1. **Problem statement:** Check if the matrix $$A = \begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & -1 \\ 1 & -1 & -1\end{bmatrix}$$ is diagonalizable. If yes, find diagonal matrix $D$ and invertible matrix $P$ such that $P^{-1}AP = D$. 2. **Step 1: Find eigenvalues of $A$.** We solve $\det(A - \lambda I) = 0$. $$A - \lambda I = \begin{bmatrix}1-\lambda & 1 & 1 \\ 1 & 1-\lambda & -1 \\ 1 & -1 & -1-\lambda\end{bmatrix}$$ Calculate determinant: $$\det(A - \lambda I) = (1-\lambda) \begin{vmatrix}1-\lambda & -1 \\ -1 & -1-\lambda\end{vmatrix} - 1 \begin{vmatrix}1 & -1 \\ 1 & -1-\lambda\end{vmatrix} + 1 \begin{vmatrix}1 & 1-\lambda \\ 1 & -1\end{vmatrix}$$ Calculate minors: $$= (1-\lambda)((1-\lambda)(-1-\lambda) - (-1)(-1)) - 1(1(-1-\lambda) - 1(-1)) + 1(1(-1) - 1(1-\lambda))$$ Simplify: $$= (1-\lambda)((1-\lambda)(-1-\lambda) - 1) - 1(-1-\lambda + 1) + 1(-1 - (1-\lambda))$$ Calculate $(1-\lambda)(-1-\lambda)$: $$= (1)(-1) + (1)(-\lambda) + (-\lambda)(-1) + (-\lambda)(-\lambda) = -1 - \lambda + \lambda + \lambda^2 = -1 + \lambda^2$$ So inside bracket: $$(-1 + \lambda^2) - 1 = \lambda^2 - 2$$ Therefore: $$= (1-\lambda)(\lambda^2 - 2) - 1(-\lambda) + 1(-1 - 1 + \lambda)$$ Simplify terms: $$= (1-\lambda)(\lambda^2 - 2) + \lambda + (-2 + \lambda) = (1-\lambda)(\lambda^2 - 2) + 2\lambda - 2$$ Expand: $$= (1)(\lambda^2 - 2) - \lambda(\lambda^2 - 2) + 2\lambda - 2 = \lambda^2 - 2 - \lambda^3 + 2\lambda + 2\lambda - 2$$ Combine like terms: $$= -\lambda^3 + \lambda^2 + 4\lambda - 4$$ Set equal to zero: $$-\lambda^3 + \lambda^2 + 4\lambda - 4 = 0$$ Multiply both sides by $-1$: $$\lambda^3 - \lambda^2 - 4\lambda + 4 = 0$$ 3. **Step 2: Find roots of characteristic polynomial:** Try rational roots $\pm1, \pm2, \pm4$. Test $\lambda=1$: $$1 - 1 - 4 + 4 = 0$$ So $\lambda=1$ is a root. Divide polynomial by $(\lambda - 1)$: $$\lambda^3 - \lambda^2 - 4\lambda + 4 = (\lambda - 1)(\lambda^2 - 4)$$ Factor $\lambda^2 - 4$: $$(\lambda - 2)(\lambda + 2)$$ Eigenvalues are: $$\lambda_1 = 1, \quad \lambda_2 = 2, \quad \lambda_3 = -2$$ 4. **Step 3: Find eigenvectors for each eigenvalue.