Question

Find the eigenvalues and eigenvectors of the following matrices.\(\left(\begin{array}{rr}3 & -2 \\ -2 & 0\end{array}\right)\)

Short answer

Eigenvalues: \(4, -1\). Eigenvectors: \(\begin{pmatrix} 2 \ -1 \end{pmatrix}, \begin{pmatrix} 1 \ 2 \end{pmatrix}\).

Step-by-step solution

- Write the Characteristic Equation

To find the eigenvalues, first form the characteristic equation. This is given by \(\det(A - \lambda I) = 0\), where \(A\) is the matrix and \(\lambda\) is the eigenvalue. Here, \(A = \left(\begin{array}{rr}3 & -2 \ -2 & 0 \end{array}\right)\). The characteristic equation is \[\det\left(\begin{array}{rr}3 - \lambda & -2\ -2 & -\lambda \end{array}\right) = 0\]\.

- Compute the Determinant

Calculate the determinant of the matrix \(A - \lambda I\). This is given by \[\det\left(\begin{array}{rr}3 - \lambda & -2\ -2 & -\lambda \end{array}\right) = (3 - \lambda)(-\lambda) - (-2)(-2)\] \[= -3\lambda + \lambda^2 - 4 = \lambda^2 - 3\lambda - 4.\] Therefore, the characteristic equation is \ \lambda^2 - 3\lambda - 4 = 0.\

- Solve the Characteristic Equation

Solve the quadratic equation \(\lambda^2 - 3\lambda - 4 = 0\) for \(\lambda\). Use the quadratic formula \(\lambda = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\), where \(a=1, b=-3, \text{and} c=-4.\)\ \[\lambda = \frac{3\pm\sqrt{9+16}}{2} = \frac{3\pm 5}{2}\]. This gives the eigenvalues \[\lambda_1 = 4 \text{ and } \lambda_2 = -1.\]

- Find Eigenvectors for \lambda_1 = 4

Substitute \(\lambda_1 = 4\) back into \(A - \lambda I\) to find the eigenvector. \[\left(\begin{array}{rr}3 - 4 & -2\ -2 & -4\end{array}\right) \left(\begin{array}{c}x\ y\end{array}\right) = 0\] \[\left(\begin{array}{rr}-1 & -2\ -2 & -4\end{array}\right) \left(\begin{array}{c}x\ y\end{array}\right) = 0\].\r\rWe solve the system of equations: \ -x - 2y = 0 \ and \ -2x - 4y = 0 \, which boils down to \ x = -2y \. Therefore, one eigenvector for \(\lambda_1 = 4\) is \(\mathbf{v}_1 = \begin{pmatrix} 2 \ -1 \end{pmatrix}\).

- Find Eigenvectors for \lambda_2 = -1

Substitute \(\lambda_2 = -1\) back into \(A - \lambda I\) to find the eigenvector. \[\left(\begin{array}{rr}3 + 1 & -2\ -2 & 1\end{array}\right) \left(\begin{array}{c}x\ y\end{array}\right) = 0\] \[\left(\begin{array}{rr}4 & -2\ -2 & 1\end{array}\right) \left(\begin{array}{c}x\ y\end{array}\right) = 0\] \[4x - 2y = 0 \text{ and } -2x + y = 0.\] Solving this, we find that \ y = 2x \. Therefore, one eigenvector for \(\lambda_2 = -1\) is \(\mathbf{v}_2 = \begin{pmatrix} 1 \ 2 \end{pmatrix}\).

Key concepts

Characteristic Equation

To understand eigenvalues and eigenvectors, we first need the characteristic equation. This equation helps us find the eigenvalues of a given matrix. For a matrix A, the characteristic equation is formed by calculating the determinant of \(A - \lambda I\), where \lambda\ is the eigenvalue we want to find and I is the identity matrix. In this context, the characteristic equation is written as det(\(A - \lambda I\)) = 0. This step is essential because it links the matrix's properties to its eigenvalues, setting the stage for finding eigenvectors later.

Determinant

Calculating the determinant is a crucial step in solving the characteristic equation. The determinant of a matrix gives crucial information about the matrix, such as whether it is invertible. For a matrix \(\begin{pmatrix} a & b \ c & d \end{pmatrix}\), the determinant is calculated as ad - bc.

In our exercise, we use the determinant to simplify the characteristic equation of \(A - \lambda I\). Once the determinant is calculated, you get a polynomial equation in \lambda\, which is also known as the characteristic polynomial. This polynomial must be solved to find the eigenvalues.

Quadratic Equation

Upon calculating the determinant for our characteristic equation, we often end up with a quadratic equation. A quadratic equation takes the form \ax^2 + bx + c = 0\. In our exercise, solving the quadratic equation \(\lambda ^2 - 3 \lambda - 4=0\) yields the eigenvalues. This equation can be solved using the quadratic formula \(\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where a, b, and c are coefficients from our polynomial.

Quadratic equations are important in finding eigenvalues because they offer a way to resolve the polynomial into specific values that \lambda\ can take. These specific values (eigenvalues) are essential to move forward and find eigenvectors.

Eigenvectors

Once we have the eigenvalues from our quadratic equation, the next step is to find the eigenvectors. An eigenvector \(v\) corresponding to an eigenvalue \(\lambda\) is a non-zero vector that satisfies the equation \(Av = \lambda v\). To find these vectors, we substitute our eigenvalues back into the equation \((A - \lambda I)v = 0\) and solve for v.

Essentially, eigenvectors reveal the directions in which a transformation acts by stretching or compressing vectors, but not changing their direction. They are pivotal in many applications such as stability analysis, vibration analysis, and even Google's PageRank algorithm.