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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?

eager2learn
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3 Answers3

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Hint: Calculate the determinant. You'll immediately see the equivalence.

Michael Hoppe
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  • I'm trying with the determinant, but I didn't find it immediate. – Abramo Dec 08 '13 at 16:47
  • Try to factorize. It starts with $\det(...)= yz^2-y^2z+ ... =yz(z-y)+ ...$. You should end up with $(z-x)(z-y)(y-z)$. – Michael Hoppe Dec 08 '13 at 16:51
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    This result is due to Alexandre-Théophile V. You may try to find out what “V.” stands for. – Michael Hoppe Dec 08 '13 at 16:57
  • Thanks for your answer. Unfortunately we haven't covered determinants yet. We'll start linear maps and then determinants in the next section. Do you have hints on how to prove this with a different approach? – eager2learn Dec 08 '13 at 17:00
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    Dear Michael, that's indeed immediate if one knows about Van Der Monde determinant. Otherwise it's not so IMMEDIATE to see that factorisation =D – Abramo Dec 08 '13 at 17:05
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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 $$ \left\{ \begin{array}{ccccccc} \alpha_1 &+& \alpha_2 &+& \alpha_3 &=& 0 \\ \alpha_1x &+& \alpha_2y &+& \alpha_3z &=& 0 \\ \alpha_1x^2 &+& \alpha_2y^2 &+& \alpha_3z^2 &=& 0 \end{array} \right. $$

Adam
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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.

Du Phan
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  • Thanks for the answer, but I'm afraid I don't understand the rationale behind your hint at all. All I know at this point is that the rank can be found out by performing the gauss algorithm and is then equal to the number of pivot elements. I dont understand how that proves the iff statement to be honest. – eager2learn Dec 08 '13 at 17:05
  • What if $x=y$? Your recommended action is then not applicable. – Michael Hoppe Dec 08 '13 at 17:07
  • @eager2learn : a 3-by-3 matrix has rank 3 if and only if its determininant is nonzero. This answer looks much easier than what you were trying to do in your question, with all those cases. – Stefan Smith Dec 08 '13 at 17:15
  • @MichaelHoppe: the $\Rightarrow$ side is implied in the statement of my hint (two row is the same if one of the equalities happens, so the rank cannot be 3 then). Of course, if one wants a nicer solution, take out $(y-x)$ in the second row, then take out $(z-x)$ in the third row, then minus the third row with the second row. All ways get to the same conclusion. – Du Phan Dec 08 '13 at 17:15
  • @Du Phan : this is how I would do the problem, but unfortunately, the OP wrote in a comment that s/he doesn't know about determinants. – Stefan Smith Dec 08 '13 at 17:17
  • @eager2learn: it is nice indeed to understand the reason behind, you may consider this wikipedia (especially the section Row operations) to understand; by the way, at first you should know that $av_1 + bv_2 + cv_3 =0$ then $(a,b,c)$ is the root of the matrix with $v_1,v_2,v_3$ are three columns; and ${v_1,v_2,v_3}$ are basis iff the only root is $(a,b,c)=0$. – Du Phan Dec 08 '13 at 17:44
  • @DuPhan You can't take out zero. – Michael Hoppe Dec 08 '13 at 17:48
  • @MichaelHoppe: the $\Leftarrow$ side acts with nonzeros. – Du Phan Dec 08 '13 at 18:18