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Chapter 4: Problem 1

In Exercises \(I-6,\) show that \(\mathbf{v}\) is an eigenvector of \(A\) and find the corresponding eigenvalue.. $$A=\left[\begin{array}{ll} 0 & 3 \\ 3 & 0 \end{array}\right], \mathbf{v}=\left[\begin{array}{l} 1 \\ 1 \end{array}\right]$$

Short Answer

Expert verified
Yes, \(\mathbf{v}\) is an eigenvector of \(A\) with eigenvalue \(\lambda = 3\).

Step by step solution

01

Understand what it means to be an eigenvector

An eigenvector of a matrix \(A\) is a nonzero vector \(\mathbf{v}\) such that when \(A\) is multiplied by \(\mathbf{v}\), the result is a scalar multiple of \(\mathbf{v}\). This means \(A \mathbf{v} = \lambda \mathbf{v}\), where \(\lambda\) is a scalar known as the eigenvalue.
02

Multiply matrix A by vector v

Compute the product of matrix \(A\) and vector \(\mathbf{v}\): \[A \mathbf{v} = \begin{bmatrix} 0 & 3 \ 3 & 0 \end{bmatrix} \begin{bmatrix} 1 \ 1 \end{bmatrix}\]This results in the vector:\[A \mathbf{v} = \begin{bmatrix} 0 \cdot 1 + 3 \cdot 1 \ 3 \cdot 1 + 0 \cdot 1 \end{bmatrix} = \begin{bmatrix} 3 \ 3 \end{bmatrix}\]
03

Check if the product is a scalar multiple of v

The result from Step 2 is \(\begin{bmatrix} 3 \ 3 \end{bmatrix}\). Check if this vector is a scalar multiple of the original vector \(\mathbf{v} = \begin{bmatrix} 1 \ 1 \end{bmatrix}\). Notice:\[\begin{bmatrix} 3 \ 3 \end{bmatrix} = 3 \begin{bmatrix} 1 \ 1 \end{bmatrix}\]This shows \(A \mathbf{v} = 3 \mathbf{v}\).
04

Conclude the eigenvalue

Since \(A \mathbf{v} = 3 \mathbf{v}\), the vector \(\mathbf{v}\) is an eigenvector of \(A\) with the eigenvalue \(\lambda = 3\). The calculation confirms that multiplying \(A\) by \(\mathbf{v}\) yields a vector that is indeed a scalar multiple (scalar is 3) of \(\mathbf{v}\).

