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Chapter 4: Problem 4

Show that \(A\) and \(B\) are not similar matrices. \(A=\left[\begin{array}{rrr}1 & 2 & 0 \\ 0 & 1 & -1 \\ 0 & -1 & 1\end{array}\right], B=\left[\begin{array}{lll}2 & 1 & 1 \\ 0 & 1 & 0 \\ 2 & 0 & 1\end{array}\right]\)

Short Answer

Expert verified
Matrices \( A \) and \( B \) are not similar as they have different eigenvalues.

Step by step solution

01

Understand Similar Matrices

Two matrices, \( A \) and \( B \), are similar if there exists an invertible matrix \( P \) such that \( P^{-1}AP = B \). Part of checking similarity involves comparing their eigenvalues as similar matrices have the same eigenvalues.
02

Determine Eigenvalues of Matrix A

The eigenvalues of a matrix can be found by solving the characteristic polynomial obtained from \( \det(A - \lambda I) = 0 \), where \( I \) is the identity matrix. For \( A \), this results in a cubic equation. Upon solving, we find the eigenvalues of \( A \) to be \( \lambda_1 = 1, \lambda_2 = 1, \lambda_3 = 1 \).
03

Determine Eigenvalues of Matrix B

Similarly, find the eigenvalues for matrix \( B \) by calculating \( \det(B - \lambda I) = 0 \). After solving the characteristic polynomial for \( B \), we find that the eigenvalues are \( \lambda_1 = 2, \lambda_2 = 1, \lambda_3 = 1 \).
04

Compare Eigenvalues of A and B

Matrix \( A \) has eigenvalues \( 1, 1, 1 \), while matrix \( B \) has eigenvalues \( 2, 1, 1 \). Since they do not have the same eigenvalues, \( A \) and \( B \) are not similar.

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Key Concepts

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

Eigenvalues
Eigenvalues are special numbers associated with a matrix that provide insight into its characteristics. To determine the eigenvalues of a matrix, you subtract the eigenvalue times the identity matrix from the original matrix and set the determinant equal to zero. This process yields a characteristic equation. Solving this equation鈥攐ften a polynomial鈥攑rovides the eigenvalues of the matrix.
  • They are solutions to the equation \( \det(A - \lambda I) = 0 \).
  • Eigenvalues indicate how a matrix acts on vectors; for example, they can show if the matrix stretches or compresses the vectors.
In our task, finding the eigenvalues is critical because similar matrices should have identical eigenvalues. This step helps in determining matrix similarity and whether two matrices can be transformed into one another by a conjugation with an invertible matrix.
Characteristic Polynomial
The characteristic polynomial is a polynomial that is derived from a square matrix and plays a central role in finding eigenvalues. For a given matrix \( A \), the characteristic polynomial is obtained by calculating \( \det(A - \lambda I) \), where \( I \) is the identity matrix of the same size as \( A \) and \( \lambda \) represents an eigenvalue.
  • A cubic characteristic polynomial will have up to three roots, which correspond to the eigenvalues of the matrix.
  • This polynomial summarizes the essential features of the matrix, such as its eigenvalues and its trace (sum of eigenvalues).
Solving the characteristic polynomial is essential for understanding the intrinsic properties of the matrix. These properties include stability in systems of linear equations and determining whether matrices are similar. In the given solution, solving the characteristic polynomial enabled us to find the eigenvalues, which are key for testing matrix similarity.
Invertible Matrix
An invertible matrix or non-singular matrix is a square matrix that has an inverse. The inverse of a matrix \( A \) is a matrix \( B \) such that when \( A \) is multiplied by \( B \), the result is the identity matrix. In terms of matrix similarity, an invertible matrix \( P \) must satisfy the equation \( P^{-1}AP = B \) for matrices \( A \) and \( B \) to be similar.
  • A matrix is invertible if and only if its determinant is non-zero.
  • Invertibility is vital for operations like solving systems of equations, finding matrix similarity, and performing matrix transformations.
  • Only invertible matrices can be used to test for similarity, meaning they can transform matrix \( A \) into matrix \( B \) by conjugation.
Understanding these concepts ensures that you grasp the role of invertibility in matrix operations and proofs of similarity.
Determinant
The determinant is a scalar value that is computed from a square matrix. It provides important properties and insight into the behavior of the matrix. The determinant helps identify whether a matrix is invertible and is closely linked to the process of finding eigenvalues.
  • The formula for a 3x3 matrix \( A \) is \( \det(A) = a(ei鈭抐h)鈭抌(di鈭抐g)+c(dh鈭抏g) \), using elements of the matrix across the first row.
  • If the determinant of a matrix is zero, the matrix is singular, or non-invertible.
  • The determinant is crucial in calculating the characteristic polynomial and ensuring matrices in linear transformations don't collapse dimensions.
When solving for similarity, the determinant provides initial insights: if the determinant of \( A - \lambda I \) is zero for an eigenvalue \( \lambda \), that eigenvalue is a valid solution. This ultimately affects the determination of matrix similarity, focusing on non-zero determinants for invertibility and transformations.

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Most popular questions from this chapter

Calculate the positive eigenvalue and \(a\) corresponding positive eigenvector of the Leslie matrix \(L\). $$L=\left[\begin{array}{ll} 1 & 1.5 \\ 0.5 & 0 \end{array}\right]$$

Prove that if \(A\) is a diagonalizable matrix such that every eigenvalue of \(A\) is either 0 or 1 , then \(A\) is idempotent (that is, \(A^{2}=A\) ).

(a) Use mathematical induction to prove that, for \(n \geq 2,\) the companion matrix \(C(p)\) of \(p(x)=x^{n}+\) \(a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}\) has characteristic polynomial \((-1)^{n} p(\lambda) .\) [Hint: Expand by cofactors along the last column. You may find it helpful to introduce the polynomial \(q(x)=\left(p(x)-a_{0}\right) / x .\) (b) Show that if \(\lambda\) is an eigenvalue of the companion matrix \(C(p)\) in Equation \((4),\) then an eigenvector corresponding to \(\lambda\) is given by $$\left[\begin{array}{c} \lambda^{n-1} \\ \lambda^{n-2} \\ \vdots \\ \lambda \\ 1 \end{array}\right]$$ If \(p(x)=x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}\) and \(A\) is a square matrix, we can define a square matrix \(p(A) b y\) $$p(A)=A^{n}+a_{n-1} A^{n-1}+\cdots+a_{1} A+a_{0} I$$ An important theorem in advanced linear algebra says that if \(c_{A}(\lambda)\) is the characteristic polynomial of the matrix \(A\), then \(c_{A}(A)=O\) (in words, every matrix satisfies its characteristic equation). This is the celebrated Cayley-Hamilton Theorem, named after Arthur Cayley \((1821-1895)\) pictured below, and sir William Rowan Hamilton (see page 2 ). Cayley proved this theorem in \(1858 .\) Hamilton discovered it, independently, in his work on quaternions, a generalization of the complex numbers.

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 \(\lambda^{2}+a \lambda+b=0\)

Solve the recurrence relation with the given initial conditions. $$y_{1}=1, y_{2}=6, y_{n}=4 y_{n-1}-4 y_{n-2} \text { for } n \geq 3$$
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