1. **Find eigenvalues, eigenvectors, and basis of eigenspaces for matrix** (i) Given $$A=\begin{bmatrix}2 & 0 & 0 \\ 0 & 4 & 5 \\ 0 & 4 & 3\end{bmatrix}$$ - Step 1: Find eigenvalues by solving $$\det(A-\lambda I)=0$$ $$\det\begin{bmatrix}2-\lambda & 0 & 0 \\ 0 & 4-\lambda & 5 \\ 0 & 4 & 3-\lambda\end{bmatrix} = (2-\lambda) \det\begin{bmatrix}4-\lambda & 5 \\ 4 & 3-\lambda\end{bmatrix}$$ - Step 2: Calculate the 2x2 determinant: $$ (4-\lambda)(3-\lambda) - 20 = (12 - 4\lambda - 3\lambda + \lambda^2) - 20 = \lambda^2 - 7\lambda - 8 $$ - Step 3: Characteristic polynomial: $$ (2-\lambda)(\lambda^2 - 7\lambda - 8) = 0 $$ - Step 4: Solve $$\lambda^2 - 7\lambda - 8=0$$ using quadratic formula: $$\lambda = \frac{7 \pm \sqrt{49 + 32}}{2} = \frac{7 \pm \sqrt{81}}{2} = \frac{7 \pm 9}{2}$$ So eigenvalues: $$\lambda_1 = 2, \quad \lambda_2 = 8, \quad \lambda_3 = -1$$ - Step 5: Find eigenvectors for each eigenvalue by solving $$(A - \lambda I)\mathbf{x} = 0$$ For $$\lambda=2$$: $$\begin{bmatrix}0 & 0 & 0 \\ 0 & 2 & 5 \\ 0 & 4 & 1\end{bmatrix} \mathbf{x} = 0$$ From first row: no restriction. From second and third rows: $$2x_2 + 5x_3 = 0$$ $$4x_2 + x_3 = 0$$ Solve: Multiply second by 2: $$4x_2 + 10x_3 = 0$$ Subtract third: $$ (4x_2 + 10x_3) - (4x_2 + x_3) = 9x_3 = 0 \Rightarrow x_3=0$$ Then from $$2x_2 + 5(0) = 0$$, $$x_2=0$$ Eigenvector: $$\mathbf{x} = \begin{bmatrix}x_1 \\ 0 \\ 0\end{bmatrix}$$ Any $$x_1 \neq 0$$, so eigenvector basis: $$\{\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}\}$$ For $$\lambda=8$$: $$A - 8I = \begin{bmatrix}-6 & 0 & 0 \\ 0 & -4 & 5 \\ 0 & 4 & -5\end{bmatrix}$$ Solve: $$-4x_2 + 5x_3 = 0$$ $$4x_2 - 5x_3 = 0$$ Both equations are the same, so: $$-4x_2 + 5x_3 = 0 \Rightarrow 5x_3 = 4x_2 \Rightarrow x_3 = \frac{4}{5} x_2$$ Eigenvector: $$\mathbf{x} = \begin{bmatrix}0 \\ x_2 \\ \frac{4}{5} x_2\end{bmatrix} = x_2 \begin{bmatrix}0 \\ 1 \\ \frac{4}{5}\end{bmatrix}$$ Basis: $$\{\begin{bmatrix}0 \\ 1 \\ \frac{4}{5}\end{bmatrix}\}$$ For $$\lambda=-1$$: $$A + I = \begin{bmatrix}3 & 0 & 0 \\ 0 & 5 & 5 \\ 0 & 4 & 4\end{bmatrix}$$ Solve: $$5x_2 + 5x_3 = 0 \Rightarrow x_2 + x_3 = 0 \Rightarrow x_3 = -x_2$$ $$4x_2 + 4x_3 = 0 \Rightarrow 4x_2 - 4x_2 = 0$$ (consistent) Eigenvector: $$\mathbf{x} = \begin{bmatrix}0 \\ x_2 \\ -x_2\end{bmatrix} = x_2 \begin{bmatrix}0 \\ 1 \\ -1\end{bmatrix}$$ Basis: $$\{\begin{bmatrix}0 \\ 1 \\ -1\end{bmatrix}\}$$ --- **Slug:** eigenvalues eigenvectors **Subject:** linear algebra **Desmos:** {"latex":"","features":{"intercepts":true,"extrema":true}} **q_count:** 6