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If a vector \(\vec{v}\) is an eigenvector of both \(A\) and \(B,\) is \(\vec{v}\) necessarily an eigenvector of \(A B ?\)

Short Answer

Expert verified
Yes, if \(\vec{v}\) is an eigenvector of both A and B, it is necessarily an eigenvector of \(AB\) as well, with the corresponding eigenvalue being the product of the eigenvalues for \(\vec{v}\) for A and B, \(\lambda_A\lambda_B\).

Step by step solution

01

Understanding Eigenvector Definition

An eigenvector of a matrix is a non-zero vector that only changes by a scalar factor when that matrix is applied to it. Mathematically, for a matrix M and a vector \(\vec{v}\), if \(\vec{v}\) is an eigenvector of M, then there exists a scalar \(\lambda\) such that \(M\vec{v} = \lambda\vec{v}\).
02

Analyzing the Eigenvector of Matrix A

Given that \(\vec{v}\) is an eigenvector of matrix A, there exists a scalar \(\lambda_A\) such that \(A\vec{v} = \lambda_A\vec{v}\).
03

Analyzing the Eigenvector of Matrix B

Similarly, since \(\vec{v}\) is also an eigenvector of matrix B, there is a scalar \(\lambda_B\) such that \(B\vec{v} = \lambda_B\vec{v}\).
04

Analyzing the Product AB Acting on \(\vec{v}\)

When the product of matrices A and B (in that order) acts on \(\vec{v}\), we find \(AB\vec{v} = A(B\vec{v}) = A(\lambda_B\vec{v}) = \lambda_B(A\vec{v}) = \lambda_B(\lambda_A\vec{v}) = (\lambda_A\lambda_B)\vec{v}\). Here, \(\lambda_A\lambda_B\) is a scalar, which means \(\vec{v}\) is also an eigenvector of the matrix product AB with the corresponding eigenvalue \(\lambda_A\lambda_B\).

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Key Concepts

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

Eigenvector Definition
In linear algebra, an eigenvector is a special vector associated with a matrix that reveals much about the matrix's properties. The defining characteristic of an eigenvector \(\vec{v}\), is that when it is transformed by a matrix \(M\), it merely scales by a particular factor and does not change its direction. This factor is known as an eigenvalue, denoted by \(\lambda\). The formal definition states: if \(\vec{v}\) is an eigenvector of the matrix \(M\), then \(M\vec{v} = \(\lambda\)\vec{v}\). This equation expresses the fundamental relationship between a matrix and its eigenvectors and eigenvalues. It is worth noting that eigenvectors are never the zero vector, since the zero vector would not fulfill the requirement of being scaled by \(M\).An easy way to remember the concept is to think of an eigenvector as a 'direction' that does not get mixed with others when a matrix transformation is applied. Imagine stretching or compressing an elastic band in one direction but not changing its overall orientation; that's akin to what a matrix does to its eigenvectors.
Matrix Multiplication and Eigenvectors
When dealing with multiple matrices, it is important to understand how matrix multiplication can affect eigenvectors. Typically, when we multiply a vector by a matrix, we can get a new vector with a different direction and magnitude. However, with eigenvectors, multiplication by their corresponding matrix only changes the vector's scale, not its direction. But what about the product of two matrices? If we are given matrices \(A\) and \(B\), and a vector \(\vec{v}\) is an eigenvector of both, things become interesting.

Intuitive Understanding of Matrix Products and Eigenvectors

When \(A\) and \(B\) both individually act on \(\vec{v}\), they scale it by their respective eigenvalues, say \(\lambda_A\) and \(\lambda_B\). When considering the product \(AB\), one can visualize this as first applying \(B\) to \(\vec{v}\), scaling it by \(\lambda_B\), and then applying \(A\) to the result, scaling it further by \(\lambda_A\). So, the action of \(AB\) on the eigenvector \(\vec{v}\) results in multiplying \(\vec{v}\) by the product of the eigenvalues, that is, \(\lambda_A\lambda_B\). This new scalar, \(\lambda_A\lambda_B\), can be considered an eigenvalue associated with the eigenvector \(\vec{v}\) with respect to the matrix product \(AB\).
Eigenvalue
An eigenvalue is a scalar that represents how much an eigenvector is stretched or squashed by a matrix transformation. It's the 'eigen' part of the term 'eigenvector' - coming from the German word for 'own' or 'characteristic'. So, when we speak of an eigenvalue, we are discussing the characteristic factor that allows an eigenvector to remain on its own path under the transformation.The concept of eigenvalues is central not only to understanding matrix transformations but also to various applications across physics and engineering. For example, in vibrations analysis, eigenvalues can represent natural frequencies, and in population studies, they can help calculate long-term growth rates.

Finding and Using Eigenvalues

To find the eigenvalues of a matrix, we solve the characteristic polynomial obtained from the equation \(|M - \(\lambda\)I| = 0\), where \(I\) is the identity matrix, and \(|\cdot|\) represents the determinant. This equation can sometimes be complex and usually requires more advanced mathematical techniques to solve. But once the eigenvalues are known, they are quite powerful. They can help us understand the behavior of systems modeled by matrices and play an essential role in procedures such as diagonalization and the determination of matrix stability.

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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 \(\vec{x}(t+1)=A \vec{x}(t)\). $$A=\left[\begin{array}{rr} 0.8 & 0.7 \\ -0.7 & 0.8 \end{array}\right]$$

Consider the matrices A in Exercises 11 through \(16 .\) For which real numbers \(k\) is the zero state a stable equilibrium of the dynamical system \(\vec{x}(t+1)=A \vec{x}(t) ?\) $$A=\left[\begin{array}{rr} 0.7 & k \\ 0 & -0.9 \end{array}\right]$$

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. $$\left[\begin{array}{rr} 3 & 1 \\ -2 & 5 \end{array}\right]$$

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.

For the matrices \(A\) in Exercises 17 through \(24,\) find real closed formulas for the trajectory \(\vec{x}(t+1)=A \vec{x}(t)\) where \(\vec{x}(0)=\left[\begin{array}{l}0 \\ 1\end{array}\right] .\) Draw a rough sketch. $$A=\left[\begin{array}{rr} 0.6 & -0.8 \\ 0.8 & 0.6 \end{array}\right]$$

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