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

**Runty** · April 13th 2010, 11:29 AM

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.

**nimon** · April 13th 2010, 12:11 PM

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 $v_{j}=0.$

**Runty** · April 13th 2010, 12:32 PM

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.

**nimon** · April 13th 2010, 12:46 PM

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.

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