Let $\displaystyle S=\{v_1,v_2,...,v_k\}$ be an orthogonal set of vectors in $\displaystyle R^n$. If $\displaystyle S$ is linearly dependent, prove that one of the $\displaystyle v_j$ must be the zero vector.
Let $\displaystyle S=\{v_1,v_2,...,v_k\}$ be an orthogonal set of vectors in $\displaystyle R^n$. If $\displaystyle S$ is linearly dependent, prove that one of the $\displaystyle v_j$ must be the zero vector.
Since $\displaystyle \{v_{1},\ldots,v_{k}\}$ is a linearly dependent set, there is a non-trivial solution to the equation
$\displaystyle \lambda_{1}v_{1}+\ldots+\lambda_{k}v_{k}=0, $
i.e. at least one $\displaystyle \lambda_{j}$ is non-zero.
Then compute
$\displaystyle \begin{array}{rcl} 0&=&\langle v_{j},0\rangle \\ &=&\langle v_{j}, \sum_{i=1}^{k} \lambda_{i}v_{i} \rangle \\ &=&\sum_{i=1}^{k} \lambda_{i}\langle v_{j},v_{i} \rangle \\ &=&\lambda_{i}\langle v_{j},v_{j} \rangle,\\ \end{array}$
which implies that $\displaystyle \|v_{j}\|^{2} = 0$ because $\displaystyle \lambda_{j}\neq 0$, so $\displaystyle v_{j}=0.$
No problem
It sounds like you already know this, but if you're using the dot product you can just rewrite $\displaystyle \langle v, w \rangle$ as $\displaystyle v\cdot w$ and $\displaystyle \| v \|$ as $\displaystyle |v|$ and everything still works.
Sorry if this is obvious! Just in case you're really unfamiliar with inner-product notation.