1. **Problem:** Find the eigenvalues and eigenvectors of a given square matrix $A$. 2. **Formula and Explanation:** - Eigenvalues $\lambda$ satisfy the characteristic equation $$\det(A - \lambda I) = 0,$$ where $I$ is the identity matrix. - For each eigenvalue $\lambda$, the eigenvector $\mathbf{v}$ satisfies $$ (A - \lambda I)\mathbf{v} = \mathbf{0}.$$ - Eigenvectors are non-zero vectors that satisfy this equation. 3. **Step-by-step solution:** - Compute the matrix $A - \lambda I$. - Find the determinant and solve $$\det(A - \lambda I) = 0$$ to get eigenvalues $\lambda$. - For each eigenvalue, substitute back into $$ (A - \lambda I)\mathbf{v} = \mathbf{0} $$ and solve the system of linear equations to find eigenvectors. 4. **Example:** Suppose $$A = \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}.$$ - Compute characteristic polynomial: $$\det\left(\begin{bmatrix}4-\lambda & 1 \\ 2 & 3-\lambda \end{bmatrix}\right) = (4-\lambda)(3-\lambda) - 2 \times 1 = \lambda^2 - 7\lambda + 10 = 0.$$ - Solve quadratic equation: $$\lambda^2 - 7\lambda + 10 = 0 \implies (\lambda - 5)(\lambda - 2) = 0,$$ so eigenvalues are $$\lambda_1 = 5, \quad \lambda_2 = 2.$$ - Find eigenvectors: For $$\lambda_1 = 5,$$ $$ (A - 5I) = \begin{bmatrix} -1 & 1 \\ 2 & -2 \end{bmatrix},$$ solve $$ (A - 5I)\mathbf{v} = \mathbf{0} $$: $$ -v_1 + v_2 = 0 \implies v_2 = v_1.$$ Eigenvector corresponding to $$\lambda_1=5$$ is any scalar multiple of $$\begin{bmatrix}1 \\ 1\end{bmatrix}.$$ For $$\lambda_2 = 2,$$ $$ (A - 2I) = \begin{bmatrix} 2 & 1 \\ 2 & 1 \end{bmatrix},$$ solve $$ (A - 2I)\mathbf{v} = \mathbf{0} $$: $$ 2v_1 + v_2 = 0 \implies v_2 = -2v_1.$$ Eigenvector corresponding to $$\lambda_2=2$$ is any scalar multiple of $$\begin{bmatrix}1 \\ -2\end{bmatrix}.$$ 5. **Summary:** - Eigenvalues are $$5$$ and $$2$$. - Corresponding eigenvectors are $$\begin{bmatrix}1 \\ 1\end{bmatrix}$$ and $$\begin{bmatrix}1 \\ -2\end{bmatrix}$$ respectively. This method applies to any square matrix to find eigenvalues and eigenvectors.