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Chapter 4: Problem 15

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
Eigenvalues: \( \lambda_1 = 1 \), \( \lambda_2 = 0 \). Eigenvectors: \( \begin{bmatrix} x \\ 0 \end{bmatrix} \) for \( \lambda = 1 \), \( \begin{bmatrix} 0 \\ y \end{bmatrix} \) for \( \lambda = 0 \).

Step by step solution

01

Understanding the Matrix

Matrix \( A \) projects any vector onto the \( x \)-axis. The effect of this matrix is to choose the \( x \)-component of the vector and make the \( y \)-component zero.
02

Formulate the Eigenvalue Equation

For any eigenvector \( \mathbf{v} \) of \( A \) with an eigenvalue \( \lambda \), the equation \( A\mathbf{v} = \lambda \mathbf{v} \) must hold, where \( A = \begin{bmatrix} 1 & 0 \ 0 & 0 \end{bmatrix} \).
03

Solve for Eigenvalues

Formulate the characteristic equation by solving \( \det(A - \lambda I) = 0 \). This becomes: \[ \begin{vmatrix} 1 - \lambda & 0 \ 0 & -\lambda \end{vmatrix} = 0 \] Simplifying gives the characteristic polynomial: \( (1-\lambda)(-\lambda) = 0 \), which results in eigenvalues \( \lambda_1 = 1 \) and \( \lambda_2 = 0 \).
04

Find Eigenvectors for \( \lambda = 1 \)

Substitute \( \lambda = 1 \) into \( (A - \lambda I)\mathbf{v} = 0 \): \( \begin{bmatrix} 1 - 1 & 0 \ 0 & 0 - 1 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = 0 \), or \( \begin{bmatrix} 0 & 0 \ 0 & -1 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \). Here, any vector of the form \( \begin{bmatrix} x \ 0 \end{bmatrix} \) with \( x eq 0 \) is an eigenvector (e.g., \( \begin{bmatrix} 1 \ 0 \end{bmatrix} \)).
05

Find Eigenvectors for \( \lambda = 0 \)

Substitute \( \lambda = 0 \) into \( (A - \lambda I)\mathbf{v} = 0 \): \( \begin{bmatrix} 1 & 0 \ 0 & 0 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \). This yields the solutions \( x = 0 \) and \( y \) can be any value, giving eigenvectors of the form \( \begin{bmatrix} 0 \ y \end{bmatrix} \), e.g., \( \begin{bmatrix} 0 \ 1 \end{bmatrix} \).

