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Prove that any orthogonal set $S$ consisting of non zero vectors is linearly independent.

My try

By contradiction we assume that the orthogonal set $S$ consisting of non zero vectors is linearly dependent.

Representing $S$ as an orthogonal set, we get

$$0=C_{1}<V_{1},V_{i}>+C_{2}<V_{2},V_{i}>+...C_{k}<V_{k},V_{i}>$$

While representing $S$ as a linearly depedent set, we get

$$0=C_{1}V_{1}+C_{2}V_{2}+...C_{k}V_{k}$$

rearranging the expression gives

$$ C_{1}=\frac{-C_{2}V_{2}+...-C_{k}V_{k}}{V_{1}} $$

Substituting it back to the orthogonal set $S$,we get

$$0=\frac{-C_{2}V_{2}+...-C_{k}V_{k}}{V_{1}}<V_{1},V_{i}>+C_{2}<V_{2},V_{i}>+...C_{k}<V_{k},V_{i}>$$

Which is a contradiction since any fraction with a denominator cant equate to zero.

Is my proof correct and is there a better way to do it?. Could anyone explain. Thanks

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  • $\begingroup$ Some MathJax advice: < and > mean "less than" and "greater than", and produce spacing correct for that meaning only; to make angle brackets, use \langle and \rangle. $\endgroup$ Jul 27, 2015 at 10:25
  • $\begingroup$ What do you means "divide vector"? $\endgroup$
    – GAVD
    Jul 27, 2015 at 10:27

2 Answers 2

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Let $v_1, \dots, v_n$ be orthogonal vectors so that $v_i \neq 0$ for all $i$. Assume that for some $c_1, \dots, c_n$ where for some $j$ holds $c_j \neq 0$ we have $$\sum_{i=1}^n c_i v_i =0.$$ Then $$\left\langle \: v_j \: \middle| \: \sum_{i=1}^n c_i v_i \: \right\rangle = \sum_{i=1}^n c_i \left\langle \: v_j \: \middle| \: v_i \: \right\rangle= c_j \, \left\| v_j\right\|^2=0,$$ which is a contradiction.

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  • $\begingroup$ Is my way of doing correct? It seems to have a similar idea to this? $\endgroup$
    – ys wong
    Jul 27, 2015 at 10:33
  • $\begingroup$ I don't think you can solve it the way you tried. I'm not sure though! $\endgroup$ Jul 27, 2015 at 10:36
  • $\begingroup$ You wrote "Representing S as an orthogonal set, we get....". In the expression after that, all but the $i$:th term will cancel due to the orthogonality of the $v$-vectors. $\endgroup$ Jul 27, 2015 at 10:38
  • $\begingroup$ Becasue i took the inner product to show that the expression equates to $0$ hence they have to cancel each other out. But after substituting the linear dependent expression which then gives a denominator. we know that it is not possible to have a demoniator and still equate to $0$, hence the contradiction $\endgroup$
    – ys wong
    Jul 27, 2015 at 10:47
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Let $S$ be a orthogonal set of non-zero vector. Denote that $S = \{v_0,v_1,\dots,v_n\}$. Assume that $$\sum \alpha_i v_i = 0.$$

Now, you check that $$\langle \sum \alpha_i v_i, v_i \rangle = 0$$ for all i.

And you will imply that $\alpha_i = 0$ for all i.

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