1. Let's start by stating the problem: Find the eigenvalues of a given square matrix $A$. 2. The eigenvalues $\lambda$ of a matrix $A$ satisfy the characteristic equation: $$\det(A - \lambda I) = 0$$ where $I$ is the identity matrix of the same size as $A$. 3. To find the eigenvalues, subtract $\lambda$ times the identity matrix from $A$, then compute the determinant of the resulting matrix. 4. Set the determinant equal to zero and solve the resulting polynomial equation for $\lambda$. 5. Each solution $\lambda$ is an eigenvalue of $A$. Example: Find the eigenvalues of $$A = \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}$$ Step 1: Write the characteristic equation: $$\det\left(\begin{bmatrix}4 & 1 \\ 2 & 3\end{bmatrix} - \lambda \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\right) = 0$$ Step 2: Compute $A - \lambda I$: $$\begin{bmatrix}4 - \lambda & 1 \\ 2 & 3 - \lambda\end{bmatrix}$$ Step 3: Calculate the determinant: $$ (4 - \lambda)(3 - \lambda) - 2 \times 1 = 0 $$ Step 4: Expand and simplify: $$ (4 - \lambda)(3 - \lambda) - 2 = (12 - 4\lambda - 3\lambda + \lambda^2) - 2 = \lambda^2 - 7\lambda + 10 = 0 $$ Step 5: Solve the quadratic equation: $$ \lambda^2 - 7\lambda + 10 = 0 $$ Using the quadratic formula: $$ \lambda = \frac{7 \pm \sqrt{(-7)^2 - 4 \times 1 \times 10}}{2} = \frac{7 \pm \sqrt{49 - 40}}{2} = \frac{7 \pm 3}{2} $$ Step 6: Find the roots: $$ \lambda_1 = \frac{7 + 3}{2} = 5 $$ $$ \lambda_2 = \frac{7 - 3}{2} = 2 $$ Therefore, the eigenvalues of matrix $A$ are $5$ and $2$.