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Suppose \(\vec{v}\) is an eigenvector of the \(n \times n\) matrix \(A\) with eigenvalue \(4 .\) Explain why \(\vec{v}\) is an eigenvector of \(A^{2}+2 A+3 I_{n} .\) What is the associated eigenvalue?

Short Answer

Expert verified
\(\vec{v}\) is an eigenvector of \(A^2 + 2A + 3I_n\) with eigenvalue 27.

Step by step solution

01

Review Eigenvector and Eigenvalue Definitions

Recall the definition of an eigenvector \(\vec{v}\) and eigenvalue \(\lambda\) for a matrix \(A\): if \(\vec{v}\) is an eigenvector of \(A\) with eigenvalue \(\lambda\), then \(A\vec{v} = \lambda\vec{v}\). In this exercise, \(\lambda = 4\), so \(A\vec{v} = 4\vec{v}\).
02

Apply Eigenvalue Property to \(A^2\)

To show that \(\vec{v}\) is also an eigenvector of \(A^2\), calculate \(A^2\vec{v}\). Knowing \(A\vec{v} = 4\vec{v}\), we can write \(A(A\vec{v}) = A(4\vec{v}) = 4(A\vec{v}) = 4(4\vec{v}) = 16\vec{v}\). This shows \(\vec{v}\) is an eigenvector of \(A^2\) with eigenvalue \(16\).
03

Apply Eigenvalue Property to \(2A\)

Next, let's find the transformation of \(\vec{v}\) by \(2A\). Since \(A\vec{v} = 4\vec{v}\), it follows that \(2A\vec{v} = 2(4\vec{v}) = 8\vec{v}\). So, \(\vec{v}\) is still an eigenvector of \(2A\) with eigenvalue \(8\).
04

Apply Eigenvalue Property to \(3I_n\)

Finally, consider the identity matrix \(I_n\). For any \(n \times n\) matrix, \(I_n\vec{v} = \vec{v}\). Multiplying by \(3\) gives \(3I_n\vec{v} = 3\vec{v}\). Here, \(\vec{v}\) is also an eigenvector of \(3I_n\) with eigenvalue \(3\).
05

Combine the Results to Find the Eigenvector of \(A^2 + 2A + 3I_n\)

Now sum the results of each part to find the action of \(A^2 + 2A + 3I_n\) on \(\vec{v}\): \[ (A^2 + 2A + 3I_n)\vec{v} = A^2\vec{v} + 2A\vec{v} + 3I_n\vec{v} = 16\vec{v} + 8\vec{v} + 3\vec{v} = (16+8+3)\vec{v} = 27\vec{v} \.\] This demonstrates that \(\vec{v}\) is an eigenvector of \(A^2 + 2A + 3I_n\) with eigenvalue \(27\).

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Key Concepts

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

Eigenvalue of Matrix
The concept of an eigenvalue of a matrix is a fundamental one in linear algebra, offering deep insights into the matrix's characteristics and the transformations it induces.

Eigenvalues are scalars associated with a particular linear transformation represented by a matrix. When a matrix, let's denote it as \(A\), acts on a vector \(\vec{v}\), the vector is either stretched, shrunk, or unchanged, and sometimes flipped. If the vector's direction doesn't change, that vector is called an eigenvector, and the factor by which its magnitude is scaled is the corresponding eigenvalue \(\lambda\). Formally, provided \(\vec{v}\) is an eigenvector, we write this transformation as \(A\vec{v} = \lambda\vec{v}\).

This property reveals a lot about the matrix. For instance, the determinant and trace of a matrix are closely tied to its eigenvalues. If you were to calculate the determinant of \(A - \lambda I_n\), where \(I_n\) is the identity matrix, it would be equal to zero when \(\lambda\) is an eigenvalue of \(A\). Understanding eigenvalues enables more complex matrix operations such as finding the power of a matrix, which as we saw in the exercise, leverages the eigenvalues to simplify computations drastically for expressions like \(A^2 + 2A + 3I_n\).
Eigenvector Properties
Eigenvectors have several intriguing properties that are useful in a wide range of applications, from differential equations to quantum mechanics. Here are some key properties:

  • Eigenvectors corresponding to distinct eigenvalues are linearly independent. This characteristic allows for the diagonalization of matrices, where a matrix can be expressed as a product of its eigenvectors and eigenvalues in a diagonal matrix.
  • The eigenvectors of a matrix form a basis for the space if and only if the matrix has a full set of linearly independent eigenvectors. This means we can describe any vector in the space as a combination of these eigenvectors.
  • Scaling an eigenvector by a non-zero scalar results in another eigenvector associated with the same eigenvalue. It's why in the exercise, multiplying \(A\vec{v}\) by 2 gave us \(2A\vec{v} = 8\vec{v}\), reinforcing that the resultant vector is also an eigenvector with the eigenvalue scaled by the same factor.
  • Regarding eigenvalues, the sum of all eigenvalues (with multiplicity) equals the trace of the matrix, and the product of the eigenvalues equals the determinant of the matrix.

