# Thread: [SOLVED] If S is linearly dependent, prove that one of its vectors must be the zero v

1. ## [SOLVED] If S is linearly dependent, prove that one of its vectors must be the zero v

Let $S=\{v_1,v_2,...,v_k\}$ be an orthogonal set of vectors in $R^n$. If $S$ is linearly dependent, prove that one of the $v_j$ must be the zero vector.

2. Since $\{v_{1},\ldots,v_{k}\}$ is a linearly dependent set, there is a non-trivial solution to the equation

$\lambda_{1}v_{1}+\ldots+\lambda_{k}v_{k}=0,$

i.e. at least one $\lambda_{j}$ is non-zero.

Then compute

$\begin{array}{rcl} 0&=&\langle v_{j},0\rangle \\ &=&\langle v_{j}, \sum_{i=1}^{k} \lambda_{i}v_{i} \rangle \\ &=&\sum_{i=1}^{k} \lambda_{i}\langle v_{j},v_{i} \rangle \\ &=&\lambda_{i}\langle v_{j},v_{j} \rangle,\\ \end{array}$

which implies that $\|v_{j}\|^{2} = 0$ because $\lambda_{j}\neq 0$, so $v_{j}=0.$

3. Originally Posted by nimon
Since $\{v_{1},\ldots,v_{k}\}$ is a linearly dependent set, there is a non-trivial solution to the equation

$\lambda_{1}v_{1}+\ldots+\lambda_{k}v_{k}=0,$

i.e. at least one $\lambda_{j}$ is non-zero.

Then compute

$\begin{array}{rcl} 0&=&\langle v_{j},0\rangle \\ &=&\langle v_{j}, \sum_{i=1}^{k} \lambda_{i}v_{i} \rangle \\ &=&\sum_{i=1}^{k} \lambda_{i}\langle v_{j},v_{i} \rangle \\ &=&\lambda_{i}\langle v_{j},v_{j} \rangle,\\ \end{array}$

which implies that $\|v_{j}\|^{2} = 0$ because $\lambda_{j}\neq 0$, so $v_{j}=0.$
Our course material didn't use inner product brackets, but your proof seems good. Thanks for the help.

4. No problem

It sounds like you already know this, but if you're using the dot product you can just rewrite $\langle v, w \rangle$ as $v\cdot w$ and $\| v \|$ as $|v|$ and everything still works.

Sorry if this is obvious! Just in case you're really unfamiliar with inner-product notation.