Let's solve this step by step: ### **Step 1: Use the Diagonalization Theorem** The matrix \(A\) is given in factored form \(A = PDP^{-1}\), where: - \(P\) is the matrix of eigenvectors, - \(D\) is the diagonal matrix of eigenvalues. From the image, we see: \[ A = PDP^{-1} = \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 0 \\ 1 & 1 & 2 \\ \end{bmatrix} \begin{bmatrix} 4 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 0 \\ 1 & 0 & -1 \\ \end{bmatrix} \] **Explanation:** The diagonal entries of \(D\) are the **eigenvalues** of \(A\), and the **columns of \(P\)** are the eigenvectors (basis for each eigenspace). --- ### **Step 2: List Eigenvalues and Their Corresponding Bases** From \(D\), the eigenvalues are: **4, 1, 1**. The corresponding eigenvectors (from columns of \(P\)): - For eigenvalue \(4\): first column of \(P\) → \([1, 1, 1]^T\) - For eigenvalue \(1\): second column of \(P\) → \([1, -1, 0]^T\) - For eigenvalue \(1\): third column of \(P\) → \([1, 0, -1]^T\) However, eigenvalue \(1\) is repeated, so it has **two linearly independent eigenvectors**, forming a basis for its eigenspace. **Explanation:** We have two distinct eigenvalues: \(4\) and \(1\). - For \(\lambda = 4\): basis is \(\{[1, 1, 1]^T\}\) - For \(\lambda = 1\): basis is \(\{[1, -1, 0]^T, [1, 0, -1]^T\}\) --- ### **Final Answer:** **Choice B is correct:** > In ascending order, the two distinct eigenvalues are \(\lambda_1 = 1\) and \(\lambda_2 = 4\). Bases for the corresponding eigenspaces are \(\{[1, -1, 0]^T, [1, 0, -1]^T\}\) and \(\{[1, 1, 1]^T\}\), respectively.