OK, I am having a real problem with this and I am desperate.
I have a set of vectors. {(1,0.-1), (2,5,1), (0,-4,3)}
How do I check is this is a basis for $R^3?$
My text says a basis B for a vector space V is a linearly independent subset of V that generates V. OK then. I need to see if these vectors are linearly independent, yes?
If that is so, then for these to be linearly independent the following must be true:
$a_1v_1 + a_2v_2 + … + a_nv_n $ ≠ 0 for any scalars $a_i$
Is this the case or not?
If it is, then I just have to see if
$a_1(1,0.-1)+ a_2(2,5,1)+ a_3(0,-4,3)$ = 0
or
$a_1 + 2a_2 + 0a_3 = 0$
$0a_1 + 5a_2 – 4a_3 = 0$
$-a_1 + a_2 + 3a_3 = 0$
has a solution.
Adding these equations up I get $8a_2 – a_3 = 0$ or $a_3 = 8a_2$
so $5a_2 – 32a_2 = 0$ which gets me $a_2 = 0$ and that implies $a_1 = 0 $and$ a_3=0$ as well.
So they are all linearly dependent and thus they are not a basis for $R^3$.
Something tells me that this is wrong. But I am having a hell of a time figuring this stuff out. Please someone help, and I ask: pretend I am the dumbest student you ever met.
Best Answer
A set of vectors $v_1, v_2, ..., v_n$ is linearly independent if and only if we have that
$$a_1v_1 + a_2v_2 + ... + a_nv_n = 0 \;\;$$
only when $ a_1 = a_2 = ... = a_n = 0 $.
(After all, any linear combination of three vectors in $\mathbb R^3$, when each is multiplied by the scalar $0$, is going to be yield the zero vector!)
So you have, in fact, shown linear independence. And any set of three linearly independent vectors in $\mathbb R^3$ spans $\mathbb R^3$. Hence your set of vectors is indeed a basis for $\mathbb R^3$.