** - For $\lambda=1$ solve $(A - I)\mathbf{v} = 0$: $$A - I = \begin{bmatrix}0 & 1 & 1 \\ 1 & 0 & -1 \\ 1 & -1 & -2\end{bmatrix}$$ From first row: $0\cdot v_1 + v_2 + v_3 = 0 \Rightarrow v_2 = -v_3$ From second row: $v_1 + 0\cdot v_2 - v_3 = 0 \Rightarrow v_1 = v_3$ From third row: $v_1 - v_2 - 2 v_3 = 0$ substitute $v_1 = v_3$, $v_2 = -v_3$: $$v_3 - (-v_3) - 2 v_3 = v_3 + v_3 - 2 v_3 = 0$$ Consistent. Eigenvector: $$\mathbf{v}_1 = v_3 \begin{bmatrix}1 \\ -1 \\ 1\end{bmatrix}$$ - For $\lambda=2$ solve $(A - 2I)\mathbf{v} = 0$: $$A - 2I = \begin{bmatrix}-1 & 1 & 1 \\ 1 & -1 & -1 \\ 1 & -1 & -3\end{bmatrix}$$ From first row: $-v_1 + v_2 + v_3 = 0 \Rightarrow v_1 = v_2 + v_3$ From second row: $v_1 - v_2 - v_3 = 0$ substitute $v_1$: $$(v_2 + v_3) - v_2 - v_3 = 0$$ True for all $v_2, v_3$. From third row: $v_1 - v_2 - 3 v_3 = 0$ substitute $v_1 = v_2 + v_3$: $$(v_2 + v_3) - v_2 - 3 v_3 = -2 v_3 = 0 \Rightarrow v_3 = 0$$ Then $v_1 = v_2$. Eigenvector: $$\mathbf{v}_2 = v_2 \begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}$$ - For $\lambda=-2$ solve $(A + 2I)\mathbf{v} = 0$: $$A + 2I = \begin{bmatrix}3 & 1 & 1 \\ 1 & 3 & -1 \\ 1 & -1 & 1\end{bmatrix}$$ From first row: $3 v_1 + v_2 + v_3 = 0$ From second row: $v_1 + 3 v_2 - v_3 = 0$ From third row: $v_1 - v_2 + v_3 = 0$ Add first and third rows: $$(3 v_1 + v_2 + v_3) + (v_1 - v_2 + v_3) = 4 v_1 + 2 v_3 = 0 \Rightarrow 2 v_1 + v_3 = 0 \Rightarrow v_3 = -2 v_1$$ From third row: $$v_1 - v_2 + v_3 = 0 \Rightarrow v_2 = v_1 + v_3 = v_1 - 2 v_1 = -v_1$$ Eigenvector: $$\mathbf{v}_3 = v_1 \begin{bmatrix}1 \\ -1 \\ -2\end{bmatrix}$$ 5. **Step 4: Form matrix $P$ with eigenvectors as columns and diagonal matrix $D$ with eigenvalues:** $$P = \begin{bmatrix}1 & 1 & 1 \\ -1 & 1 & -1 \\ 1 & 0 & -2\end{bmatrix}, \quad D = \begin{bmatrix}1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & -2\end{bmatrix}$$ 6. **Step 5: Verify $P$ is invertible (determinant nonzero).** Calculate $\det(P)$: $$= 1 \begin{vmatrix}1 & -1 \\ 0 & -2\end{vmatrix} - 1 \begin{vmatrix}-1 & -1 \\ 1 & -2\end{vmatrix} + 1 \begin{vmatrix}-1 & 1 \\ 1 & 0\end{vmatrix}$$ $$= 1(1 \cdot -2 - 0) - 1((-1)(-2) - (-1)(1)) + 1((-1)(0) - 1(1))$$ $$= 1(-2) - 1(2 - (-1)) + 1(0 - 1) = -2 - 1(3) - 1 = -2 - 3 - 1 = -6 \neq 0$$ So $P$ is invertible. **Answer for (a):** $A$ is diagonalizable with $$D = \begin{bmatrix}1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & -2\end{bmatrix}, \quad P = \begin{bmatrix}1 & 1 & 1 \\ -1 & 1 & -1 \\ 1 & 0 & -2\end{bmatrix}$$ --- **Slug:** diagonalizable matrix **Subject:** linear algebra **Desmos:** {"latex":"y=","features":{"intercepts":true,"extrema":true}} **q_count:** 3