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)

Answer 1

(1)

Multiplying by 3 both sides of the equality you get that

`3u = 3c_1v_1+3c_2v_2+3c_3v_3`

3u is in the Span of the vectors
`\{v_1,v_2,v_3\}`.

(2)

That's not true, consider the following counter example.

`v_1 = (0,0,0,1)\\v_2 = (0,0,1,0)\\v_3 = (0,1,0,0)\\v_4 = (1,0,0,0)\\u = (0,1,1,1)`

`u` is a linear combination of
`v_1,v_2,v_3` but is NOT a linear combination of
`v_1,v_2,v_3,v_4.`

Explanation:

(1)

As the hint indicates, you know that

`u = c_1 v_1 + c_2v_2+c_3v_3`

Then, if you multiply both sides of the equality by 3, you get that

`3u = 3c_1v_1+3c_2v_2+3c_3v_3`

And that's it. 3u is in the Span of the vectors
`\{v_1,v_2,v_3\}`

(2)

That's not true, consider the following counter example.

`v_1 = (0,0,0,1)\\v_2 = (0,0,1,0)\\v_3 = (0,1,0,0)\\v_4 = (1,0,0,0)\\u = (0,1,1,1)`

`u` is a linear combination of
`v_1,v_2,v_3` but is NOT a linear combination of
`v_1,v_2,v_3,v_4.`