Answer:
See the proof below.
Step-by-step explanation:
For this proof we need to begin with the assumption, on this case the vectors u and v are different [tex] u \neq v [/tex]
We are assuming that {u,v} is a basis of V and distinct, so then we have that V is a 2 dimensional vector space and we have the following condition:
[tex] a_1 u + a_2 v =0[/tex] if [tex] a_1 =a_2 = 0[/tex] (1)
We need to show that {u+v,au} is also a basis for V, so then we need to show this:
[tex] a_1 (u+v) +a_2 (au) =0[/tex]
That is equivalent to:
[tex] a_1 u + a_1 v + a_2 au =0[/tex]
We can take common factor u and we got this:
[tex] (a_1 +a_2 a)u + a_1 v=0[/tex]
from condition 1 we need to have this:
[tex] a_1 +a_2 a =0 , a \neq 0[/tex]
So then [tex] a_2 a =0[/tex] and [tex]a_2 =0[/tex]
And we will see that [tex] a_1 = a_2 =0[/tex] so then we can conclude that {u+v, au} is a linearly independent set of two vectors and are a basis for V.
{u+v, au} is a basis for the space V.