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Key Concepts

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

Projection Matrix
A projection matrix is a special type of matrix that "projects" vectors onto a certain subspace. In the case of matrix \( A = \begin{bmatrix} 1 & 0 \ 0 & 0 \end{bmatrix} \), it projects any vector onto the x-axis.
This means it takes any vector, extracts the x-component, and makes the y-component zero.
  • For example, if you have a vector \( \begin{bmatrix} x \ y \end{bmatrix} \), after multiplication with matrix \( A \), it becomes \( \begin{bmatrix} x \ 0 \end{bmatrix} \).
  • This operation is significant in various applications, including computer graphics and data analysis, where projections are used to reduce dimensions or simplify data.
Understanding projection matrices helps in visualizing transformations applied to vectors, making it easier to anticipate the outcome of matrix operations.
Characteristic Polynomial
The characteristic polynomial is a key concept used to find eigenvalues of a matrix.
For matrix \( A \), we form the characteristic polynomial by calculating the determinant of \( A - \lambda I \), where \( I \) is the identity matrix.
  • First, subtract \( \lambda \) from the diagonal elements of matrix \( A \), giving \( \begin{bmatrix} 1-\lambda & 0 \ 0 & -\lambda \end{bmatrix} \).
  • Then, find the determinant: \((1-\lambda)(-\lambda) = 0\).
This results in a polynomial that when solved, gives the eigenvalues of the matrix. The solutions for this polynomial are \( \lambda_1 = 1 \) and \( \lambda_2 = 0 \).
These eigenvalues provide insights into the matrix's transformative properties, such as stretching or compressing along specific directions.
Eigenvalue Equation
An essential step in solving for eigenvalues and eigenvectors is the eigenvalue equation, given as \( A\mathbf{v} = \lambda \mathbf{v} \).
  • Here, \( A \) is the matrix, \( \lambda \) is the eigenvalue, and \( \mathbf{v} \) is the eigenvector.
  • This equation states that when a matrix \( A \) acts on an eigenvector \( \mathbf{v} \), the output is a scalar multiple of the original vector.
To find eigenvectors, substitute each eigenvalue into the equation \((A - \lambda I) \mathbf{v} = 0\).
  • For \( \lambda = 1 \), it simplifies to find vectors in form \( \begin{bmatrix} x \ 0 \end{bmatrix} \).
  • For \( \lambda = 0 \), it gives vectors in form \( \begin{bmatrix} 0 \ y \end{bmatrix} \).
Understanding this equation helps in identifying directions in which a matrix stretches or compresses vectors.
Geometric Interpretation
Eigenvectors and eigenvalues have a fascinating geometric interpretation.
Considering matrix \( A \), it's clear that:
  • Eigenvalue \( \lambda = 1 \) with eigenvectors such as \( \begin{bmatrix} 1 \ 0 \end{bmatrix} \) means any vector on the x-axis remains unchanged by the transformation.
  • Eigenvalue \( \lambda = 0 \) with vectors such as \( \begin{bmatrix} 0 \ 1 \end{bmatrix} \) means any vector on the y-axis collapses to zero.
In essence, the matrix projects all vectors onto the x-axis, demonstrating the concept of a linear transformation visually.
This geometric view gives insight into how matrices can be used to simplify complex physical or graphical systems by focusing on specific directions or dimensions of interest.

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Most popular questions from this chapter

\(P\) is the transition matrix of a regular Markov chain. Find the long range transition matrix \(L\) of \(P\). $$P=\left[\begin{array}{ll} \frac{1}{3} & \frac{1}{6} \\ \frac{2}{3} & \frac{5}{6} \end{array}\right]$$

In general, it is difficult to show that two matrices are similar. However, if two similar matrices are diagonalizable, the task becomes easier. Show that \(A\) and \(B\) are similar by showing that they are similar to the same diagonal matrix. Then find an invertible matrix P such that \(P^{-1} A P=B\). $$A=\left[\begin{array}{ll} 5 & -3 \\ 4 & -2 \end{array}\right], B=\left[\begin{array}{ll} -1 & 1 \\ -6 & 4 \end{array}\right]$$

Write out the first six terms of the sequence defined by the recurrence relation with the given initial conditions. $$a_{1}=128, a_{n}=a_{n-1} / 2 \text { for } n \geq 2$$

Two species, \(X\) and \(Y\), live in a symbiotic relationship. That is, neither species can survive on its own and each depends on the other for its survival. Initially, there are 15 of \(X\) and 10 of \(Y\). If \(x=x(t)\) and \(y=y(t)\) are the sizes of the populations at time \(t\) months, the growth rates of the two populations are given by the system $$\begin{array}{l} x^{\prime}=-0.8 x+0.4 y \\ y^{\prime}=0.4 x-0.2 y \end{array}$$ Determine what happens to these two populations.

Species \(X\) preys on species \(Y .\) The sizes of the populations are represented by \(x=x(t)\) and \(y=y(t) .\) The growth rate of each population is governed by the system of differential equations \(\mathbf{x}^{\prime}=A \mathbf{x}+\mathbf{b},\) where \(\mathbf{x}=\left[\begin{array}{l}x \\\ y\end{array}\right]\) and \(\mathbf{b}\) is a constant vector. Determine what happens to the two populations for the given \(A\) and b and initial conditions \(\mathbf{x}(0) .\) (First show that there are constants a and \(b\) such that the substitutions \(x=u+\) a and \(y=v+b\) convert the system into an equivalent one with no constant terms.) $$A=\left[\begin{array}{rr} -1 & 1 \\ -1 & -1 \end{array}\right], \mathbf{b}=\left[\begin{array}{r} 0 \\ 40 \end{array}\right], \mathbf{x}(0)=\left[\begin{array}{l} 10 \\ 30 \end{array}\right]$$
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