Let's break down the solution into **two steps** as requested: --- ### **Step 1: Find the Eigenvectors for Each Eigenvalue** Given matrix \(A\): \[ A = \begin{bmatrix} 4 & 4 & 2 \\ 4 & 2 & 4 \\ 2 & 4 & 4 \end{bmatrix} \] The eigenvalues are given: **10, -2, 2**. #### **Eigenvector for λ = 10:** Solve \( (A - 10I)\vec{x} = 0 \): \[ A - 10I = \begin{bmatrix} -6 & 4 & 2 \\ 4 & -8 & 4 \\ 2 & 4 & -6 \end{bmatrix} \] Let \( \vec{x} = (x, y, z) \). The system is: \[ \begin{cases} -6x + 4y + 2z = 0 \\ 4x - 8y + 4z = 0 \\ 2x + 4y - 6z = 0 \\ \end{cases} \] Let's try \( x = y = z \): \[ -6x + 4x + 2x = 0 \rightarrow 0 = 0 \] \[ 4x - 8x + 4x = 0 \rightarrow 0 = 0 \] \[ 2x + 4x - 6x = 0 \rightarrow 0 = 0 \] So, **eigenvector for λ=10 is (1, 1, 1)**. --- #### **Explanation:** The eigenvector for λ=10 is found to be (1, 1, 1) because substituting this into the equations yields 0 for all, confirming it's correct. --- ### **Step 2: Find Eigenvectors for λ = -2 and λ = 2, Orthonormalize, and Write P and D** #### **Eigenvector for λ = -2:** \[ A + 2I = \begin{bmatrix} 6 & 4 & 2 \\ 4 & 4 & 4 \\ 2 & 4 & 6 \end{bmatrix} \] Solve \( (A + 2I)\vec{x} = 0 \): Let’s try \( x = 1, y = -2, z = 1 \): \[ 6*1 + 4*(-2) + 2*1 = 6 - 8 + 2 = 0 \\ 4*1 + 4*(-2) + 4*1 = 4 - 8 + 4 = 0 \\ 2*1 + 4*(-2) + 6*1 = 2 - 8 + 6 = 0 \\ \] So, eigenvector is **(1, -2, 1)**. --- #### **Eigenvector for λ = 2:** \[ A - 2I = \begin{bmatrix} 2 & 4 & 2 \\ 4 & 0 & 4 \\ 2 & 4 & 2 \end{bmatrix} \] Let’s try \( x = 1, y = 0, z = -1 \): \[ 2*1 + 4*0 + 2*(-1) = 2 - 2 = 0 \\ 4*1 + 0*0 + 4*(-1) = 4 - 4 = 0 \\ 2*1 + 4*0 + 2*(-1) = 2 - 2 = 0 \\ \] So, eigenvector is **(1, 0, -1)**. --- #### **Orthonormalize the Vectors:** Since these are already orthogonal, just normalize: - \((1,1,1)\) norm: \(\sqrt{3}\) - \((1,-2,1)\) norm: \(\sqrt{1+4+1}= \sqrt{6}\) - \((1,0,-1)\) norm: \(\sqrt{1+1}= \sqrt{2}\) So the orthonormalized vectors are: \[ \frac{1}{\sqrt{3}}(1,1,1),\quad \frac{1}{\sqrt{6}}(1,-2,1),\quad \frac{1}{\sqrt{2}}(1,0,-1) \] --- #### **Matrix P (columns are orthonormal eigenvectors) and D (diagonal eigenvalues):** \[ P = \left[ \frac{1}{\sqrt{3}},\ \frac{1}{\sqrt{6}},\ \frac{1}{\sqrt{2}},\ \frac{1}{\sqrt{3}},\ -\frac{2}{\sqrt{6}},\ 0,\ \frac{1}{\sqrt{3}},\ \frac{1}{\sqrt{6}},\ -\frac{1}{\sqrt{2}} \right] \] (row-wise, 3x3) \[ D = [10, 0, 0, 0, -2, 0, 0, 0, 2] \] --- ## **Final Answer:** **P:** \[ \frac{1}{\sqrt{3}},\ \frac{1}{\sqrt{6}},\ \frac{1}{\sqrt{2}},\ \frac{1}{\sqrt{3}},\ -\frac{2}{\sqrt{6}},\ 0,\ \frac{1}{\sqrt{3}},\ \frac{1}{\sqrt{6}},\ -\frac{1}{\sqrt{2}} \] **D:** \[ 10,\ 0,\ 0,\ 0,\ -2,\ 0,\ 0,\ 0,\ 2 \] --- **Explanation:** In Step 2, we found the orthonormal eigenvectors for each eigenvalue and constructed the diagonal matrix D (with eigenvalues on the diagonal) and matrix P (with orthonormal eigenvectors as columns).