Question

Find the eigenvalues and eigenvectors of the matrices in the following problems. $$\left(\begin{array}{ll}5 & 1 \\\4 & 2\end{array}\right)$$

Short answer

Eigenvalues: \(\lambda_1 = 6\), \(\lambda_2 = 1\). Eigenvectors: \mathbf{v_1} = \begin{pmatrix} 1 \ 1 \end{pmatrix}, \mathbf{v_2} = \begin{pmatrix} 1 \ -4 \end{pmatrix}.

Step-by-step solution

- Find the Characteristic Equation

The characteristic equation is found using the determinant of the matrix \(A - \lambda I\). For the given matrix \(A = \begin{pmatrix} 5 & 1 \ 4 & 2 \end{pmatrix}\), compute the determinant of \(A - \lambda I\): \[ \text{det}(A - \lambda I) = \begin{vmatrix} 5 - \lambda & 1 \ 4 & 2 - \lambda \end{vmatrix} = (5 - \lambda)(2 - \lambda) - 4 \].

- Simplify the Determinant

Expand and simplify the determinant to find the characteristic polynomial: \[(5 - \lambda)(2 - \lambda) - 4 = 10 - 5\lambda - 2\lambda + \lambda^2 - 4 = \lambda^2 - 7\lambda + 6 \].

- Solve the Characteristic Equation

Set the characteristic polynomial equal to zero to find the eigenvalues: \[\lambda^2 - 7\lambda + 6 = 0 \]. Factor the quadratic equation: \[(\lambda - 6)(\lambda - 1) = 0 \]. Hence, the eigenvalues are \(\lambda_1 = 6\) and \(\lambda_2 = 1\).

- Find Eigenvectors for \(\lambda_1 = 6\)

Substitute \(\lambda_1 = 6\) back into \(A - \lambda I\) to find the eigenvector: \[ \begin{pmatrix} 5 - 6 & 1 \ 4 & 2 - 6 \end{pmatrix} \begin{pmatrix} x_1 \ x_2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \Rightarrow \begin{pmatrix} -1 & 1 \ 4 & -4 \end{pmatrix} \begin{pmatrix} x_1 \ x_2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \].

- Solve for Eigenvector \(\mathbf{v_1}\)

Solving the system, we get \( -x_1 + x_2 = 0 \) or \x_2 = x_1 \. The eigenvector for \(\lambda_1 = 6\) is any scalar multiple of \( \mathbf{v_1} = \begin{pmatrix} 1 \ 1 \end{pmatrix} \).

- Find Eigenvectors for \(\lambda_2 = 1\)

Substitute \(\lambda_2 = 1\) back into \(A - \lambda I\) to find the eigenvector: \[ \begin{pmatrix} 5 - 1 & 1 \ 4 & 2 - 1 \end{pmatrix} \begin{pmatrix} y_1 \ y_2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \Rightarrow \begin{pmatrix} 4 & 1 \ 4 & 1 \end{pmatrix} \begin{pmatrix} y_1 \ y_2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \].

- Solve for Eigenvector \(\mathbf{v_2}\)

Solving the system, we get \( 4y_1 + y_2 = 0 \) or \y_2 = -4y_1 \. The eigenvector for \(\lambda_2 = 1\) is any scalar multiple of \(\begin{pmatrix} 1 \ -4 \end{pmatrix} \).

Key concepts

Characteristic Equation

In linear algebra, the characteristic equation is a crucial step in finding eigenvalues of a matrix. For a given matrix, say, \(A\), we derive the characteristic equation by calculating the determinant of \(A - \lambda I\) where \(\lambda\) is a scalar and \(I\) is the identity matrix of the same dimension as \(A\). The determinant must be set to zero, as follows:
\[ \text{{det}}(A - \lambda I) = 0 \]
In our given example matrix, \[A = \begin{pmatrix} 5 & 1 \ 4 & 2 \end{pmatrix}, alchemy.\] we calculate the determinant of \(A - \lambda I\):
\[ \text{{det}}(\begin{pmatrix} 5 - \lambda & 1\ \ 4 & 2 - \lambda \end{pmatrix}) = (5 - \lambda)(2 - \lambda) - 4 = \lambda^2 - 7\lambda + 6 \]
This polynomial equation is the characteristic equation. It must be solved to find the eigenvalues \(\lambda\).

Determinant

The determinant is a special scalar value that can be computed from the elements of a square matrix. It provides essential properties such as whether the matrix is invertible. To calculate the determinant of a 2x2 matrix \(\begin{pmatrix}a & b \ c & d\end{pmatrix}\), use the formula:
\[ \det \begin{pmatrix} a & b \ c & d \end{pmatrix} = ad - bc \]
For the matrix \(\begin{pmatrix} 5 & 1 \ 4 & 2 \end{pmatrix}\), the determinant calculation steps are:
\[ \text{{det}}\begin{pmatrix} 5 - \lambda & 1\ \ 4 & 2 - \lambda \end{pmatrix} = (5 - \lambda)(2 - \lambda) - 4 \]
This becomes our characteristic polynomial \(\lambda^2 - 7\////////////lambda + 6 \).

Eigenvalues

Eigenvalues are the special set of scalars associated with a linear system of equations, i.e., a matrix equation. They are found by solving the characteristic equation. For our example, the characteristic polynomial is:
\[ \lambda^2 - 7\lambda + 6 = 0 \]
Solving this quadratic equation by factoring, we get:
\[ (\lambda - 6)(\lambda - 1) = 0 \]
Thus, the eigenvalues are \(\lambda_1 = 6\) and \(\lambda_2 = 1\).

Eigenvectors

Once eigenvalues are determined, eigenvectors can be found. These are vectors that, when multiplied by the matrix, yield a scalar multiple of themselves—i.e., they do not change direction. For each eigenvalue, say, \(\lambda_i\), the eigenvector \(\text{{v}}_i\) can be found using the equation:
\[ (A - \lambda_i I)\text{{v}}_i = 0 \]
For our example, for the eigenvalue \(\lambda_1 = 6\) we solve:
\[ \begin{pmatrix} -1 & 1 \ 4 & -4 \end{pmatrix} \begin{pmatrix} x_1 \ x_2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \]
Similarly, for \(\lambda_2 = 1\),
\[ \begin{pmatrix} 4 & 1 \ 4 & 1 \end{pmatrix} \begin{pmatrix} y_1 \ y_2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \]
Both systems yield solutions for the respective eigenvectors.

Quadratic Equation

Quadratic equations are polynomial equations of the form \(\text{{ax}}^2 + bx + c = 0\). They can be solved using factoring, completing the square, or the quadratic formula. In our context, the characteristic equation is a quadratic equation, \[ \lambda^2 - 7\lambda + 6 = 0 \]
We solve this by factoring:
\[ (\lambda - 6)(\lambda - 1) = 0 \implies \] \( \lambda_1 = 6\) and \(\lambda_2 = 1\).