That statement doesn't even make sense. It talks about "the vector" but "independent" and "dependent" apply tosetsof vectors, not individual vectors (a set containingonevector is "linearly independent" if and only that vector is not the 0 vector.)

Okay, I didn't notice the title specified "a set containing single vector" so my parenthetical statement applies.

No, it doesnt. Look at the set {<1, 1, 1>, <2,2,2>}. -2<1, 1, 1>+ 1<2, 2, 2>= <0, 0, 0>.b) Every linear dependent set conatins the zero vector?

this means that c1, c2, and c3 can be multiplied by the v1,v2,v3.... and get the0with c1,c2,c3.... not having to be zero right? That means that this statement is True.

Thanks for your help