Question

Find the eigenvalues and eigenvectors of the following matrices.\(\left(\begin{array}{ll}1 & 3 \\ 2 & 2\end{array}\right)\)

Short answer

Eigenvalues: \(4, -1\). Eigenvectors: \[ \begin{pmatrix} 1 \ 1 \end{pmatrix} \] for \(\lambda = 4\) and \[ \begin{pmatrix} -3 \ 2 \end{pmatrix} \] for \(\lambda = -1\).

Step-by-step solution

- Write the characteristic equation

To find the eigenvalues, set up the characteristic equation by finding the determinant of the matrix \(A - \lambda I\), where \(A\) is your matrix and \(I\) is the identity matrix:\[ A = \left(\begin{array}{ll}1 & 3 \ 2 & 2\end{array}\right) \] Subtract \(\lambda\) times the identity matrix \(I\) from \(A\) and set the determinant to zero:\[ \left| \begin{array}{cc} 1 - \lambda & 3 \ 2 & 2 - \lambda \end{array} \right| = 0 \]

- Solve the determinant

Calculate the determinant of the matrix \[ \left(\begin{array}{cc} 1 - \lambda & 3 \ 2 & 2 - \lambda \end{array}\right) \]: Using the formula for the determinant of a 2x2 matrix, \(det(A) = a*d - b*c\):\[ (1-\lambda)(2-\lambda) - (3)(2) = 0 \] Simplify the above equation:\[ (1-\lambda)(2-\lambda) - 6 = 0 \] \[ \lambda^2 - 3\lambda - 4 = 0 \]

- Find the eigenvalues

Solve the quadratic equation to find the eigenvalues \(\lambda\):\[ \lambda^2 - 3\lambda - 4 = 0 \] Factoring the quadratic equation:\[ (\lambda - 4)(\lambda + 1) = 0 \] The solutions to the equation are:\[ \lambda = 4, \lambda = -1 \]

- Find the eigenvectors for \( \lambda = 4 \)

Substitute \(\lambda = 4\) back into the equation \(A - \lambda I = 0\) to find the eigenvector. Set up the matrix:\[ \left(\begin{array}{cc} 1-4 & 3 \ 2 & 2-4 \end{array}\right) = \left(\begin{array}{cc} -3 & 3 \ 2 & -2 \end{array}\right) \] Find the vector \(v\) that satisfies the equation \[ \left(\begin{array}{cc} -3 & 3 \ 2 & -2 \end{array}\right) \left( \begin{array}{c} x \ y \end{array} \right) = \left(\begin{array}{c} 0 \ 0 \end{array} \right) \] Simplifying the system, we get \-3x + 3y = 0\ and \2x - 2y = 0\, which simplifies to \x = y\. So, an eigenvector corresponding to \(\lambda = 4\) is \[ \begin{pmatrix} 1 \ 1 \end{pmatrix} \] (or any scalar multiple).

- Find the eigenvectors for \( \lambda = -1 \)

Substitute \(\lambda = -1\) back into the equation \(A - \lambda I = 0\) to find the eigenvector. Set up the matrix:\[ \left(\begin{array}{cc} 1-(-1) & 3 \ 2 & 2-(-1) \end{array}\right) = \left(\begin{array}{cc} 2 & 3 \ 2 & 3 \end{array}\right) \] Find the vector \(v\) that satisfies the equation \[ \left(\begin{array}{cc} 2 & 3 \ 2 & 3 \end{array}\right) \left( \begin{array}{c} x \ y \end{array} \right) = \left(\begin{array}{c} 0 \ 0 \end{array} \right) \] Simplifying the system, we get \2x + 3y = 0\ and \2x + 3y = 0\, which simplifies to \x = -\frac{3}{2}y\. So, an eigenvector corresponding to \(\lambda = -1\) is \[ \begin{pmatrix} -3 \ 2 \end{pmatrix} \] (or any scalar multiple).

Key concepts

Characteristic Equation

The characteristic equation is crucial when finding eigenvalues and eigenvectors of a matrix. It involves setting up an equation derived from the matrix, which helps identify the unique values associated with the matrix.
To start, you need the given matrix, let's call it \(A\). For example, if \(A = \begin{pmatrix}1 & 3 \ 2 & 2\thumb\right)\), we aim to find eigenvalues by solving the characteristic equation. First, we subtract \(λ\) times the identity matrix \(I\) from our matrix \(A\). This becomes \(A - λI\). The identity matrix \(I\) for a 2x2 matrix looks like this: \(\begin{pmatrix}1 & 0 \ 0 & 1\face_with_monocle \right)\).
Now, subtracting \(λI\) from \(A\) gives us \(\begin{pmatrix}1-λ & 3 \ 2 & 2-λ\right)\). The characteristic equation is then formed by setting the determinant of \(A - λI\) to zero: \(det(A - λI) = 0\). This forms the quadratic equation we solve to find eigenvalues.

Determinant

The determinant is a scalar value that can be computed from the elements of a square matrix. It provides important properties regarding the matrix, including whether it is invertible.
For a 2x2 matrix \(\begin{pmatrix}a & b \ c & d\right)\), the determinant is calculated as \(det(A) = ad - bc\). This helps in finding the characteristic equation by setting \(det(A - λI) = 0\).
Let's take our example, \(A = \begin{pmatrix}1 & 3 \ 2 & 2\right)\). After subtracting \(λI\), we have \(A - λI = \begin{pmatrix}1-λ & 3 \ 2 & 2-λ\right)\). The determinant is \((1-λ)(2-λ) - (3)(2) = λ^2 - 3λ - 4 = 0\). This algebraic expression, when solved, gives us the eigenvalues.

Eigenvalues

Eigenvalues are found from the characteristic equation and are key to understanding a matrix's properties. They are the values of \(λ\) for which \(Av = λv\), where \(v\) is an eigenvector.
From our quadratic equation \(λ^2 - 3λ - 4 = 0\), solving this gives the eigenvalues. Factoring the equation, we get \((λ - 4)(λ + 1) = 0\). Therefore, the eigenvalues are \(λ = 4\) and \(λ = -1\).
These values are critical as they indicate particular scaling factors in the transformation described by the matrix.

Eigenvectors

Eigenvectors are non-zero vectors that change at most by a scalar factor when that linear transformation is applied. They correspond to eigenvalues and provide insights into the transformation represented by the matrix.
For \(λ = 4\), substitute back into \(A - λI = 0\): \(\begin{pmatrix}1-4 & 3 \ 2 & 2-4\right) = \begin{pmatrix}-3 & 3 \ 2 & -2\right)\). Solving \((-3x + 3y = 0\) and \(2x - 2y = 0)\), we find \(x = y\). Thus, an eigenvector is \(v = \begin{pmatrix}1 \ 1\right)\).
For \(λ = -1\), substitute back: \(\begin{pmatrix}1-(-1) & 3 \ 2 & 2-(-1)\right) = \begin{pmatrix}2 & 3 \ 2 & 3\right)\). Solving \((2x + 3y = 0\)), we get \(x = -3/2y\). Hence, an eigenvector is \(v = \begin{pmatrix}-3 \ 2\right)\). Eigenvectors offer direction and scale to the transformation, making them vital in data analysis, computer graphics, and more.