# Step-by-Step Solution Let \( \{v_1, v_2, \dots, v_k\} \) be an **orthogonal set** in a vector space \( V \) with inner product \( \langle \cdot, \cdot \rangle \), and let \( \alpha_1, \alpha_2, \dots, \alpha_k \) be scalars. We want to prove: \[ \left\| \sum_{i=1}^k \alpha_i v_i \right\|^2 = \sum_{i=1}^k |\alpha_i|^2 \|v_i\|^2 \] where \( \|x\|^2 = \langle x, x \rangle \). --- ## Step 1: Write the Norm Squared \[ \left\| \sum_{i=1}^k \alpha_i v_i \right\|^2 = \left\langle \sum_{i=1}^k \alpha_i v_i, \sum_{j=1}^k \alpha_j v_j \right\rangle \] --- ## Step 2: Expand the Inner Product By bilinearity (or sesquilinearity, in complex spaces) and linearity of inner product: \[ = \sum_{i=1}^k \sum_{j=1}^k \alpha_i \overline{\alpha_j} \langle v_i, v_j \rangle \] --- ## Step 3: Use Orthogonality Orthogonality means \( \langle v_i, v_j \rangle = 0 \) for \( i \ne j \), and \( \langle v_i, v_i \rangle = \|v_i\|^2 \): \[ = \sum_{i=1}^k \alpha_i \overline{\alpha_i} \langle v_i, v_i \rangle \] --- ## Step 4: Simplify \[ = \sum_{i=1}^k |\alpha_i|^2 \|v_i\|^2 \] --- ## Final Answer \[ \boxed{ \left\| \sum_{i=1}^k \alpha_i v_i \right\|^2 = \sum_{i=1}^k |\alpha_i|^2 \|v_i\|^2 } \] This is a generalization of the **Pythagorean theorem** for orthogonal vectors in an inner product space. --- ## Additional: Definition of a Norm A **norm** on a vector space \( V \) over a field \( F \) is a function \( \|\cdot\| : V \to \mathbb{R} \) such that for all \( x, y \in V \) and \( \alpha \in F \): 1. **(N1) Non-negativity and definiteness:** \( \|x\| \geq 0 \) and \( \|x\| = 0 \) iff \( x = 0 \) 2. **(N2) Absolute scalability:** \( \|\alpha x\| = |\alpha| \|x\| \) 3. **(N3) Triangle inequality:** \( \|x + y\| \leq \|x\| + \|y\| \) This definition is independent of any inner product.