Question

Find all eigenvalues and eigenvectors of the given matrix. $$ \left(\begin{array}{rr}{5} & {-1} \\ {3} & {1}\end{array}\right) $$

Short answer

Based on the given solution, the matrix has two eigenvalues, λ = 2 and λ = 4. The eigenvector corresponding to eigenvalue λ = 2 is v = (1, 3), and the eigenvector corresponding to eigenvalue λ = 4 is v = (1, 1).

Step-by-step solution

Determine the characteristic equation

To find the characteristic equation, we need to compute the determinant of the matrix (A - λI): $$ \begin{vmatrix} 5- \lambda & -1 \\ 3 & 1 - \lambda \end{vmatrix} $$ Now calculate the determinant: $$ (5 - \lambda)(1 - \lambda) - (-1)(3) = \lambda^2 - 6\lambda + 8 $$

Find the eigenvalues

Solve the quadratic equation we found in Step 1 to determine the eigenvalues: $$ \lambda^2 - 6\lambda + 8 = 0 $$ Factoring the quadratic, we get: $$ (\lambda - 2)(\lambda - 4) = 0 $$ So, the eigenvalues are λ = 2 and λ = 4.

Find the eigenvectors

Let's find the eigenvectors corresponding to each eigenvalue. First, for λ = 2, plug this value into the equation (A - λI)v = 0: $$ \begin{pmatrix} 5 - 2 & -1 \\ 3 & 1 - 2 \end{pmatrix} v = \begin{pmatrix} 0 \\ 0 \end{pmatrix} $$ Which is equivalent to the following system of linear equations: $$ \begin{cases} 3v_1 - v_2 = 0\\ 3v_1 - v_2 = 0 \end{cases} $$ This system has infinitely many solutions, but we look for a basic eigenvector. So we can set v₁ = 1 and obtain v₂ = 3. The eigenvector for λ = 2 is v = (1, 3). Now, for λ = 4, plug this value into the equation (A - λI)v = 0: $$ \begin{pmatrix} 5 - 4 & -1 \\ 3 & 1 - 4 \end{pmatrix} v = \begin{pmatrix} 0 \\ 0 \end{pmatrix} $$ Which is equivalent to the following system of linear equations: $$ \begin{cases} v_1 - v_2 = 0\\ 3v_1 + 3v_2 = 0 \end{cases} $$ This system has infinitely many solutions, but we look for a basic eigenvector. So we can set v₁ = 1 and obtain v₂ = 1. The eigenvector for λ = 4 is v = (1, 1).

State the eigenvalues and eigenvectors

In conclusion, the eigenvalues and their corresponding eigenvectors of the given matrix are: Eigenvalue λ = 2, Eigenvector v = (1, 3) Eigenvalue λ = 4, Eigenvector v = (1, 1)