1. **State the problem:** We are given a transformation defined by the matrix $$A = \begin{bmatrix}4 & 2 \\ -1 & 1\end{bmatrix}$$ and need to find its eigenvectors. 2. **Recall the formula:** Eigenvectors $\mathbf{v}$ satisfy $$A\mathbf{v} = \lambda \mathbf{v}$$ where $\lambda$ is an eigenvalue. 3. **Find eigenvalues:** Solve $$\det(A - \lambda I) = 0$$ where $$I = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$$. Calculate: $$\det\left(\begin{bmatrix}4-\lambda & 2 \\ -1 & 1-\lambda\end{bmatrix}\right) = (4-\lambda)(1-\lambda) - (-1)(2) = (4-\lambda)(1-\lambda) + 2$$ Expand: $$(4-\lambda)(1-\lambda) + 2 = 4 - 4\lambda - \lambda + \lambda^2 + 2 = \lambda^2 - 5\lambda + 6$$ Set equal to zero: $$\lambda^2 - 5\lambda + 6 = 0$$ 4. **Solve quadratic:** $$\lambda = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot 6}}{2} = \frac{5 \pm \sqrt{25 - 24}}{2} = \frac{5 \pm 1}{2}$$ So eigenvalues are: $$\lambda_1 = 3, \quad \lambda_2 = 2$$ 5. **Find eigenvectors for each eigenvalue:** For $\lambda_1 = 3$: Solve $$ (A - 3I)\mathbf{v} = 0 $$ $$A - 3I = \begin{bmatrix}4-3 & 2 \\ -1 & 1-3\end{bmatrix} = \begin{bmatrix}1 & 2 \\ -1 & -2\end{bmatrix}$$ Set system: $$1 \cdot x + 2 \cdot y = 0$$ $$-1 \cdot x - 2 \cdot y = 0$$ From first equation: $$x = -2y$$ Eigenvector corresponding to $\lambda_1=3$ is any scalar multiple of: $$\mathbf{v}_1 = \begin{bmatrix}-2 \\ 1\end{bmatrix}$$ 6. For $\lambda_2 = 2$: $$A - 2I = \begin{bmatrix}4-2 & 2 \\ -1 & 1-2\end{bmatrix} = \begin{bmatrix}2 & 2 \\ -1 & -1\end{bmatrix}$$ Set system: $$2x + 2y = 0$$ $$-x - y = 0$$ From first equation: $$2x = -2y \implies x = -y$$ Eigenvector corresponding to $\lambda_2=2$ is any scalar multiple of: $$\mathbf{v}_2 = \begin{bmatrix}-1 \\ 1\end{bmatrix}$$ **Final answer:** Eigenvalues: $3$ and $2$. Corresponding eigenvectors: $$\mathbf{v}_1 = \begin{bmatrix}-2 \\ 1\end{bmatrix}, \quad \mathbf{v}_2 = \begin{bmatrix}-1 \\ 1\end{bmatrix}$$