linear algebra
$$\{(1,0,0)^T, (0,1,1)^T, (1,0,1)^T, (1,2,3)^T\}$$
What I did first was write the linear combination of these vectors which I do not know how to format on here but is simply:
$$c_1(1,0,0) + c_2(0,1,1) + c_3(1,0,1) + c_4(1,2,3) = (x, y, z)$$
where the vectors are column vectors.
Then I get the equations $c_1 + c_3 + c_4 = x, c_2 + 2c_4 = y, c_2 + c_3 + 3c_4 = z$.
And this is where I get stuck. How do I prove that the vectors span $\Bbb R^3$ from here?
Two sets of vectors in the same vector space, $S_1$ and $S_2$, span the same subspace if and only if:
There may be ways of inferring these properties without actually showing them directly (such as dimension arguments like user6312 suggests), but it really ends up coming down to showing this holds or that this cannot hold.
Note that these conditions are not intrinsic: it's almost never enough to just look at $S_1$ without thinking about $S_2$, and to then look at $S_2$ without looking at $S_1$; it is only in very extreme circumstances that this suffices (when you can prove that the "sizes" of the spans don't match, a dimension argument) that this can prove to be enough.
So just figuring out if the sets are linearly dependent or independent is not enough in this case.
So: are $(-2,-6,0)$ and $(1,1,-2)$ each linear combinations of $(1,2,-1)$, $(0,1,1)$, and $(2,5,-1)$? Yes: you can try solving the two systems of linear equations: $$\begin{align*} \alpha_1 \left(\begin{array}{r}1\\2\\-1\end{array}\right) + \beta_1\left(\begin{array}{c}0\\1\\1\end{array}\right) + \gamma_1\left(\begin{array}{r}2\\5\\-1\end{array}\right) &= \left(\begin{array}{r}-2\\-6\\0\end{array}\right)\\ \alpha_2 \left(\begin{array}{r}1\\2\\-1\end{array}\right) + \beta_2\left(\begin{array}{c}0\\1\\1\end{array}\right) + \gamma_2\left(\begin{array}{r}2\\5\\-1\end{array}\right) &= \left(\begin{array}{r}1\\1\\-2\end{array}\right) \end{align*}$$ and see if they each have solutions. (It can even be done both at once, by doing row reduction on $$\left(\begin{array}{rrr|rr} 1 & 0 & 2 & -2 & 1\\ 2 & 1 & 5 & -6 & 1\\ -1 & 1 & -1 & 0 & -2 \end{array}\right).$$ If either system has no solutions, then you know that not every vector in $S_2$ is in the span of $S_1$ and you are done; if both systems have solutions, then every vector in $S_2$ is in the span of $S_1$, so $\mathrm{span}(S_2)\subseteq \mathrm{span}(S_1)$.
Then you need to see if the converse inclusion holds: is every vector in $S_1$ a linear combination of the vectors in $S_2$? That is, can we solve the three systems of linear equations? $$\begin{align*} \rho_1\left(\begin{array}{r}-2\\-6\\0\end{array}\right) + \sigma_1 \left(\begin{array}{r}1\\1\\-2\end{array}\right) &= \left(\begin{array}{r}1\\2\\-1\end{array}\right)\\ \rho_2\left(\begin{array}{r}-2\\-6\\0\end{array}\right) + \sigma_2 \left(\begin{array}{r}1\\1\\-2\end{array}\right) &= \left(\begin{array}{r}0\\1\\1\end{array}\right)\\ \rho_3\left(\begin{array}{r}-2\\-6\\0\end{array}\right) + \sigma_3 \left(\begin{array}{r}1\\1\\-2\end{array}\right) &= \left(\begin{array}{r}2\\5\\-1\end{array}\right) \end{align*}$$ If we can solve all, then each vector in $S_1$ is in the span of $S_2$, so $\mathrm{span}(S_1)\subseteq \mathrm{span}(S_2)$; together with the previous inclusion, this would show the spans are equal. If some equation cannot be solved, then not every vector in $S_1$ is in the span of $S_2$, so the spans are different.
(There is one thing that you can rescue from your efforts: since you proved that the set $S_1$ is linearly dependent, you can extract from it a linearly independent set (in this case, for instance, the first two vectors), and replace $S_1$ with that set of two vectors (because the third vector is a linear combination of the first two). That will mean that checking that "every vector in $S_1$ is a linear combination of the vectors in $S_2$" and checking that "every vector in $S_2$ is a linear combination of the vectors in $S_1$" will be simpler: instead of checking five things, you only need to check four.)
What you would do is to take the determinant of a $3\times 3$ matrix whose columns are the vectors that you conjecture to be independent (thus they span $\mathbb R^3$). Notice that since you have $4$ of these vectors you select three. The procedure is to take the determinant. If it is different from $0$ then the $3$ vectors you selected are independent and they span $\mathbb R^3$. If the $\det$ is $0$ select other $3$ vectors... If all of the combinations of vectors that form a $3\times 3$ matrix have determinant $0$, then the vectors are dependent and they don't span $\mathbb R^3$. If any of these matrices formed by putting $3$ vectors as its columns has determinant different from $0$, then the other vector can be written as a combination of the three independent vectors you've found.
For instance:
$$\det\left[\begin{array}{r,r,r}1&0&0\\0&1&4\\0&-1&-3\end{array}\right]=-3+4=1\neq0.$$
Therefore the first three vectors you mentioned are independent and span $\mathbb R^3$.
Moreover, we know that if there are constants $k_1,k_2,k_3,k_4$, at least one of them different from $0$, such that for every $c_1,c_2,c_3,c_4\in\mathbb R$ (these come from the vector solution $\mathbf c$) we have that (if we name the columns $C_1,C_2,C_3,C_4$) $k_1C_1+k_2C_2+k_3C_3+k_4C_4=\boldsymbol0$, then the columns are dependent.
So let's assume that the columns are dependent and we will try to reach a contradiction. If $k_1\neq 0$ the contradiction is trivial since the first column of your matrix is $\left[\begin{array}{c}1\\0\\0\end{array}\right]\Longrightarrow k_1c_1\left[\begin{array}{c}1\\0\\0\end{array}\right]+\boldsymbol \xi\neq \left[\begin{array}{c}0\\0\\0\end{array}\right]\text{for some } c_1\in\mathbb R.$ The same happens if we select $k_4\neq0$. So the thing turns interesting when $k_1,k_4=0$. Then we are left with the two columns of the middle. Of course if additionally $k_2=0$ or $k_3=0$ then the contradiction is trivial as well. But if $k_2,k_3\neq 0$, then we have the system: \begin{align} c_2+4c_3=u_2, && -c_2-3c_3=u_3. \end{align} $\text{That yields } c_3=u_2+u_3.$ Therefore $c_3$ does not depend on $c_2$ and it is impossible that $k_2(c_2+4c_3)+k_3(-c_2-3c_3)=0,\,\forall c_2,c_3\in\mathbb R$.
Best Answer
Hint: If you can generate $[1,0,0]^T$, $[0,1,0]^T$ and $[0,0,1]^T$ with these vectors, then you can span the whole space.
Using this it's easy to see that just the first three vectors span the space.