Is {v1,...,vk} Linearly Independent Given vi.vj=0 When i≠j?

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SUMMARY

The discussion confirms that the set of nonzero vectors {v1,...,vk} is linearly independent if the dot product vi.vj equals zero for all i ≠ j. This property indicates that no vector in the set can be expressed as a linear combination of the others. The proof involves taking the dot product of the linear combination equation 0 = α1 v1 + ... + αk vk with each vector vi, leading to the conclusion that all coefficients αi must be zero.

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suppose v1,...,vk are nonzero vectors with the property that vi.vj=0 whenever i is not equal to j. Prove that {v1,...,vk} is linearly independent.
 
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Suppose [tex]\alpha_1, \ldots, \alpha_k[/tex] are such that [tex]0 = \alpha_1 v_1 + \ldots + \alpha_k v_k[/tex]. Try taking the dot product of this equation with each of the [tex]v_i[/tex]s and see what it tells you about the [tex]\alpha_i[/tex]s.
 
thank u, i got it now...
 

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