(1) Let { v 1 , v 2 , v 3 } be a set of vectors in R n . If u is Span { v 1 , v

Question

(1) 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}

(2) Let u, v1,v2,v3 and v4 be vectors in the Rn . If u can be written as a linear combination of v1,v2, and v3 show that u can also be written as a linear combination of v1,v2,v3 and v4 .

(Hint: Look at the previous question and remember you can use 0 as a coefficient in your equation.)

(3) Let u, v, w1, w2, and w3 be vectors in the Rn. If u and v can both be written as linear combinations of w1, w2 and w3 show that u+v can also be written as a linear combination of w1, w2, and w3.

(hint:this is similar to the previous ones. Remember to choose different letters for your coefficients for u and v)

Explanation / Answer

1.

Let u = c1v1+c2v2+c3v3

=>

3u = (3c1)*v1 +(3c2)*v2 + (3c3)*v3

=>
3u belongs to span {v1,v2,v3}

2.

given that u can be written as a linear combination of v1,v2,v3

=>

there exists scalars c1,c2,c3 such that

u = c1v1+c2v2+c3v3

= c1v1+c2v2+c3v3 +0*v4

=>
u can be written as a linear combination of v1,v2,v3,v4

thus proved

3)

given that u,v can be written as a linear combination of w1,w2,w3

=>

there exists scalars c1,c2,c3, d1,d2,d3 such that

u = c1w1+c2w2+c3w3

v = d1w1+d2w2+d3w3

=>

u+v = (c1+d1)w1 + (c2+d2)w2+ (c3+d3)w3

=>

u+v can also be written as a linear combination of w1,w2,w3

thus proved

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