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}.

Answer 1

Answer:

(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 $v_1,v_2,v_3,v_4.$

Step-by-step 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 $v_1,v_2,v_3,v_4.$