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For each of the matrices find all real eigenvalues, with their algebraic multiplicities. Show your work. Do not use technology. $$\left[\begin{array}{ll} 5 & -4 \\ 2 & -1 \end{array}\right]$$

Short Answer

Expert verified
The matrix has two real eigenvalues: \(\lambda = 1\) and \(\lambda = 3\), each with an algebraic multiplicity of 1.

Step by step solution

01

Write down the characteristic polynomial

To find the eigenvalues of a matrix, we first write down its characteristic polynomial, which is given by the determinant of \(A - \lambda I\), where \(A\) is the matrix and \(I\) is the identity matrix. For the given matrix A, the characteristic polynomial is calculated as \(\det(\begin{bmatrix} 5 - \lambda & -4 \ 2 & -1 - \lambda \end{bmatrix})\).
02

Calculate the determinant

Calculate the determinant of the matrix obtained in Step 1: \(\det(\begin{bmatrix} 5 - \lambda & -4 \ 2 & -1 - \lambda \end{bmatrix}) = (5 - \lambda)(-1 - \lambda) - (-4)(2)\).
03

Simplify and find the roots of the polynomial

Expand and simplify the expression obtained to form a quadratic equation: \(\lambda^2 - 4\lambda + 3 = 0\). We then find the roots of this quadratic equation, which are the eigenvalues of the matrix A.
04

Calculate eigenvalues

Solving the quadratic equation \(\lambda^2 - 4\lambda + 3 = 0\) by factoring, we get \(\lambda = 1\) and \(\lambda = 3\).
05

Determine algebraic multiplicities

The algebraic multiplicity of an eigenvalue is the number of times it is a root of the characteristic polynomial. Here, each eigenvalue \(\lambda = 1\) and \(\lambda = 3\) appears exactly once, so each has an algebraic multiplicity of 1.

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Key Concepts

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

Characteristic Polynomial
To begin understanding eigenvalues, it's essential to grasp what a characteristic polynomial is. In essence, this polynomial is your golden ticket to finding eigenvalues. When scrutinizing a matrix, like the one given in our exercise, unleashing its characteristic polynomial involves a bit of matrix algebra magic.

We take the original matrix and perform a slight alteration: for each element on the main diagonal, we subtract a variable often denoted as \(\text{\lambda}\), representing our eventual eigenvalues. This concoction results in a new matrix, \(A - \lambda I\), where \(I\) stands for the identity matrix, and \(A\) is our original matrix. The determinant of this modified matrix is our coveted characteristic polynomial.

Tackling this step-by-step means calculating the determinant of a two-by-two matrix, like so: \(\det(\begin{bmatrix} 5 - \lambda & -4 \ 2 & -1 - \lambda \end{bmatrix})\), which later transforms into a polynomial equation. The roots of this polynomial, deduced by setting it equal to zero, are the matrix's eigenvalues, the very core of our quest.
Determinant Calculation
Pondering how to calculate the determinant? It's not as daunting as it seems. When faced with a two-by-two matrix, as seen in our exercise, the formula is straightforward: \(\det(\begin{bmatrix} a & b \ c & d \end{bmatrix}) = ad - bc\). Now, let's apply this to our specific case: \(\det(\begin{bmatrix} 5 - \lambda & -4 \ 2 & -1 - \lambda \end{bmatrix}) = (5 - \lambda)(-1 - \lambda) - (-4)(2)\).

Ensuring You're on the Right Path

As we follow through with the determinant calculation, we're led to a quadratic equation. It's essential to expand meticulously and to simplify thoroughly, at which point our equation will be ready for factorization or for the quadratic formula - whichever you prefer. Whichever the approach, the goal is to excavate the eigenvalues, those mystical numbers that signify so much in the realm of linear algebra.
Algebraic Multiplicity
Stumbling upon the term 'algebraic multiplicity' might seem like encountering a mythical beast, but fear not. It's simply a name for the number of times an eigenvalue appears as a root of the characteristic polynomial. Think of it like counting how many times a specific number shows up in our polynomial puzzle.

Eigenvalues: The Repeat Offenders

Upon solving the polynomial and finding that our eigenvalues are \(\lambda = 1\) and \(\lambda = 3\), we determine that each appears exactly once. Their algebraic multiplicity is therefore 1. In more complex scenarios, an eigenvalue could repeat, indicating a higher multiplicity. Multiplicity isn't just a headcount — it has profound implications on the geometric nature of our matrix and can signal deeper patterns at play within the linear transformation it represents.

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Most popular questions from this chapter

For the matrices \(A\) in Exercises 1 through \(10,\) determine whether the zero state is a stable equilibrium of the dynamical system \(\vec{x}(t+1)=A \vec{x}(t)\). $$A=\left[\begin{array}{rr} 2.4 & -2.5 \\ 1 & -0.6 \end{array}\right]$$

Consider the dynamical system \\[ \vec{x}(t+1)=A \vec{x}(t), \quad \text { where } \quad A=\left[\begin{array}{ccc} 0.4 & 0.1 & 0.5 \\ 0.4 & 0.3 & 0.1 \\ 0.2 & 0.6 & 0.4 \end{array}\right] \\] perhaps modeling the way people surf a mini-Web, as in Exercise 7.4 .1 a. Using technology, compute a high power of \(A,\) such as \(A^{20}\). What do you observe? Make a conjecture for \(\lim _{t \rightarrow \infty} A^{t} .\) (In part e, you will prove this conjecture.) b. Use technology to find the complex eigenvalues of A. Is matrix \(A\) diagonalizable over \(\mathbb{C} ?\) c. Find the equilibrium distribution \(\vec{x}_{e q u}\) for \(A,\) that is, the unique distribution vector in the eigenspace \(E_{1}\) d. Without using Theorem 7.4 .1 (which was proven only for matrices that are diagonalizable over \(\mathbb{R}\) ), show that \(\lim _{t \rightarrow \infty}\left(A^{t} \vec{x}_{0}\right)=\vec{x}_{e q u}\) for any distribution vector \(\vec{x}_{0} .\) Hint: Adapt the proof of Theorem 7.4 .1 to the complex case. e. Find \(\lim _{t \rightarrow \infty} A^{t},\) proving your conjecture from part a.

a. Consider a real \(n \times n\) matrix with \(n\) distinct real eigenvalues \(\lambda_{1}, \ldots, \lambda_{n},\) where \(\left|\lambda_{i}\right| \leq 1\) for all \(i=\) \(1, \ldots, n,\) Let \(\vec{x}(t)\) be a trajectory of the dynamical system \(\vec{x}(t+1)=A \vec{x}(t) .\) Show that this trajectory is bounded; that is, there is a positive number \(M\) such that \(\|\vec{x}(t)\| \leq M\) for all positive integers \(t\) b. Are all trajectories of the dynamical system \\[ \vec{x}(t+1)=\left[\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right] \vec{x}(t) \\] bounded? Explain.

Consider the dynamical system $$\vec{x}(t+1)=\left[\begin{array}{cc} 1.1 & 0 \\ 0 & \lambda \end{array}\right] \vec{x}(t)$$ Sketch a phase portrait of this system for the given values of \(\lambda:\) $$\lambda=1.2$$

For the matrices \(A\) listed in Exercises 13 through 17 find an invertible matrix \(S\) such that \(S^{-1} A S=\left[\begin{array}{rr}a & -b \\ b & a\end{array}\right]\) where a and b are real numbers. $$\left[\begin{array}{rr} 3 & 1 \\ -2 & 5 \end{array}\right]$$

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