Let's solve each part step by step, with explanations after each step. --- ### (a) **Show that \( Px = \sum_{i=1}^k \langle x, u_i \rangle u_i \)** **Step 1:** Recall that \( U \) is the \( n \times k \) matrix whose columns are the orthonormal basis vectors \( u_1, u_2, \ldots, u_k \) for the subspace \( U \subseteq \mathbb{R}^n \). So, \( U = [u_1\ u_2\ ...\ u_k] \), and \( P = UU^T \). For any \( x \in \mathbb{R}^n \): \[ Px = UU^T x \] Here, \( U^T x \) gives a \( k \times 1 \) vector whose \( i \)-th entry is \( u_i^T x = \langle u_i, x \rangle \). **Explanation:** \( U^T x = \begin{bmatrix} u_1^T x \\ u_2^T x \\ \vdots \\ u_k^T x \end{bmatrix} \) This vector contains the projections of \( x \) onto each basis vector \( u_i \). --- **Step 2:** \( U \) times this vector reconstructs the linear combination: \[ U (U^T x) = \sum_{i=1}^k (u_i^T x) u_i = \sum_{i=1}^k \langle x, u_i \rangle u_i \] So, \[ Px = \sum_{i=1}^k \langle x, u_i \rangle u_i \] **Explanation:** This shows that \( Px \) is the orthogonal projection of \( x \) onto the subspace \( U \), written as a linear combination of orthonormal basis vectors. --- **Final Answer for (a):** \[ Px = \sum_{i=1}^k \langle x, u_i \rangle u_i \] where \( \langle x, u_i \rangle \) is the inner product of \( x \) and \( u_i \). --- Would you like to proceed to part (b)?

### (b) **Prove that \( P \) is idempotent and symmetric, i.e., \( P^2 = P \) and \( P^T = P \).** **Step 1:** Recall that \( P = U U^T \), where \( U \) has orthonormal columns. Compute \( P^2 \): \[ P^2 = (U U^T)(U U^T) = U (U^T U) U^T \] **Explanation:** Since \( U^T U \) is a \( k \times k \) matrix, and because the columns of \( U \) are orthonormal, we have: \[ U^T U = I_k \] the \( k \times k \) identity matrix. --- **Step 2:** Substitute back: \[ P^2 = U I_k U^T = U U^T = P \] which shows that \( P \) is **idempotent**. To show symmetry: \[ P^T = (U U^T)^T = (U^T)^T U^T = U U^T = P \] since transpose reverses the order and \( (U^T)^T = U \). Thus, \( P \) is **symmetric**. **Explanation:** Idempotency indicates that projecting twice is the same as projecting once, and symmetry shows that the projection matrix is equal to its transpose. **Final answer:** \[ \boxed{ P^2 = P \quad \text{and} \quad P^T = P } \] Would you like to continue to part (c)?

### (c) **Show that \( x - Px \) is orthogonal to every vector in \( U \).** **Step 1:** Recall that \( Px \) is the projection of \( x \) onto \( U \), and \( x - Px \) is the residual vector. To show that \( x - Px \) is orthogonal to every vector \( u \in U \), it suffices to show that their inner product is zero for all \( u \in U \). Since the basis vectors \( u_i \) span \( U \), it suffices to check for each basis vector \( u_i \): \[ \langle x - Px, u_i \rangle = 0 \] **Explanation:** If the residual vector is orthogonal to each basis vector, it is orthogonal to the entire subspace \( U \). --- **Step 2:** Calculate \( \langle x - Px, u_i \rangle \): \[ \langle x - Px, u_i \rangle = \langle x, u_i \rangle - \langle Px, u_i \rangle \] Recall from part (a): \[ Px = \sum_{j=1}^k \langle x, u_j \rangle u_j \] So, \[ \langle Px, u_i \rangle = \left\langle \sum_{j=1}^k \langle x, u_j \rangle u_j, u_i \right\rangle = \sum_{j=1}^k \langle x, u_j \rangle \langle u_j, u_i \rangle \] Since \( u_j \) are orthonormal: \[ \langle u_j, u_i \rangle = \delta_{ji} \] where \( \delta_{ji} \) is the Kronecker delta, equal to 1 if \( j = i \), and 0 otherwise. Therefore, \[ \langle Px, u_i \rangle = \langle x, u_i \rangle \] Putting it all together: \[ \langle x - Px, u_i \rangle = \langle x, u_i \rangle - \langle x, u_i \rangle = 0 \] Since this is true for all basis vectors \( u_i \), the residual \( x - Px \) is orthogonal to all vectors in \( U \). **Final conclusion:** \[ \boxed{ \text{For all } u \in U,\quad \langle x - Px, u \rangle = 0 } \] which means \( x - Px \) is orthogonal to \( U \). --- ### **Summary of answers:** - **(a)** \( Px = \sum_{i=1}^k \langle x, u_i \rangle u_i \) - **(b)** \( P \) is idempotent and symmetric: \( P^2 = P \) and \( P^T = P \) - **(c)** \( x - Px \) is orthogonal to every vector in \( U \) Let me know if you'd like any further clarification!

Certainly! Here's the final summarized answer: **(a)** The orthogonal projection of \( x \) onto \( U \) can be expressed as: \[ \boxed{ Px = \sum_{i=1}^k \langle x, u_i \rangle u_i } \] where \( u_i \) are the orthonormal basis vectors of \( U \). **(b)** The projection matrix \( P \) is both idempotent and symmetric: \[ \boxed{ P^2 = P \quad \text{and} \quad P^T = P } \] **(c)** The residual vector \( x - Px \) is orthogonal to every vector in \( U \). This completes the proof of all parts.