Let's solve the problem step by step: --- ## (a) **Show that \(Ax = \operatorname{proj}_V x\)** **Step 1:** Given \(A = uu^T\) and \(V = \text{Span}\{u\}\). For any \(x \in \mathbb{R}^n\), \[ A x = u(u^T x) \] - \(u^T x\) is a scalar (dot product). - So, \(A x\) is the vector \(u\) scaled by \(u^T x\). **Step 2:** Recall the formula for the projection of \(x\) onto the subspace \(V\) spanned by a unit vector \(u\): \[ \operatorname{proj}_V x = (u^T x)u \] (Since \(u\) is a unit vector, \(u^T u = 1\).) **Step 3 (Explanation):** Comparing the two: \[ A x = u(u^T x) = (u^T x)u = \operatorname{proj}_V x \] **Conclusion for (a):** \[ \boxed{A x = \operatorname{proj}_V x} \] So, multiplying by \(A\) projects any vector \(x\) onto the span of \(u\). --- ## (b) **Show that \(A\) has eigenvalue 1 and determine a corresponding eigenspace.** **Step 1:** Let’s look for eigenvectors \(v\) such that \(A v = \lambda v\). **Step 2:** Try \(v = u\): \[ A u = u(u^T u) = u \cdot 1 = u \] So, \(\lambda = 1\). **Step 3 (Explanation):** Any scalar multiple of \(u\) also works, so the eigenspace corresponding to eigenvalue 1 is \(\text{Span}\{u\}\). **Conclusion for (b):** \[ \boxed{ \text{Eigenvalue: } 1, \quad \text{Eigenspace: } \text{Span}\{u\} } \] --- ## (c) **Show that \(A^2 = A\) (idempotent).** **Step 1:** Compute \(A^2\): \[ A^2 = (uu^T)(uu^T) \] **Step 2:** Use associativity: \[ A^2 = u(u^T u)u^T = u(1)u^T = uu^T = A \] (since \(u^T u = 1\) for a unit vector) **Step 3 (Explanation):** So, \(A^2 = A\), which means \(A\) is idempotent. **Conclusion for (c):** \[ \boxed{A^2 = A} \] --- ## **Final Answers:** **(a)** \(Ax = \operatorname{proj}_V x\) **(b)** Eigenvalue 1, eigenspace \(\text{Span}\{u\}\) **(c)** \(A^2 = A\) (idempotent)