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Use hand calculations to find the characteristic polynomial and eigenvalues for each of the matrices. \(A=\left(\begin{array}{rr}5 & 3 \\ -6 & -4\end{array}\right)\)

Short Answer

Expert verified
Eigenvalues are \( 2 \) and \( -1 \).

Step by step solution

01

Set Up the Characteristic Equation

The characteristic polynomial of a matrix is determined by the determinant of the matrix subtracted by \( \lambda I \), where \( I \) is the identity matrix. Let \( A = \begin{pmatrix} 5 & 3 \ -6 & -4 \end{pmatrix} \) and \( I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \). The matrix \( A - \lambda I \) is \( \begin{pmatrix} 5-\lambda & 3 \ -6 & -4-\lambda \end{pmatrix} \).
02

Calculate the Determinant of \( A - \lambda I \)

The determinant of a 2x2 matrix \( \begin{pmatrix} a & b \ c & d \end{pmatrix} \) is \( ad - bc \). Here, \( a = 5-\lambda \), \( b = 3 \), \( c = -6 \), and \( d = -4-\lambda \). So, the determinant is: \[(5-\lambda)(-4-\lambda) - (3)(-6) = \lambda^2 - \lambda - 2.\]
03

Find the Characteristic Polynomial

The characteristic polynomial is the expression we obtained from the determinant: \[\lambda^2 - \lambda - 2.\]
04

Solve the Characteristic Polynomial to Find Eigenvalues

Solve the quadratic equation \( \lambda^2 - \lambda - 2 = 0 \) using the factorization method. Factoring the equation, we get:\[(\lambda - 2)(\lambda + 1) = 0.\]The solutions to this equation are the eigenvalues, given by:\( \lambda = 2 \) and \( \lambda = -1 \).

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Key Concepts

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

Eigenvalues
Eigenvalues are essential in understanding the properties of matrices. They are special numbers associated with a square matrix that reflect how the matrix acts on space. When you apply a matrix to a vector, it often stretches, shrinks, or reverses directions. An eigenvector doesn't change direction when transformed, and its associated eigenvalue tells us how much it stretches or shrinks. In this context, finding eigenvalues involves solving the characteristic polynomial of a matrix. The roots of this polynomial are the eigenvalues.

For our specific matrix, these eigenvalues were found to be \( \lambda = 2 \) and \( \lambda = -1 \). This means that when the matrix is applied to its eigenvectors, one gets multiplied by 2 and the other by -1.

These numbers help in many applications, from stability analysis in systems to principal component analysis in statistics. Understanding them is key to harnessing the full power of linear algebra.
Matrix Determinant
The determinant is a scalar value that provides important information about a matrix. For any square matrix, like our 2x2 example, the determinant gives us insights into the matrix's properties and its invertibility.

To calculate the determinant for a 2x2 matrix \( \begin{pmatrix} a & b \ c & d \end{pmatrix} \), use the formula \( ad - bc \). This simple multiplication and subtraction provide a quick measure of certain matrix features. When the determinant is zero, the matrix is not invertible, meaning it doesn't have an inverse.
  • The determinant of \( A - \lambda I \), derived while finding the eigenvalues, was \( \lambda^2 - \lambda - 2 \).
  • This expression helped us set up the characteristic equation, which ultimately leads to solving for eigenvalues.
Understanding determinants is vital in solving problems involving matrix equations and transformations.
Quadratic Equation
Quadratic equations often appear in solving characteristic polynomials in linear algebra. These equations are typically in the form \( ax^2 + bx + c = 0 \). They arise naturally in the process of finding eigenvalues of matrices.

In step 4, the characteristic polynomial \( \lambda^2 - \lambda - 2 \) was solved as a quadratic equation. Factoring it, we found \((\lambda - 2)(\lambda + 1) = 0\). Solving these factors gives \( \lambda = 2 \) and \( \lambda = -1 \), the eigenvalues.
  • The solutions to these factored expressions are valuable as they confirm the behavior of the matrix in its transformation space.
  • Quadratic equations provide a systematic approach to finding critical numbers in many areas of mathematics and applied sciences.
Mastering them involves recognizing patterns and efficiently factorizing them, paving the way to deeper insights into matrix behavior.
Linear Algebra
Linear algebra is a branch of mathematics focused on vectors, vector spaces, and linear mappings. It is foundational in exploring and understanding the interactions between linear equations and matrices.

In this exercise, many linear algebra principles blend together, helping unravel the story a matrix tells through its eigenvalues.
  • Key linear algebra concepts like **characteristic polynomial**, **eigenvectors**, and **matrix operations** are at play.
  • It emphasizes understanding spatial interpretations of algebraic equations, making it hugely significant in fields like computer science, physics, and economics.
Recognizing patterns, solving for eigenvalues, and applying matrix operations, all fall under linear algebra's umbrella.

With linear algebra, one can not only solve complex equations involving matrices but also visualize transformations in high-dimensional spaces, offering a holistic view of data and systems.

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Most popular questions from this chapter

Suppose that \(A\) is a real \(2 \times 2\) matrix with one eigenvalue \(\lambda\) of multiplicity two. Show that the solution to the initial value problem \(\mathbf{y}^{\prime}=A \mathbf{y}\) with \(\mathbf{y}(0)=\mathbf{v}\) is given by $$ \mathbf{y}(t)=e^{\lambda t}[\mathbf{v}+t(A-\lambda I) \mathbf{v}] . $$

Consider the system $$ \mathbf{y}^{\prime}=\left(\begin{array}{rrr} 1 & -1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{array}\right) \mathbf{y} . $$ (a) Find the eigenvalues and eigenvectors. Use your numerical solver to sketch the half-lines generated by the exponential solutions \(\mathbf{y}(t)=C_{i} e^{\lambda_{i} t} \mathbf{v}_{i}\) for \(i=1,2\), and 3 . (b) The six half-line solutions found in part (a) come in pairs that form straight lines. These three lines, taken two at a time, generate three planes. In turn these planes divide phase space into eight octants. Use your numerical solver to add solution trajectories with initial conditions in each of the eight octants. (c) Based on your phase portrait, what would be an appropriate name for the equilibrium point in this case?

For the \(2 \times 2\) matrices, use \(p(\lambda)=\) \(\lambda^{2}-T \lambda+D\), where \(T=\operatorname{tr}(A)\) and \(D=\operatorname{det}(A)\), to compute the characteristic polynomial. Then, use \(p(\lambda)=\operatorname{det}(A-\lambda I)\) to calculate the characteristic polynomial a second time and compare the results. \(A=\left(\begin{array}{rr}-10 & -25 \\ 5 & 10\end{array}\right)\)

Find the general solution of the system \(\mathbf{y}^{\prime}=A \mathbf{y}\) for the given matrix \(A\). \(A=\left(\begin{array}{rr}-4 & -5 \\ 2 & 2\end{array}\right)\)

Provides a general solution of \(\mathbf{y}^{\prime}=A \mathbf{y}\), for some \(A\). Without the help of a computer or a calculator, sketch the half-line solutions generated by each exponential term of the solution. Then, sketch a rough approximation of a solution in each region determined by the half-line solutions. Use arrows to indicate the direction of motion on all solutions. Classify the equilibrium point as a saddle, a nodal sink, or a nodal source. \(\mathbf{y}(t)=C_{1} e^{-t}\left(\begin{array}{l}2 \\\ 1\end{array}\right)+C_{2} e^{-2 t}\left(\begin{array}{r}-1 \\\ 1\end{array}\right)\)

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