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**Unformatted text preview: **is -1-1 1 . 3. (4 pts) Suppose that V is a vector space with a basis B = { ~v 1 ,~v 2 } . Suppose that ~v V . State what vector space the vector [ ~v ] B is in. Suppose that [ ~v ] B is nonzero. Show that ~v would have to be nonzero as well. [ ~v ] B is, by denition, the column vector of weights required to make ~v as a linear combination of the vectors in B . Since B has two vectors in it, this column vector has two entries, so it is in R 2 . Suppose that [ ~v ] B = [ a b ]. Then, ~v = a~v 1 + b~v 2 . Since B is linearly independent, this linear combination can only be zero if a = b = 0. This was assumed not to be the case (since [ ~v ] B is nonzero), so ~v 6 = ~ 0....

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