Question

Find the eigenvalues of the given matrix. For each eigenvalue, give an eigenvector. $$\left[\begin{array}{ll}3 & 12 \\ 1 & -1\end{array}\right]$$

Short answer

Eigenvalues are 5 and -3; eigenvectors are \(\begin{bmatrix} 6 \\ 1 \end{bmatrix}\) and \(\begin{bmatrix} -2 \\ 1 \end{bmatrix}\).

Step-by-step solution

Identify the Matrix

The given matrix is a 2x2 matrix: \(A = \begin{bmatrix} 3 & 12 \ 1 & -1 \end{bmatrix} \). To find the eigenvalues and eigenvectors, we'll follow the standard procedure for solving these.

Find the Characteristic Equation

To find eigenvalues, solve the characteristic equation \(det(A - \lambda I) = 0 \), where \(I\) is the identity matrix. The matrix \(A - \lambda I\) is \(\begin{bmatrix} 3-\lambda & 12 \ 1 & -1-\lambda \end{bmatrix} \). The determinant is \(\det(\begin{bmatrix} 3-\lambda & 12 \ 1 & -1-\lambda \end{bmatrix}) = (3-\lambda)(-1-\lambda) - 12\cdot 1 = \lambda^2 - 2\lambda - 15 \). This needs to be set to zero for eigenvalue computation.

Solve the Characteristic Equation

The characteristic equation is \(\lambda^2 - 2\lambda - 15 = 0\). To solve this quadratic equation, apply the quadratic formula: \(\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \(a = 1, b = -2, c = -15\). Calculate the eigenvalues: \(\lambda = \frac{2 \pm \sqrt{4 + 60}}{2} = \frac{2 \pm 8}{2} \). Therefore, the eigenvalues are \(\lambda_1 = 5\) and \(\lambda_2 = -3\).

Find Eigenvector for \\(\lambda_1 = 5\\)

To find the eigenvector associated with \(\lambda_1 = 5\), solve the equation \((A - 5I)\mathbf{v} = 0\). The matrix \(A - 5I\) is \(\begin{bmatrix} -2 & 12 \ 1 & -6 \end{bmatrix} \). Solving \(-2x + 12y = 0\) gives \(x = 6y\). Choose \(y = 1\), then \(x = 6\). Thus, one eigenvector is \(\begin{bmatrix} 6 \ 1 \end{bmatrix} \).

Find Eigenvector for \\(\lambda_2 = -3\\)

To find the eigenvector associated with \(\lambda_2 = -3\), solve the equation \((A + 3I)\mathbf{v} = 0\). The matrix \(A + 3I\) is \(\begin{bmatrix} 6 & 12 \ 1 & 2 \end{bmatrix} \). Solving \(6x + 12y = 0\) gives \(x = -2y\). Choose \(y = 1\), then \(x = -2\). Thus, one eigenvector is \(\begin{bmatrix} -2 \ 1 \end{bmatrix} \).

Key concepts

Eigenvectors

Eigenvectors are fundamental to understanding how a matrix operates on a vector. When you multiply a matrix by one of its eigenvectors, it merely stretches or shrinks the eigenvector without changing its direction. This is incredibly useful in areas like physics and engineering, where we often want to see how a system behaves after transformation.

• An eigenvector, denoted as \(\mathbf{v}\), satisfies the equation \(A\mathbf{v} = \lambda\mathbf{v}\).
• Here, \(A\) is our matrix, and \(\lambda\) is the eigenvalue corresponding to \(\mathbf{v}\).

In the example provided, once we found the eigenvalues, we solved \((A - \lambda I)\mathbf{v} = 0\). This means we only need to work with linear equations to find these vectors.
They're typically found by setting one of the vector components free, like choosing \(y=1\) to solve for \(x\). This gives us a direction vector, not something with a fixed length. These vectors are used extensively in simplifying complex systems.

Characteristic Equation

The characteristic equation is key to finding the eigenvalues of a matrix. By definition, eigenvalues are the solutions to this equation. We derive it by taking the determinant of \(A - \lambda I\) and setting it to zero.

• \(I\) is the identity matrix of the same size as \(A\).
• For a 2x2 matrix, this involves some straightforward calculations with determinants.

By setting the determinant of \(A - \lambda I\) to zero, we get a polynomial equation in terms of \(\lambda\), usually quadratic for a 2x2 matrix.
In our specific case, solving \(\lambda^2 - 2\lambda - 15 = 0\) provided us the eigenvalues for the given matrix. These eigenvalues are critical for understanding the stretching/shrinking actions the matrix performs on its vectors.

2x2 Matrix

Mathematically and practically, a 2x2 matrix is one of the simplest kinds of matrices. Its accessibility makes it a perfect entry point into the study of linear transformations. These matrices consist of two rows and two columns, making calculations relatively manageable.

• An example of a 2x2 matrix is \(\begin{bmatrix} 3 & 12 \ 1 & -1 \end{bmatrix}\).
• From such matrices, we can determine transformations, like scaling or rotating vectors.

Two critical aspects of working with 2x2 matrices involve finding their eigenvalues and eigenvectors. Since our original exercise involved such a matrix, we used it to craft the characteristic equation, solved it using the quadratic formula, and subsequently derived the eigenvectors. These small steps simplify larger concepts, building a solid foundation for linear algebra.

Quadratic Formula

The quadratic formula is a crucial tool for extracting solutions from quadratic equations. Used extensively in math, particularly in algebra and calculus, it provides the solutions for equations of the form \(ax^2 + bx + c = 0\).

• The formula is \(\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
• It finds the roots of any quadratic equation, irrespective of how it was derived.

For the task of finding eigenvalues of a 2x2 matrix, the determinant equation effectively becomes a quadratic equation in \(\lambda\). In this exercise, by substituting the values \(a = 1\), \(b = -2\), and \(c = -15\) into the formula, we calculated the eigenvalues \(\lambda_1 = 5\) and \(\lambda_2 = -3\). Each solution is pivotal in identifying how a matrix interacts with its eigenvectors.