Chapter 8: Problem 7
Given \(\mathbf{A}\) in a deformation \(\mathbf{y}=\mathbf{A x},\) find the principal directions and corresponding factors of extension or contraction, Show the details. $$\left[\begin{array}{ll} 3 & 5 \\ 5 & 3 \end{array}\right]$$
Short Answer
Expert verified
Principal directions: \( [1,1] \) and \( [1,-1] \); factors: 8 and -2.
Step by step solution
01
Define the matrix A
The given matrix \( \mathbf{A} \) is \( \begin{bmatrix} 3 & 5 \ 5 & 3 \end{bmatrix} \). This matrix will be analyzed to find the principal directions and their corresponding factors of extension or contraction.
02
Find the eigenvalues of matrix A
To find the eigenvalues, we solve the characteristic equation \( \text{det}( \mathbf{A} - \lambda \mathbf{I} ) = 0 \). For matrix \( \mathbf{A} = \begin{bmatrix} 3 & 5 \ 5 & 3 \end{bmatrix} \), the characteristic equation becomes: \[\text{det}\left(\begin{bmatrix} 3 & 5 \ 5 & 3 \end{bmatrix} - \lambda\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\right) = \text{det}\begin{bmatrix} 3 - \lambda & 5 \ 5 & 3 - \lambda \end{bmatrix} = 0\]Calculating the determinant gives us: \((3-\lambda)(3-\lambda) - 25 = \lambda^2 - 6\lambda - 16 = 0\).
03
Solve the characteristic equation
The characteristic equation is \( \lambda^2 - 6\lambda - 16 = 0 \). Solving this quadratic equation using the quadratic formula: \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -6 \), \( c = -16 \), we find: \[\lambda = \frac{6 \pm \sqrt{36 + 64}}{2} = \frac{6 \pm \sqrt{100}}{2} = \frac{6 \pm 10}{2}\]. Thus, the eigenvalues are \( \lambda_1 = 8 \) and \( \lambda_2 = -2 \).
04
Find the eigenvectors for each eigenvalue
For \( \lambda_1 = 8 \), we solve \( \begin{bmatrix} 3 & 5 \ 5 & 3 \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \end{bmatrix} = 8\begin{bmatrix} x_1 \ x_2 \end{bmatrix} \) leading to the system: \[\begin{align*}3x_1 + 5x_2 &= 8x_1 \5x_1 + 3x_2 &= 8x_2\end{align*}\], reducing to: \[\begin{align*}-5x_1 + 5x_2 &= 0 \5x_1 - 5x_2 &= 0\end{align*}\]. Simplifying gives \( x_1 = x_2 \), so the eigenvector is a scalar multiple of \( \begin{bmatrix} 1 \ 1 \end{bmatrix} \). Similarly, for \( \lambda_2 = -2 \), solving yields \(-3x_1 + 5x_2 = 5x_1 - 3x_2\), resulting in \( x_1 = -x_2 \), thus the eigenvector is a scalar multiple of \( \begin{bmatrix} 1 \ -1 \end{bmatrix} \).
05
Interpret the results
The principal directions of the matrix \( \mathbf{A} \) are given by the eigenvectors \( \begin{bmatrix} 1 \ 1 \end{bmatrix} \) and \( \begin{bmatrix} 1 \ -1 \end{bmatrix} \). Corresponding factors of extension or contraction are the eigenvalues \( 8 \) (extension) and \( -2 \) (contraction).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Principal Directions
Principal directions are crucial in understanding how a matrix transformation affects vectors. These directions are determined by the eigenvectors of a matrix. An eigenvector points in a principal direction, describing the direction in which the transformation interprets as a pure scaling, without any rotation or distortion.
When we analyze the matrix \( \mathbf{A} \), its principal directions are captured by its eigenvectors. For our matrix \( \mathbf{A} = \begin{bmatrix} 3 & 5 \ 5 & 3 \end{bmatrix} \), the eigenvectors are \( \begin{bmatrix} 1 \ 1 \end{bmatrix} \) and \( \begin{bmatrix} 1 \ -1 \end{bmatrix} \). Each of these describes a direction through the plane.
The principal direction \( \begin{bmatrix} 1 \ 1 \end{bmatrix} \) means that any vector aligned with this direction gets stretched by the factor of the corresponding eigenvalue \( 8 \), an extension. Conversely, the direction \( \begin{bmatrix} 1 \ -1 \end{bmatrix} \) leads to contraction along its vector by the factor \( -2 \). Understanding these principal directions helps predict how transformations behave geometrically.