The understanding of these properties allows for powerful shorthand calculations and transformations, as experienced in the exercise with the matrix \(A^2 + 2A + 3I_n\).
Matrix Transformations
Matrix transformations represent one of the most essential concepts in linear algebra. They provide a framework for understanding how spaces and vectors within them are altered when multiplied by matrices.

A matrix can be perceived as a function that takes a vector and transforms it into another vector. Specifically, a transformation alters a vector's length (scale), direction (rotate), or both. Transformations like rotation, reflection, and shearing are represented by specific matrices. For example, rotation matrices can rotate vectors in a plane around the origin, and reflection matrices can flip vectors across a specified axis.

Understanding how matrices acts on vectors is crucial in a variety of applications. For instance, in the context of systems of linear equations, each equation can be viewed as a transformation, and the solution to the system is the vector that remains invariant under these transformations.

The exercise provided a great example of a more complex matrix transformation by combining square, scaling, and identity operations on the vector \(\vec{v}\), resulting in \(A^2\vec{v} + 2A\vec{v} + 3I_n\vec{v}\). By understanding the basic transformations performed by \(A\), \(A^2\), and \(3I_n\), we were able to determine the nature of the combined transformation and the resulting eigenvalue of 27. Through such exercises, we develop a clearer grasp of the mechanisms at work within matrix algebra and their influence on vector spaces.

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Most popular questions from this chapter

For which values of the real constant a are the matrices in Exercises 45 through 50 diagonalizable over \(\mathbb{C} ?\) $$\left[\begin{array}{ll} 1 & 1 \\ a & 1 \end{array}\right]$$

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 \(\lambda:\) $$\lambda=0.9$$

Find all the eigenvalues and "eigenvectors" of the linear transformations. \(L(A)=A+A^{T}\) from \(\mathbb{R}^{2 \times 2}\) to \(\mathbb{R}^{2 \times 2}\). Is \(L\) diagonalizable?

In his high school final examination (Aarau, Switzerland, 1896 , young Albert Einstein \((1879-1955)\) was given the following problem: In a triangle \(A B C\), let \(P\) be the center of the inscribed circle. We are told that \(\overline{A P}=1, \overline{B P}=\frac{1}{2},\) and \(\overline{C P}=\frac{1}{3} .\) Find the radius \(\rho\) of the inscribed circle. Einstein worked through this problem as follows: \\[ \begin{array}{l} \sin \left(\frac{\alpha}{2}\right)=\rho \\ \sin \left(\frac{\beta}{2}\right)=2 \rho \\ \sin \left(\frac{\gamma}{2}\right)=3 \rho \end{array} \\] For every triangle the following equation holds: \\[ \begin{array}{l} \sin ^{2}\left(\frac{\alpha}{2}\right)+\sin ^{2}\left(\frac{\beta}{2}\right)+\sin ^{2}\left(\frac{\gamma}{2}\right) \\ +2 \sin \left(\frac{\alpha}{2}\right) \sin \left(\frac{\beta}{2}\right) \sin \left(\frac{\gamma}{2}\right)=1 \end{array} \\] In our case \\[ 14 \rho^{2}+12 \rho^{3}-1=0 \\] Now let \\[ \rho=\frac{1}{x} \\]. At this point we interrupt Einstein's work and ask you to finish the job. Hint: Exercise 40 is helpful. Find the exact solution (in terms of trigonometric and inverse trigonometric functions \(),\) and give a numerical approximation as well. (By the way, Einstein, who was allowed to use a logarithm table, solved the problem correctly.) Source: The Collected Papers of Albert Einstein, Vol. 1 Princeton University Press, 1987.

If \(A\) is a matrix of rank \(1,\) show that any nonzero vector in the image of \(A\) is an eigenvector of \(A\)

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