Find the eigenvalues and corresponding eigenvectors of the matrix \( A = \begin{bmatrix} 4 & 1 \ 2 & 3 \end{bmatrix} \).
Solution
Step 1 ：First we find the eigenvalues. They are the roots of the characteristic equation, \( det(A - \lambda I) = 0 \), where \( I \) is the identity matrix and \( \lambda \) are the eigenvalues. So, we have \( \begin{vmatrix} 4 - \lambda & 1 \ 2 & 3 - \lambda \end{vmatrix} = 0 \), which simplifies to \( (4 - \lambda)(3 - \lambda) - 2 = 0 \). This equation simplifies to \( \lambda^2 - 7\lambda + 10 = 0 \), whose roots are \( \lambda_1 = 2 \) and \( \lambda_2 = 5 \).
Step 2 ：Next, we find the eigenvectors. We solve the system of linear equations \( (A - \lambda I)v = 0 \) for each eigenvalue, where \( v \) is the corresponding eigenvector. For \( \lambda_1 = 2 \), we get the system \( \begin{bmatrix} 2 & 1 \ 2 & 1 \end{bmatrix} \begin{bmatrix} v_1 \ v_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \). This gives us \( v = \begin{bmatrix} 1 \ -2 \end{bmatrix} \). Similarly, for \( \lambda_2 = 5 \), we get the system \( \begin{bmatrix} -1 & 1 \ 2 & -2 \end{bmatrix} \begin{bmatrix} v_1 \ v_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \). This gives us \( v = \begin{bmatrix} 1 \ 1 \end{bmatrix} \).
