Let {v1,v2,v3} be a set of vectors in Rn . If u is Span {v1,v2,v3}, show that 3u is in Span {v1,v2,v3}.(Hint: since you are told that u is in the span {v1,v2,v3}, you can automatically say that there are scalars c1,c2 and c3 so that u = c1v1+c2v2+c3v3. Your goal now is to find a way to write 3u as a linear combination of {v1,v2,v3}

See proof below

Step-by-step explanation:

We will use the hint. The statement of the hint holds true, as the linear span of a set of vectors T is equal to the set of linear combinations of vectors in T.

Denote the linear span of vectors with the curly brackets < >, that is,

Let , then u is a linear combination of v1,v2,v3, that is, there exist scalars such that . Multiply by 3 in both sides to get , with

Since , as real numbers are closed under multiplication. Therefore 3u is a linear combination of the vectors , that is,  

answer: -1

step-by-step explanation:

make sure you have version 2.50

try looking in the very back of the book : )

step-by-step explanation:

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