1. **Problem Statement:** Prove that a set of mutually orthogonal non-zero vectors is always linearly independent. 2. **Definitions and Formula:** - Vectors $\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n$ are mutually orthogonal if $\mathbf{v}_i \cdot \mathbf{v}_j = 0$ for all $i \neq j$. - A set of vectors is linearly independent if the only solution to $c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + \cdots + c_n \mathbf{v}_n = \mathbf{0}$ is $c_1 = c_2 = \cdots = c_n = 0$. 3. **Proof:** - Assume $c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + \cdots + c_n \mathbf{v}_n = \mathbf{0}$. - Take the dot product of both sides with $\mathbf{v}_k$ for some $k$: $$\left(c_1 \mathbf{v}_1 + \cdots + c_n \mathbf{v}_n\right) \cdot \mathbf{v}_k = \mathbf{0} \cdot \mathbf{v}_k = 0$$ - By distributivity and orthogonality: $$c_1 (\mathbf{v}_1 \cdot \mathbf{v}_k) + \cdots + c_k (\mathbf{v}_k \cdot \mathbf{v}_k) + \cdots + c_n (\mathbf{v}_n \cdot \mathbf{v}_k) = 0$$ - Since $\mathbf{v}_i \cdot \mathbf{v}_k = 0$ for $i \neq k$, this reduces to: $$c_k \|\mathbf{v}_k\|^2 = 0$$ - Because $\mathbf{v}_k$ is non-zero, $\|\mathbf{v}_k\|^2 > 0$, so: $$c_k = 0$$ - This holds for all $k = 1, 2, \ldots, n$, so all coefficients $c_k$ are zero. 4. **Conclusion:** The only solution to the linear combination equaling zero is the trivial solution, so the set of mutually orthogonal non-zero vectors is linearly independent.