Let's work through each part step by step. Given: \[ A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 1 & 2 \\ \end{bmatrix} \] --- ### (a) **Determine the eigenvalues of A.** The eigenvalues \(\lambda\) satisfy: \[ \det(A - \lambda I) = 0 \] Compute \(A - \lambda I\): \[ A - \lambda I = \begin{bmatrix} 1-\lambda & 0 & 0 \\ 0 & 1-\lambda & 0 \\ 1 & 1 & 2-\lambda \\ \end{bmatrix} \] The determinant is: \[ \det(A - \lambda I) = \begin{vmatrix} 1-\lambda & 0 & 0 \\ 0 & 1-\lambda & 0 \\ 1 & 1 & 2-\lambda \\ \end{vmatrix} \] Since the first two columns are upper triangular with zeros below, expand along the first two rows: \[ = (1-\lambda) \cdot \begin{vmatrix} 1-\lambda & 0 \\ 1 & 2-\lambda \end{vmatrix} \] Now expand the \(2 \times 2\) determinant: \[ = (1-\lambda) \left[ (1-\lambda)(2-\lambda) - (1)(0) \right] = (1-\lambda)^2 (2-\lambda) \] Set equal to zero: \[ (1-\lambda)^2 (2-\lambda) = 0 \] So the eigenvalues are: \[ \boxed{1, 1, 2} \] (1 with multiplicity 2, 2 with multiplicity 1) --- ### (b) **Find a basis of each eigenspace of A. Describe the eigenspaces geometrically.** #### **For \(\lambda = 1\):** Solve \( (A - I)\vec{v} = 0 \): \[ A - I = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 1 & 1 \\ \end{bmatrix} \] Let \(\vec{v} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}\). So: \[ \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 1 & 1 \\ \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ x + y + z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \] So \(x + y + z = 0\). The general solution is: \[ \vec{v}_1 = \begin{bmatrix} x \\ y \\ z \end{bmatrix} = x\begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} + y\begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix} \] **Basis for eigenspace \(\lambda=1\):** \[ \left\{ \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix} \right\} \] **Geometric description:** It's a plane through the origin in \(\mathbb{R}^3\) defined by \(x + y + z = 0\). --- #### **For \(\lambda = 2\):** Solve \( (A - 2I)\vec{v} = 0 \): \[ A - 2I = \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 1 & 1 & 0 \\ \end{bmatrix} \] Let \(\vec{v} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}\). First two rows: \[ -1 \cdot x = 0 \implies x = 0\\ -1 \cdot y = 0 \implies y = 0 \] Third row: \[ 1 \cdot x + 1 \cdot y + 0 \cdot z = 0 \implies 0 + 0 = 0 \] So \(z\) is free. So: \[ \vec{v}_2 = \begin{bmatrix} 0 \\ 0 \\ z \end{bmatrix} = z\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \] **Basis for eigenspace \(\lambda=2\):** \[ \left\{ \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \right\} \] **Geometric description:** It's the \(z\)-axis in \(\mathbb{R}^3\). --- ### (c) **Find three linearly independent eigenvectors of A. (Do they form a basis of \(\mathbb{R}^3\)?)** From above, take: - For \(\lambda=1\): \(\vec{v}_1 = \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}\), \(\vec{v}_2 = \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix}\) - For \(\lambda=2\): \(\vec{v}_3 = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}\) These are linearly independent. To check, suppose \[ a\vec{v}_1 + b\vec{v}_2 + c\vec{v}_3 = 0 \] \[ a\begin{bmatrix}1\\0\\-1\end{bmatrix} + b\begin{bmatrix}0\\1\\-1\end{bmatrix} + c\begin{bmatrix}0\\0\\1\end{bmatrix} = \begin{bmatrix}a\\b\\-a-b+c\end{bmatrix} = \begin{bmatrix}0\\0\\0\end{bmatrix} \] So \(a=0\), \(b=0\), and \(-a-b+c=0 \implies c=0\). **Do they form a basis?** Yes, they are three linearly independent vectors in \(\mathbb{R}^3\), so they form a basis. --- ### (d) **Let \(k \in \mathbb{N}\). What is \(A^k \vec{v}_1\)? What is \(A^k \vec{v}_2\)? What is \(A^k \vec{v}_3\)?** Recall: If \(\vec{v}\) is an eigenvector with eigenvalue \(\lambda\), then \(A^k \vec{v} = \lambda^k \vec{v}\). So: - \(\vec{v}_1\) and \(\vec{v}_2\) have eigenvalue \(1\) - \(\vec{v}_3\) has eigenvalue \(2\) Thus: \[ A^k \vec{v}_1 = 1^k \vec{v}_1 = \vec{v}_1 \] \[ A^k \vec{v}_2 = 1^k \vec{v}_2 = \vec{v}_2 \] \[ A^k \vec{v}_3 = 2^k \vec{v}_3 \] --- ### (e) **Suppose \(v = c_1\vec{v}_1 + c_2\vec{v}_2 + c_3\vec{v}_3\). Express \(Av\) as a linear combination of \(\vec{v}_1, \vec{v}_2, \vec{v}_3\).** We know: \[ Av = c_1A\vec{v}_1 + c_2A\vec{v}_2 + c_3A\vec{v}_3 \] But \(A\vec{v}_1 = 1\vec{v}_1\), \(A\vec{v}_2 = 1\vec{v}_2\), \(A\vec{v}_3 = 2\vec{v}_3\): \[ Av = c_1 \vec{v}_1 + c_2 \vec{v}_2 + 2c_3 \vec{v}_3 \] Or, explicitly, \[ \boxed{ Av = c_1\vec{v}_1 + c_2\vec{v}_2 + 2c_3\vec{v}_