Show that $(1,1,1)$, $(a,b,c)$, $(a^2, b^2, c^2)$ are linearly indepdenent for distinct $a,b,c$ Show that $(1,1,1)$, $(a,b,c)$, $(a^2, b^2, c^2)$ are linearly indepdenent, where $a,b,$ and $c$ are distinct real numbers.
I will show my attempt and then state where I get stuck.
Suppose $c_1(1,1,1) + c_2(a,b,c) + c_3(a^2,b^2,c^2) = 0$
This leads to the three equations:
$c_1+c_2a+c_3a^2 = c_1+c_2b+c_3b^2 = c_1+c_2c+c_3c^2 = 0$. Now I am not sure how to sure from here that each $c_i$ must be 0. Hints appreciated.
 A: Note: the three vectors are not linearly independent if $a = b$, or if $a = c$, or if $b = c$. So you will need to assume $a, b, c$ are distinct.
Hint: Define the quadratic polynomial $f(x) = c_1 + c_2 x + c_3 x^2$. What are $f(a), f(b), f(c)$? What can you conclude about the roots of $f$?
A: You have a linear system of three equations in three unknowns, which you can write in matrix form as
$$
 A\mathbf{c} = \mathbf{0},
$$
where
$$
 A = \begin{pmatrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{pmatrix} \in \mathbb{R}^{3 \times 3}
$$
is your coefficient matrix and
$$
 \mathbf{c} = \begin{pmatrix}c_1\\c_2\\c_3\end{pmatrix} \in \mathbb{R}^3
$$
is your unknown vector. In these terms, then, $(1,1,1)$, $(a,b,c)$, and $(a^2,b^2,c^2)$ are linearly independent if and only if the system $A\mathbf{c} = \mathbf{0}$ has a unique solution, if and only if $A$ is invertible, if and only if $\det(A) \neq 0$. But now,
$$
 \det(A) = \begin{vmatrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{vmatrix} = \begin{vmatrix}1&a&a^2\\0&b-a&b^2-a^2\\0&c-a&c^2-a^2 \end{vmatrix}=(b-a)(c^2-a^2)-(b^2-a^2)(c-a)\\=(b-a)(c-a)\left((c+a)-(b+a) \right) = (b-a)(c-a)(c-b),
$$
from which it follows that $(1,1,1)$, $(a,b,c)$, and $(a^2,b^2,c^2)$ are linearly independent if and only if $a$, $b$, and $c$ are all distinct.
