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Chapter 4: Problem 13

In Exercises \(13-18\), find the eigenvalues and eigenvectors of A geometrically. \(A=\left[\begin{array}{rr}-1 & 0 \\ 0 & 1\end{array}\right]\) (reflection in the \(y\) -axis)

Short Answer

Expert verified
Eigenvalues: 1, -1; Eigenvectors: \(\begin{bmatrix} 0 \\ 1 \end{bmatrix}\), \(\begin{bmatrix} 1 \\ 0 \end{bmatrix}\).

Step by step solution

01

Understand the Geometric Interpretation

The matrix \(A = \begin{bmatrix} -1 & 0 \ 0 & 1 \end{bmatrix}\) represents a reflection transformation in the \(y\)-axis. In simple terms, it changes the sign of the \(x\)-coordinate while keeping the \(y\)-coordinate unchanged when applied to a vector.
02

Identify the Eigenvalues

To find the eigenvalues geometrically, examine how a vector is affected by reflection. A vector along the \(y\)-axis (e.g., \(\begin{bmatrix} 0 \ 1 \end{bmatrix}\)) remains unchanged, thus produces an eigenvalue of 1. A vector along the \(x\)-axis (e.g., \(\begin{bmatrix} 1 \ 0 \end{bmatrix}\)) gets inverted, yielding an eigenvalue of -1.
03

Calculate the Eigenvectors

For \(\lambda = 1\), the eigenvector is any vector of the form \(\begin{bmatrix} 0 \ y \end{bmatrix}\), meaning any scalar multiple of \(\begin{bmatrix} 0 \ 1 \end{bmatrix}\). For \(\lambda = -1\), the eigenvector is any vector of the form \(\begin{bmatrix} x \ 0 \end{bmatrix}\), meaning any scalar multiple of \(\begin{bmatrix} 1 \ 0 \end{bmatrix}\).
04

Present the Results

The eigenvalues of \(A\) are 1 and -1. The corresponding eigenvectors for \(\lambda = 1\) are \(\begin{bmatrix} 0 \ 1 \end{bmatrix}\) and for \(\lambda = -1\) are \(\begin{bmatrix} 1 \ 0 \end{bmatrix}\). Geometrically, these vectors indicate directions on the \(y\)-axis and \(x\)-axis where transformations under \(A\) act as scalar multiplications by 1 or -1, respectively.

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Key Concepts

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

Geometric Interpretation
Understanding the geometric interpretation of matrices can provide meaningful insights into their transformations. For the matrix \( A = \begin{bmatrix} -1 & 0 \ 0 & 1 \end{bmatrix} \), it acts as a reflection across the \(y\)-axis.
This means every point or vector impacted by this matrix will have its \(x\)-coordinate inverted, while the \(y\)-coordinate remains the same.

The reflection matrix essentially flips a vector over the \(y\)-axis. Visualize how a point on the right of the \(y\)-axis would move to the same distance on the left side, mirroring its position. This affects its orientation but not its magnitude or length.
  • Changes sign of the \(x\)-coordinate
  • Keeps the \(y\)-coordinate unchanged
  • Acts as a mirror parallel to the \(y\)-axis
Reflection Matrix
The particular type of transformation described by the matrix \( A \) is fundamentally a reflection. This kind of transformation is recognized by its action of reversing a component's direction without altering its magnitude.

Talking about a reflection matrix, it typically contains "-1" and "1" along the diagonal, as seen here, aligning with the plane or axis mirrored.

The left diagonal element, "-1" in this case, represents the axis of inversion (here, on the \(x\)-axis). The right element, "1", reveals nothing changes along the \(y\)-axis.
  • Matrix diagonal has entries \
Transformation in the Y-axis
When transformations occur in the \(y\)-axis due to reflections, they produce distinct outcomes based on vector placement. Here, the interesting part is how it subtly defines eigenvalues and eigenvectors.

In cases where a whole vector remains stable, free of change, like along \(y\)-axis for our matrix, it confirms as one of the eigenvectors. Similarly, any changes (here reverse the \(x\) component) ascertain another eigenvector. Therefore, by inspecting how directions transform (or don't), both eigenvalues and eigenvectors are unveiled in simplistic intuitive ways:
  • Vectors along the \(y\)-axis, e.g., \(\begin{bmatrix} 0 \ 1 \end{bmatrix}\), yield eigenvalue 1.
  • Vectors along the \(x\)-axis, e.g., \(\begin{bmatrix} 1 \ 0 \end{bmatrix}\), yield eigenvalue -1.
Thus, the transformation points out how reflections unlock the mystery of matrices' core attributes effortlessly.

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Most popular questions from this chapter

Shows that the long-run behavior of a population can be determined directly from the entries of its Leslie matrix. The net reproduction rate of a population is defined as \\[ r=b_{1}+b_{2} s_{1}+b_{3} s_{1} s_{2}+\cdots+b_{n} s_{1} s_{2} \cdots \cdot s_{n-1} \\] where the \(b_{i}\) are the birth rates and the \(s_{j}\) are the survival rates for the population. (a) Explain why \(r\) can be interpreted as the average number of daughters born to a single female over (b) Show that \(r=1\) if and only if \(\lambda_{1}=1\). (This represents zero population growth.) [Hint: Let \\[ g(\lambda)=\frac{b_{1}}{\lambda}+\frac{b_{2} s_{1}}{\lambda^{2}}+\frac{b_{3} s_{1} s_{2}}{\lambda^{3}}+\dots+\frac{b_{n} s_{1} s_{2} \cdots s_{n-1}}{\lambda^{n}} \\] Show that \(\lambda\) is an eigenvalue of \(L\) if and only if \(g(\lambda)=1 .]\) (c) Assuming that there is a unique positive eigenvalue \(\lambda_{1}\), show that \(r<1\) if and only if the population is decreasing and \(r>1\) if and only if the population is increasing.

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.

Let \(x=x(t)\) be a twice-differentiable function and consider the second order differential equation \\[ x^{\prime \prime}+a x^{\prime}+b x=0 \\] (a) Show that the change of variables \(y=x^{\prime}\) and \(z=\) \(x\) allows equation (11) to be written as a system of two linear differential equations in \(y\) and \(z\) (b) Show that the characteristic equation of the system in part (a) is \(\lambda^{2}+a \lambda+b=0\)

Find an invertible matrix Pand a matrix \(C\) of the form \(C=\left[\begin{array}{cr}a & -b \\ b & a\end{array}\right]\) such that \(A=P C P^{-1}\) Sketch the first six points of the trajectory for the dynamical system \(\mathbf{x}_{\mathrm{k}+1}=A \mathbf{x}_{\mathrm{k}}\) with \(\mathbf{x}_{0}=\left[\begin{array}{l}1 \\ 1\end{array}\right]\) and classify the origin as \(a\) spiral attractor, spiral repeller, or orbital center $$A=\left[\begin{array}{rr} 0.1 & -0.2 \\ 0.1 & 0.3 \end{array}\right]$$

Write out the first six terms of the sequence defined by the recurrence relation with the given initial conditions. $$b_{0}=1, b_{1}=1, b_{n}=2 b_{n-1}+b_{n-2} \text { for } n \geq 2$$
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