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 $\mathbb{R}^3?$
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:
$a_1v_1 + a_2v_2 + ... + a_nv_n \neq 0$ 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 $\mathbb{R}^3$.
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.