Mathematics: Let u and v be distinct vectors of a vector spacev. show that if {u,v} is a basis for v anda is a non-zero scalar, then {u+v,au} is also abasis for v.

exceptStep-by-step explanation:what are the following are you talk in about???...

Let u and v be distinct vectors of a vector spacev.

jimena15

14.01.2022

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 $u \neq v$

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:

a_1 u + a_2 v =0 if a_1 =a_2 = 0 (1)

We need to show that {u+v,au} is also a basis for V, so then we need to show this:

a_1 (u+v) +a_2 (au) =0

That is equivalent to:

a_1 u + a_1 v + a_2 au =0

We can take common factor u and we got this:

(a_1 +a_2 a)u + a_1 v=0

from condition 1 we need to have this:

$a_1 +a_2 a =0 , a \neq 0$

So then a_2 a =0 and a_2 =0

And we will see that a_1 = a_2 =0 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.

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