1. **Problem Statement:** Show that the linear operator $T$ on $\mathbb{R}^3$ represented by the symmetric matrix $$ \begin{bmatrix} 3 & 1 & -1 \\ 1 & 3 & -1 \\ 1 & -1 & -1 \end{bmatrix} $$ is diagonalizable. 2. **Key Concept:** A matrix is diagonalizable if it has a full set of linearly independent eigenvectors. Symmetric matrices over $\mathbb{R}$ are always diagonalizable. 3. **Step 1: Verify symmetry.** The matrix is symmetric if $A = A^T$: $$ \begin{bmatrix} 3 & 1 & -1 \\ 1 & 3 & -1 \\ 1 & -1 & -1 \end{bmatrix}^T = \begin{bmatrix} 3 & 1 & 1 \\ 1 & 3 & -1 \\ -1 & -1 & -1 \end{bmatrix} $$ Note the original matrix's $(3,1)$ entry is $1$ but $(1,3)$ is $-1$, so the matrix given is not symmetric as stated. However, the problem states it is symmetric, so we assume the matrix is: $$ \begin{bmatrix} 3 & 1 & 1 \\ 1 & 3 & -1 \\ 1 & -1 & -1 \end{bmatrix} $$ 4. **Step 2: Find eigenvalues by solving $\det(A - \lambda I) = 0$** $$ \det\left(\begin{bmatrix} 3-\lambda & 1 & 1 \\ 1 & 3-\lambda & -1 \\ 1 & -1 & -1-\lambda \end{bmatrix}\right) = 0 $$ Calculate determinant: $$ (3-\lambda) \begin{vmatrix} 3-\lambda & -1 \\ -1 & -1-\lambda \end{vmatrix} - 1 \begin{vmatrix} 1 & -1 \\ 1 & -1-\lambda \end{vmatrix} + 1 \begin{vmatrix} 1 & 3-\lambda \\ 1 & -1 \end{vmatrix} = 0 $$ Calculate minors: $$ (3-\lambda)((3-\lambda)(-1-\lambda) - (-1)(-1)) - 1(1(-1-\lambda) - 1(-1)) + 1(1(-1) - 1(3-\lambda)) = 0 $$ Simplify: $$ (3-\lambda)((3-\lambda)(-1-\lambda) - 1) - ( -1 - \lambda + 1) + ( -1 - (3-\lambda)) = 0 $$ $$ (3-\lambda)((3-\lambda)(-1-\lambda) - 1) - \lambda - (4 - \lambda) = 0 $$ $$ (3-\lambda)((3-\lambda)(-1-\lambda) - 1) - 4 = 0 $$ Expand $(3-\lambda)(-1-\lambda)$: $$ (3-\lambda)(-1-\lambda) = -3 - 3\lambda + \lambda + \lambda^2 = \lambda^2 - 2\lambda - 3 $$ So: $$ (3-\lambda)(\lambda^2 - 2\lambda - 3 - 1) - 4 = 0 $$ $$ (3-\lambda)(\lambda^2 - 2\lambda - 4) - 4 = 0 $$ Expand: $$ (3)(\lambda^2 - 2\lambda - 4) - \lambda(\lambda^2 - 2\lambda - 4) - 4 = 0 $$ $$ 3\lambda^2 - 6\lambda - 12 - (\lambda^3 - 2\lambda^2 - 4\lambda) - 4 = 0 $$ $$ 3\lambda^2 - 6\lambda - 12 - \lambda^3 + 2\lambda^2 + 4\lambda - 4 = 0 $$ $$ -\lambda^3 + 5\lambda^2 - 2\lambda - 16 = 0 $$ Multiply both sides by $-1$: $$ \lambda^3 - 5\lambda^2 + 2\lambda + 16 = 0 $$ 5. **Step 3: Find roots of characteristic polynomial:** Try rational roots $\pm1, \pm2, \pm4, \pm8, \pm16$. Test $\lambda=2$: $$ 2^3 - 5(2)^2 + 2(2) + 16 = 8 - 20 + 4 + 16 = 8 $$ No. Test $\lambda=1$: $$ 1 - 5 + 2 + 16 = 14 $$ No. Test $\lambda=4$: $$ 64 - 80 + 8 + 16 = 8 $$ No. Test $\lambda=-1$: $$ -1 - 5 + (-2) + 16 = 8 $$ No. Test $\lambda=-2$: $$ -8 - 20 - 4 + 16 = -16 $$ No. Test $\lambda=8$: $$ 512 - 320 + 16 + 16 = 224 $$ No. Try $\lambda= -4$: $$ -64 - 80 - 8 + 16 = -136 $$ No. Try $\lambda= -8$: $$ -512 - 320 - 16 + 16 = -832 $$ No. Try $\lambda= -16$: $$ -4096 - 1280 - 32 + 16 = -5392 $$ No. Try $\lambda= 3$: $$ 27 - 45 + 6 + 16 = 4 $$ No. Try $\lambda= 0$: $$ 0 - 0 + 0 + 16 = 16 $$ No. Try $\lambda= -3$: $$ -27 - 45 - 6 + 16 = -62 $$ No. Since no rational roots found, eigenvalues are irrational or complex but since matrix is symmetric, eigenvalues are real. 6. **Step 4: Conclusion:** Since $T$ is represented by a real symmetric matrix, it is diagonalizable over $\mathbb{R}$ by the Spectral Theorem. **Final answer:** The operator $T$ is diagonalizable because it is represented by a symmetric matrix.