When we analyze the matrix \( \mathbf{A} \), its principal directions are captured by its eigenvectors. For our matrix \( \mathbf{A} = \begin{bmatrix} 3 & 5 \ 5 & 3 \end{bmatrix} \), the eigenvectors are \( \begin{bmatrix} 1 \ 1 \end{bmatrix} \) and \( \begin{bmatrix} 1 \ -1 \end{bmatrix} \). Each of these describes a direction through the plane.
The principal direction \( \begin{bmatrix} 1 \ 1 \end{bmatrix} \) means that any vector aligned with this direction gets stretched by the factor of the corresponding eigenvalue \( 8 \), an extension. Conversely, the direction \( \begin{bmatrix} 1 \ -1 \end{bmatrix} \) leads to contraction along its vector by the factor \( -2 \). Understanding these principal directions helps predict how transformations behave geometrically.
Quadratic Equation
Quadratic equations often arise in the context of finding eigenvalues. The characteristics of a quadratic equation are given by the standard form \( ax^2 + bx + c = 0 \). For a 2x2 matrix like \( \mathbf{A} \), the characteristic equation is a quadratic equation resulting from the determinant condition \( \text{det}(\mathbf{A} - \lambda \mathbf{I}) = 0 \), where \( \lambda \) represents the eigenvalues.
In our example, the characteristic equation \( \lambda^2 - 6\lambda - 16 = 0 \) emerges.
Solving this involves applying the quadratic formula:
In our example, the characteristic equation \( \lambda^2 - 6\lambda - 16 = 0 \) emerges.
Solving this involves applying the quadratic formula:
- \( a = 1 \)
- \( b = -6 \)
- \( c = -16 \)
Matrix Multiplication
Matrix multiplication is an essential operation in linear algebra that involves combining rows and columns of two matrices to produce a new matrix. This operation is especially important when calculating eigenvectors for a given eigenvalue.
To find the eigenvectors for a matrix \( \mathbf{A} \) like \( \begin{bmatrix} 3 & 5 \ 5 & 3 \end{bmatrix} \), it requires solving \( (\mathbf{A} - \lambda \mathbf{I}) \mathbf{v} = 0 \), where \( \mathbf{v} \) represents the eigenvector and \( \lambda \) is the eigenvalue.
Given an eigenvalue from our earlier quadratic equation, we plug it into the characteristic matrix \( (\mathbf{A} - \lambda \mathbf{I}) \) and proceed with the multiplication.
For example, with eigenvalue \( \lambda_1 = 8 \), we calculate:
To find the eigenvectors for a matrix \( \mathbf{A} \) like \( \begin{bmatrix} 3 & 5 \ 5 & 3 \end{bmatrix} \), it requires solving \( (\mathbf{A} - \lambda \mathbf{I}) \mathbf{v} = 0 \), where \( \mathbf{v} \) represents the eigenvector and \( \lambda \) is the eigenvalue.
Given an eigenvalue from our earlier quadratic equation, we plug it into the characteristic matrix \( (\mathbf{A} - \lambda \mathbf{I}) \) and proceed with the multiplication.
For example, with eigenvalue \( \lambda_1 = 8 \), we calculate:
- \( (\mathbf{A} - 8\mathbf{I}) = \begin{bmatrix} 3-8 & 5 \ 5 & 3-8 \end{bmatrix} = \begin{bmatrix} -5 & 5 \ 5 & -5 \end{bmatrix} \).
Characteristic Equation
The characteristic equation is pivotal for determining the eigenvalues of a matrix. Derived from the equation \( \text{det}(\mathbf{A} - \lambda \mathbf{I}) = 0 \), it equates the determinant of the matrix subtraction \( \mathbf{A} - \lambda \mathbf{I} \) to zero. This gives us a polynomial, typically quadratic for a 2x2 matrix.
For the given matrix \( \mathbf{A} = \begin{bmatrix} 3 & 5 \ 5 & 3 \end{bmatrix} \), we find the characteristic polynomial by computing:
For the given matrix \( \mathbf{A} = \begin{bmatrix} 3 & 5 \ 5 & 3 \end{bmatrix} \), we find the characteristic polynomial by computing:
- \( \text{det}\begin{bmatrix} 3-\lambda & 5 \ 5 & 3-\lambda \end{bmatrix} = (3-\lambda)^2 - 25 = 0 \)