Answer:
(1)
Multiplying by 3 both sides of the equality you get that
[tex]3u = 3c_1v_1+3c_2v_2+3c_3v_3[/tex]
3u is in the Span of the vectors [tex]\{v_1,v_2,v_3\}[/tex].
(2)
That's not true, consider the following counter example.
[tex]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)[/tex]
[tex]u[/tex] is a linear combination of [tex]v_1,v_2,v_3[/tex] but is NOT a linear combination of [tex]v_1,v_2,v_3,v_4.[/tex]
Step-by-step explanation:
(1)
As the hint indicates, you know that
[tex]u = c_1 v_1 + c_2v_2+c_3v_3[/tex]
Then, if you multiply both sides of the equality by 3, you get that
[tex]3u = 3c_1v_1+3c_2v_2+3c_3v_3[/tex]
And that's it. 3u is in the Span of the vectors [tex]\{v_1,v_2,v_3\}[/tex]
(2)
That's not true, consider the following counter example.
[tex]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)[/tex]
[tex]u[/tex] is a linear combination of [tex]v_1,v_2,v_3[/tex] but is NOT a linear combination of [tex]v_1,v_2,v_3,v_4.[/tex]