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Key Concepts

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

Eigenvalues
An eigenvalue is a special number associated with a square matrix and its corresponding eigenvector. It appears in the context of linear transformations, where the effect of the transformation stretches or shrinks the vector by some factor. This factor is the eigenvalue.
  • For a matrix \( A \), an eigenvalue \( \lambda \) is a scalar such that \( A \mathbf{v} = \lambda \mathbf{v} \), where \( \mathbf{v} \) is a non-zero vector called the eigenvector.
  • The equation \( A \mathbf{v} = \lambda \mathbf{v} \) signifies that applying the matrix transformation to \( \mathbf{v} \) results in a vector in the same direction, just scaled by \( \lambda \).
Understanding eigenvalues is crucial in various fields:
  • They help in simplifying matrix operations, particularly when raising matrices to powers.
  • They're used in systems of differential equations, stability analysis, and even quantum mechanics.
To find an eigenvalue, typically, you must solve the characteristic equation \( \det(A - \lambda I) = 0 \). Here, \( I \) is the identity matrix of the same size as \( A \). In our exercise, we found the eigenvalue to be \( 3 \).
Matrix Multiplication
To understand eigenvectors and eigenvalues, grasping matrix multiplication is vital. Matrix multiplication involves multiplying rows by columns.
  • The product of an \( m \times n \) matrix \( A \) with an \( n \times p \) matrix \( B \) results in an \( m \times p \) matrix \( C \).
  • Element \( c_{ij} \) of the resulting matrix is computed by summing the products of the elements from the \( i \)-th row of \( A \) and the \( j \)-th column of \( B \).
For our specific example:
  • The matrix \( A = \begin{bmatrix} 0 & 3 \ 3 & 0 \end{bmatrix} \) was multiplied by the vector \( \mathbf{v} = \begin{bmatrix} 1 \ 1 \end{bmatrix} \).
  • This resulted in \( A \mathbf{v} = \begin{bmatrix} 3 \ 3 \end{bmatrix} \), showing how matrix multiplication applies to transforming vectors.
Remember:- The multiplication is not necessarily commutative, meaning \( AB eq BA \).
- The operation is crucial in transforming and analyzing vector spaces.
Linear Transformations
In mathematics, linear transformations are represented by matrices. They depict how vectors change when subjected to transformations like rotations, scaling, or reflections.
  • Linear transformations preserve vector addition and scalar multiplication, meaning the form a straight line or flat plane.
  • For a given transformation \( f: \mathbb{R}^n \rightarrow \mathbb{R}^m \), if \( f(\mathbf{u} + \mathbf{v}) = f(\mathbf{u}) + f(\mathbf{v}) \) and \( f(c \mathbf{u}) = c f(\mathbf{u}) \), it's linear.
When dealing with a matrix \( A \):
  • Each column of matrix \( A \) acts as a basis vector for transforming \( \mathbf{v} \).
  • For eigenvectors, this specific transformation is just a scaling by the eigenvalue from the original direction.
In our exercise, the matrix \( A \) transforms the vector \( \mathbf{v} \) into a new vector simply scaled by \( 3 \), showing a precise linear transformation representing stretching.
Scalar Multiplication
Scalar multiplication is a fundamental operation in linear algebra. It involves multiplying a vector by a scalar \( \lambda \), resulting in a new vector.
  • For a vector \( \mathbf{v} = \begin{bmatrix} v_1 \ v_2 \ \vdots \end{bmatrix} \) and scalar \( \lambda \), the product is \( \lambda \mathbf{v} = \begin{bmatrix} \lambda v_1 \ \lambda v_2 \ \vdots \end{bmatrix} \).
  • This operation scales the magnitude of the vector without changing its direction if \( \lambda > 0 \).
Within our problem, after multiplying the matrix by the vector, we get \( A \mathbf{v} = 3 \mathbf{v} \), illustrating scalar multiplication at work.
  • This demonstrates that the new vector \( A \mathbf{v} \) is purely a scaled version of the original vector \( \mathbf{v} \).
  • It confirms \( \mathbf{v} \) as an eigenvector with \( \lambda = 3 \) as the eigenvalue, showcasing how scalar multiplication crucially connects with eigenvectors and values.
Scalar multiplication helps in understanding how transformations affect vectors in terms of stretching or shrinking them.

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Most popular questions from this chapter

Species \(X\) preys on species \(Y\). The sizes of the populations are represented by \(x=x(t)\) and \(y=y(t)\) The growth rate of each population is governed by the system of differential equations \(\mathbf{x}^{\prime}=A \mathbf{x}+\mathbf{b},\) where \(\mathbf{x}=\left[\begin{array}{l}x \\\ y\end{array}\right]\) and \(\mathbf{b}\) is a constant vector. Determine what happens to the two populations for the given A and b and initial conditions \(\mathbf{x}(0)\). (First show that there are constants a and \(b\) such that the substitutions \(x=u+a\) and \(y=v+b\) convert the system into an equivalent one with no constant terms.) $$A=\left[\begin{array}{rr} 1 & 1 \\ -1 & 1 \end{array}\right], \mathbf{b}=\left[\begin{array}{r} -30 \\ -10 \end{array}\right], \mathbf{x}(0)=\left[\begin{array}{l} 20 \\ 30 \end{array}\right]$$

Let \(A\) be an invertible matrix. Prove that if \(A\) is diagonalizable, so is \(A^{-1}\).

It can be shown that a nonnegative \(n \times n\) matrix is irreducible if and only if \((I+A)^{n-1}>0 .\) Use this criterion to determine whether the matrix \(A\) is irreducible. If \(A\) is reducible, find a permutation of its rows and columns that puts \(A\) into the block form \\[ \left[\begin{array}{ll} B & C \\ O & D \end{array}\right] \\] $$A=\left[\begin{array}{lllll} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 0 & 0 \end{array}\right]$$

Prove that if \(A\) is diagonalizable, so is \(A^{T}\)

Find the general solution to the given system of differential equations. Then find the specific solution that satisfies the initial conditions. (Consider all functions to be functions of \(t .\) $$\begin{array}{l} x^{\prime}=2 x-y, \quad x(0)=1 \\ y^{\prime}=-x+2 y, \quad y(0)=1 \end{array}$$
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