Chapter 4: Problem 15
In Exercises \(13-18\), find the eigenvalues and eigenvectors of A geometrically. $$A=\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right]$$ (projection onto the \(x\) -axis)
Short Answer
Expert verified
The eigenvalues are 1 and 0, with eigenvectors along the x-axis and y-axis, respectively.
Step by step solution
01
Brief Overview of the Matrix
The given matrix is \( A = \begin{bmatrix} 1 & 0 \ 0 & 0 \end{bmatrix} \). It represents a linear transformation that projects any vector onto the x-axis in \(\mathbb{R}^2\). This geometric interpretation will help in finding the eigenvalues and eigenvectors.
02
Identify the Transformation's Impact
The transformation \( A \) retains the x-component of any vector and reduces the y-component to zero. For example, \( A \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} x \ 0 \end{bmatrix} \). This means vectors along the x-axis are unchanged (eigenvectors with \( \lambda = 1 \)), while vectors along the y-axis become zero vectors.
03
Determine Eigenvalues
For a 2x2 matrix \( A \), the eigenvalues \( \lambda \) satisfy \( \det(A - \lambda I) = 0 \). Calculate: \[ A - \lambda I = \begin{bmatrix} 1 - \lambda & 0 \ 0 & 0 - \lambda \end{bmatrix} \] \[ \det(A - \lambda I) = (1-\lambda)(0-\lambda) = 0 \] This gives eigenvalues \( \lambda_1 = 1 \) and \( \lambda_2 = 0 \).
04
Find Eigenvectors for \(\lambda = 1\)
Substitute \( \lambda = 1 \) into \( A - \lambda I \): \[ \begin{bmatrix} 1 - 1 & 0 \ 0 & 0 - 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & -1 \end{bmatrix} \] The corresponding eigenvector satisfies: \[ \begin{bmatrix} 0 & 0 \ 0 & -1 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \] The solutions are all vectors of the form \( \begin{bmatrix} x \ 0 \end{bmatrix} \), representing vectors along the x-axis.
05
Find Eigenvectors for \(\lambda = 0\)
Substitute \( \lambda = 0 \) into \( A - \lambda I \): \[ \begin{bmatrix} 1 & 0 \ 0 & 0 \end{bmatrix} \] The corresponding eigenvector satisfies \[ \begin{bmatrix} 1 & 0 \ 0 & 0 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \], leading to \( x = 0 \), so all vectors of form \( \begin{bmatrix} 0 \ y \end{bmatrix} \) represent the eigenvectors, lying along the y-axis.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Geometric Interpretation
Eigenvalues and eigenvectors provide a rich geometric insight into linear transformations through matrices. In simple terms, eigenvectors point in directions that are unchanged by a transformation, even though the vector itself might be stretched or compressed by some scalar factor, known as the eigenvalue. Regarding our exercise, the matrix provided is an example of a projection onto the x-axis. This means that any vector in the plane is transformed such that its y-component becomes zero, relegating it to a vector purely on the x-axis.
Understanding this geometry, the eigenvectors for the eigenvalue \(\lambda = 1\) lie on the x-axis because they remain unchanged except for a scalar multiplication by 1. On the other hand, the eigenvectors for \(\lambda = 0\) lie on the y-axis, which are "flattened" to zero by projection.
Understanding this geometry, the eigenvectors for the eigenvalue \(\lambda = 1\) lie on the x-axis because they remain unchanged except for a scalar multiplication by 1. On the other hand, the eigenvectors for \(\lambda = 0\) lie on the y-axis, which are "flattened" to zero by projection.
Matrix Transformation
A matrix transformation is a function that maps vectors to other vectors through multiplication. It can often be visualized as a series of geometric alterations, like rotating, scaling, or reflecting vectors. The given matrix \( A = \begin{bmatrix} 1 & 0 \ 0 & 0 \end{bmatrix} \) performs a specific transformation by projecting vectors onto the x-axis. The transformation's effect is observable as it keeps the x-component while annulling the y-component of any vector.
This means, if you input a vector \(\begin{bmatrix} x \ y \end{bmatrix}\), the output is \(\begin{bmatrix} x \ 0 \end{bmatrix}\).
This means, if you input a vector \(\begin{bmatrix} x \ y \end{bmatrix}\), the output is \(\begin{bmatrix} x \ 0 \end{bmatrix}\).
- The input vector doesn't lose its x-component.
- The y-component becomes zero, as if squeezed flat onto the x-axis.
Linear Algebra
Linear Algebra deals with linear mappings represented through matrices, often described as transformations of vectors in a space. In this context, eigenvalues and eigenvectors serve as important tools to analyze such transformations.
When we discuss finding eigenvalues and eigenvectors, we refer to solving the equation \( A\mathbf{v} = \lambda \mathbf{v} \), where \( A \) is the matrix, \( \lambda \) is the eigenvalue, and \( \mathbf{v} \) is the eigenvector. Our matrix \( A \) transforms every other vector to the x-axis, and through calculation of its eigenvalues and eigenvectors, we see it's quite a contrived operation.
When we discuss finding eigenvalues and eigenvectors, we refer to solving the equation \( A\mathbf{v} = \lambda \mathbf{v} \), where \( A \) is the matrix, \( \lambda \) is the eigenvalue, and \( \mathbf{v} \) is the eigenvector. Our matrix \( A \) transforms every other vector to the x-axis, and through calculation of its eigenvalues and eigenvectors, we see it's quite a contrived operation.
- Eigenvalues \( \lambda = 1\) and \( \lambda = 0\) tell us about scalar effects.
- Vectors parallel to the x-axis are unchanged.
- Vectors along the y-axis vanish.