Show that $(1,x,x^2),(1,y,y^2),(1,z,z^2)$ form a basis of $\mathbb{R}^3$ iff $x\neq y, x \neq z, y \neq z$ I'm having some trouble with this one because I always get negated statements. If I try to prove both direction directly I get that three elements are all not equal to each other and the three vectors form a basis, or if I proof by contrapositive I get the statement that the vectors do not from a basis and therefore do not span $\mathbb{R}^3$ and are not linearly independent. Either way I end up with negated statements, which I find hard to use in proofs.
What I have come up with thus far:
$=>$ Proof by contraposition: Suppose $(1,x,x^2),(1,y,y^2),(1,z,z^2)$ form a basis of $\mathbb{R}^3$ and assume that $x=y \vee x=z \vee y=z$. 
In each case it follows that at least two of the three vectors are equal and thus all three vectors are not linearly independent, therefore $(1,x,x^2),(1,y,y^2),(1,z,z^2)$ cannot be a basis of $\mathbb{R}^3$.
$<=$ Proof by contraposition: Suppose $(1,x,x^2),(1,y,y^2),(1,z,z^2)$ are not a basis of $\mathbb{R}^3$, then the vectors are not linearly independent or do not span $\mathbb{R}^3$.
Case 1: One of the vectors is in the span of the other two. Then it follows that there is a linear combination of the following form : $1x + 0y = z$, where x is the vector $(1,x,x^2)$ and y and z the corresponding other two vectors. Then it follows that $(1,x,x^2) = (1,z,z^2)$ and therefore $x=z$ with $x,z \in \mathbb{R}$. The same result would hold in the other cases.
Case 2: No idea.
Honestly I don't really feel as if those proofs are correct. For the second direction in case 1, I'm not sure whether the fact that the vectors are not linearly independent, implies the fact that there is a linear combination of the form $1x + 0y = z$.
Can anybody tell me what I have done wrong thus far and give some hints to help me prove this?
 A: Hint: Calculate the determinant.  You'll immediately see the equivalence. 
A: Let $v_x=(1, x, x^2)^T$. Your proof for $\implies$ is ok, although i would rephrase it a little: Assumption that $v_x$, $v_y$, $v_z$ form a basis of $\mathbb R^3$ is only confusing because you don't use it anywhere and you are proving converse of this statement directly from assumption that at least two of $x$, $y$, $z$ are equal.
As of $\impliedby$: it's wrong. You can't assume that $1\cdot v_x + 0\cdot v_y = 1\cdot v_z$. You only know that $$\alpha_1v_x + \alpha_2v_y + \alpha_3v_z = 0,$$ for some $\alpha_i\in\mathbb R$ with at least one of $\alpha_i$s different than $0$. But after all I recommend proving it directly: assume that $x$, $y$, $z$ are pairwise nonequal and solve system of equations
$$
\left\{ 
\begin{array}{ccccccc}
\alpha_1 &+& \alpha_2 &+& \alpha_3 &=& 0 \\ 
\alpha_1x &+& \alpha_2y &+& \alpha_3z &=& 0 \\ 
\alpha_1x^2 &+& \alpha_2y^2 &+& \alpha_3z^2 &=& 0
\end{array}
\right.
$$
A: Hint: All you need to prove is that they form a basis for $\mathbb{R}^3$ iff the matrix composed from them has rank $3$.
Edit: Clearer,
$\left(\begin{matrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z &z^2 \end{matrix}\right)\sim \left(\begin{matrix} 1 & x & x^2 \\ 0 & y-x & (y-x)(y+x) \\ 0 & z-x & (z-x)(z+x) \end{matrix}\right)$
Let the third row minus the product of second row with $\frac{z-x}{y-x}$, one conclude the $\Leftarrow$ side.
