Question

A = {(a,b,c) where a + b = 2c}
Find a basis for A

Answer 1

But then that would make c always 0, wouldn't it? That doesn't seem right...

Answer 2

(a,0,1/2a) (0,b,1/2b)

Answer 3

i think this should do

Answer 4

yep

Answer 5

you could make use of the standard basis vectors e1=(1,0,0) e2=(0,1,0) e3=(0,0,0).
now first we need to prove this will span A...
so we need to prove that a linear combination of these vectors will be in A...
so suppose we have ANY vector u which is in A where u=(a,b,c), then we know we could rewrite this as u=ae1+be2+ce3
u=ae1+be2+ce3
u=a(1,0,0)+b(0,1,0)+c(0,0,1)
u=(a,0,0)+(0,b,0)+(0,0,c)
u=(a,b,c)
So we could write any vector u from A as a linear combination of A. so e1 and e2 and e3 spans A. now we only need to show that...
c1e1+c2e2+c3e3=0
has only trivial solution... so if we write this in a matrix we'll have
\[\left[\begin{matrix}1 & 0 & 0 \\ 0 & 1&0\\0&0&1\end{matrix}\right]\left(\begin{matrix}c1\\ c2\\c3\end{matrix}\right)=\left(\begin{matrix}0\\0\\0\end{matrix}\right)\]
now since the determinant of the coefficient matrix is not equal to zero then the system will only have exactly one solution which is the trivial solution. so the vectors e1 and e2 and e3 are linearly independent.

Therefore, e1, e2, and e3 will span A

Answer 6

@anonymoustwo44, this does,t span the desired space

Answer 7

@myko please prove that as well as you answer

Answer 8

(a,0,1/2a) (0,b,1/2b)
or if you like:
(1,0,1/2), (0,1,1/2) is the needed base

Answer 9

@anonymoustwo44
u=ae1+be2+ce3
u=a(1,0,0)+b(0,1,0)+c(0,0,1)
u=(a,0,0)+(0,b,0)+(0,0,c)
u=(a,b,c)
it's wrong!!!
your e3 as stated befor is (0,0,0)
and even more, (a,b,c) the way you constract it with your base do not necesarly belong to A, since A={(a,b,c)} | c =(a+b)/2 !!!!!!!!!!!!!

Answer 10

my bad... my e3 is (0,0,1)

Answer 11

but even this way it doesn't work....
since A={(a,b,c)} | c =(a+b)/2 !!!!!!!!!!!!!

Answer 12

it will work even if c=(a+b)/2
(a,b,c)=(a,b,(a+b)/2)=(a,0,0)+(0,b,0)+(0,0,(a+b)/2))
=a(1,0,0)+b(0,1,0)+((a+b)/2)(0,0,1)
=ae1+be2+((a+b)/2)e3 <--- see, even though c=(a+b)/2 we could still write it in as a linear combination of e1, e2, and e3 so e1 e2 and e3 spans A

Answer 13

why mad?