Question

True/False Question: If sppan{u,v}=W where u not equal to v. Then dim(W)=2.

Answer: False

Reasoning let u=1, v=2 then span(1,2}=R but dim(R)=1 not 2.

I know the answer is false. please tell me whether my reasoning is correct.

Answer 1

your reasoning is wrong

how

span{u,v} means set of all linear combinations of u and v

​​​And span{u,v}=W

then dimensions of W can b 1 or 2

depending on u and v

and u can not say dim(W)= 2 that's false

now two cases arise

if u is not multiple of v

(or v is not multiple of u)

i.e, if u and v are not multiples of each other in that case u and v become linearly independent and hence the set which is spanned by u and v has dimension equal to 2

now however if u and v are multiples of each other then in that case u and v become linearly dependent and the set which is spanned by u and v has dimension equal to 1

in your case u took u =1 v=2

But here u is multiple of v

as v=2u

So they are actually dependent

so dimension is equal to 1 not 2

...

remember a set (say) S is a basis of V

if it

(1) spans V

(2) is linearly independent

...

also note that in case of two vectors

if vectors are not multiples of each other then they are Linearly independent

if multiples of each other then they are Linearly dependent

..

hope this will help you

thank you