Do these vectors form a basis? I'm researching a potential algorithm, and I'm hoping that someone can verify my calculations.
I have sets of vectors in $\mathbb{R}^6$ that I can use.  They have a corresponding value associated with them, but this relation is not necessarily a simple one.  I'd like to find the value associated with the vector $(0,0,1,0,0,0)$.  So I'm wondering if I can perform linear algebra, using the vectors that I can create, to do so.
One vector I can create is $(a,a,a,0,0,0)$ for any real $a$.  A second is $(b,b,b,b,b,b)$.
I can also create vectors of the form $(c^0,c^1,c^2,c^0,c^1,c^2)$ for some real number $c$, where $c^k$ is just simply $c$ taken to the $k$th power.  A second form that I can create is $(0, d, 2d, 0, 0, 0)$ for real $d$.
I am wondering if these vectors form a complete basis for $\mathbb{R}^6$. 
In case it helps, I can use as many vectors as I want, adding and/or subtracting them, as long as they are of the forms above.
My Question
What I really want to know is, can I find the corresponding value for $(0,0,1,0,0,0)$?
 A: The vectors $(a,a,a,0,0,0)$ are all multiples of $(1,1,1,0,0,0)$; the vectors $(b,b,b,b,b,b)$ are all multiples of $(1,1,1,1,1,1)$. The vectors $(0,d,2d,0,0,0)$ are all multiples of $(0,1,2,0,0,0)$. The span of all these vectors give you only a 3-dimensional subspace. 
So the key lies in the vectors $(1,c,c^2,1,c,c^2)$; all these vectors span at most a $3$-dimensional subspace, since they all lie in the subspace of vectors $(x_1,x_2,x_3,x_4,x_5,x_6)$ with $x_1=x_4$, $x_2=x_5$, and $x_3=x_6$, which is $3$-dimensional. But they include $(1,1,1,1,1,1)$ (obtained with $c=1$); so you will get at best a 5-dimensional subspace from taking all these vectors together with the previously considered one; you cannot get a all of $\mathbb{R}^6$. 
In fact, you get exactly a $5$-dimensional subspace: the vectors 
$$\begin{align*}
&(1,c,c^2,1,c,c^2)\\
&(1,k,k^2,1,k,k^2)\\
&(1,\ell,\ell^2,1,\ell,\ell^2)
\end{align*}$$
are linearly independent if and only if the vectors $(1,c,c^2)$, $(1,d,d^2)$, and $(1,\ell,\ell^2)$ are linearly independent. This occurs if and only if
$$\left|\begin{array}{ccc}
1 & c & c^2\\
1 & d & d^2\\
1 & \ell & \ell^2
\end{array}\right| = (d-c)(\ell-c)(\ell-d)$$
is nonzero (this is a Vandermonde matrix); so distinct values of $c$, $d$, and $\ell$ will give you three linearly independent vectors, which therefore span
$$\mathbf{W} = \bigl\{(x_1,x_2,x_3,x_4,x_5,x_6)\in\mathbb{R}^5\mid x_1=x_4, x_2=x_5, x_3=x_6\bigr\}.$$
So your vectors span exactly a five-dimensional subspace of $\mathbb{R}^6$, and not all of $\mathbb{R}^6$ 
In fact, $(0,0,1,0,0,0)$ will be one of the vectors that does not lie in the span: if it lay in the span, then so would $(0,1,0,0,0,0) = (0,1,2,0,0,0)-2(0,0,1,0,0,0)$; hence also $(1,0,0,0,0,0)=(1,1,1,0,0,0)-(0,1,0,0,0,0)-(0,0,1,0,0,0)$.  Since you can get any vector of the form $(x,y,z,x,y,z)$, this would also allow you to obtain the other three standard basis vectors, and you would have a span equal to the entire space, which is not the case.
So, no, you cannot obtain $(0,0,1,0,0,0)$ with the vectors described.
