(I already have the answer to this question but i just need someone to explain the concept behind the logic being used here.)
v1 = (1,0,2)
v2 = (3,-1,1)
v3 = (2, -1,-1)
v4 = (4,-1, 3)
So my professor told us to write the vectors above in the equation below.
$$\pmatrix{1&3&2&4\\0&-1&-1&-1\\2&1&-1&3}\cdot\pmatrix{x\\y\\z\\w}=\pmatrix{b1\\b2\\b3}$$
(b1, b2, and b3 are arbitrary and can equal ANY vector in R^3)
he then used row reduction to get the solutions for x, y, z and w and we got the matrix below
$$
A= \begin{pmatrix} 1 & 3 & 2 & 4 & |b1\\ 0 & -1 & -1 & -1 & |b2\\0 & 0&0&0 &| b3 – 2b1 – 5b2
\end{pmatrix}
$$
so it is obvious that there is an inconsistency on the last row of the matrix above,
0 = b3 – 2b1 -5b3
but my question is, why does this inconsistency tell me that the vectors do NOT span R^3?
can someone explain to me why the vectors do not span R^3?
Best Answer
The inconsistency in that system shows that there is no solution to $xv_1+yv_2+zv_3+wv_4=b$ provided $b_3-2b_1-5b_2\neq 0$. Since $b\in\Bbb R^3$, can we have that $v_1,v_2,v_3,v_4$ span $\Bbb R^3$?
Addendum: Recall that $$\operatorname{span}(v_1,v_2,v_3,v_4)=\{v\in\Bbb R^3:v=c_1v_1+c_2v_2+c_3v_3+c_4v_4,~c_i\in\Bbb R\}$$ That is, the span of a collection of vectors is the set of linear combinations of those vectors. So the inconsistency in the system you have shows us that there is no solution to $xv_1+yv_2+zv_3+wv_4=b$ for an arbitrary vector $b\in\Bbb R$. Hence, $b$ is not a linear combination of $v_1,v_2,v_3,v_4$. So can we say that $v_1,v_2,v_3,v_4$ span $\Bbb R^3$?
In general, to show some vectors do not span a vector space, we can just show that there is a vector in the space which is not a linear combination of those vectors. Linear dependence does not imply that they do not span $\Bbb R^3$. For example, $e_1,e_2,e_3,e_1+e_2$ span $\Bbb R^3$ however they are clearly linearly dependent. In fact, any collection of more than $3$ vectors will be linearly dependent in $\Bbb R^3$, however they may or may not span $\Bbb R^3$.