_3 \end{pmatrix} = t \begin{pmatrix} 1 \ 1 \ 0 \end{pmatrix} + s \begin{pmatrix} 1 \ 0 \ 1 \end{pmatrix}\).
05

Describe geometrically

The nullspace is spanned by the vectors \(\begin{pmatrix} 1 \ 1 \ 0 \end{pmatrix}\) and \(\begin{pmatrix} 1 \ 0 \ 1 \end{pmatrix}\), so it's a plane through the origin in \(\mathbf{R}^3\). This plane contains all linear combinations of the two spanning vectors.

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Key Concepts

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

Parametric Vector Form
When describing the nullspace of a matrix, the parametric vector form is a convenient mathematical expression. It provides a pathway to visualize infinite solutions when finding vectors that satisfy the equation \( A\mathbf{x} = \mathbf{0} \). For the given matrix:
  • The equation simplifies to a single equation \( x_1 - x_2 - x_3 = 0 \).
  • By expressing \( x_1 \) in terms of \( x_2 \) and \( x_3 \), we consider them as free variables, say \( t \) and \( s \).
  • We find \( x_1 = t + s \), \( x_2 = t \), and \( x_3 = s \).
  • This leads to a solution in parametric vector form: \(\begin{pmatrix} x_1 \ x_2 \ x_3 \end{pmatrix} = t \begin{pmatrix} 1 \ 1 \ 0 \end{pmatrix} + s \begin{pmatrix} 1 \ 0 \ 1 \end{pmatrix}\).
This form helps express the entire nullspace, describing all vectors that satisfy the original matrix equation.
Geometric Interpretation
The nullspace can be understood visually through its geometric properties. In this case, the nullspace of the matrix is a plane in \( \mathbf{R}^{3} \). Here's why:
  • The matrix determines equations that express certain dependencies between variables.
  • The parametric solution \( t \begin{pmatrix} 1 \ 1 \ 0 \end{pmatrix} + s \begin{pmatrix} 1 \ 0 \ 1 \end{pmatrix} \) indicates the nullspace is spanned by two vectors.
  • These vectors lie in a plane because they are linearly independent and form a basis for the nullspace.
  • Every linear combination of these vectors remains within this plane.
This plane contains all solutions, meaning any vector in the nullspace corresponds to a point on this plane through the origin of \( \mathbf{R}^{3} \).
Linear Dependency
Understanding linear dependency is key when finding the nullspace. Here's what it entails in this scenario:
  • The system of equations derived from the matrix share a common form, indicating some rows are multiples of others.
  • In this case, the second equation is twice the first, and the third is \( -4 \) times the first.
  • This means these equations are linearly dependent and effectively redundant, leading to a singular matrix.
  • Being linearly dependent allows reduction of the system, simplifying the problem to a single key equation.
This simplification shows that despite starting with three equations, only one defines the dependency among variables, revealing the structure of the nullspace.

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Most popular questions from this chapter

Show that the system $$ \left(\begin{array}{rrr} 3 & 3 & -3 \\ -1 & -1 & 1 \\ 3 & 5 & -1 \end{array}\right)\left(\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}\right)=\left(\begin{array}{l} b_{1} \\ b_{2} \\ b_{3} \end{array}\right) $$ has solutions only if \(\mathbf{b}=\left(b_{1}, b_{2}, b_{3}\right)^{T}\) lies on the plane \(b_{1}+3 b_{2}=0\). Is the coefficient matrix singular or nonsingular? Explain.

Compute the determinant of the matrix. In each case, decide if there is a nonzero vector in the nullspace. \(\left(\begin{array}{rrrrrr}556 & 65 & -91 & 52 & 416 & -143 \\ 550 & 60 & -90 & 50 & 410 & -140 \\ -169 & -22 & 26 & -18 & -131 & 38 \\ -96 & -13 & 14 & -7 & -69 & 27 \\ 550 & 60 & -90 & 50 & 410 & -140 \\ 825 & 97 & -137 & 80 & 617 & -211\end{array}\right)\)

Find the values of \(x\) for which the indicated matrix has a nontrivial nullspace. \(\left(\begin{array}{cc}-1-x & 0 \\ 3 & 2-x\end{array}\right)\)

Consider the following collection of vectors, which you are to use. $$ \begin{array}{lll} \mathbf{u}_{1}=(1,-2)^{T}, & \mathbf{u}_{2}=(3,0)^{T}, & \mathbf{u}_{3}=(2,-4)^{T} \\ \mathbf{v}_{1}=(1,-4,4)^{T}, & \mathbf{v}_{2}=(0,-2,1)^{T}, & \mathbf{v}_{3}=(1,-2,3)^{T} \end{array} $$ In each exercise, if the given vector \(w\) lies in the span, provide a specific linear combination of the spanning vectors that equals the given vector; otherwise, provide a specific numerical argument why the given vector does not lie in the span. Is the vector \(\mathbf{w}=(-7,22,-25)^{T}\) in the \(\operatorname{span}\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right\\}\) ?

Sketch the nullspace of the given matrix in \(\mathbf{R}^{2}\). \(A=\left(\begin{array}{ll}2 & -1 \\ 4 & -2\end{array}\right)\)
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