Show that $(1,1,1)$, $(a,b,c)$, $(a^2, b^2, c^2)$ are linearly indepdenent for distinct $a,b,c$

Question

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.

This is obviously not true if $(a, b, c) = (1,1,1)$ – Aug 26 '16 at 02:56 — , Aug 26 '16 at 02:56
$a, b, c$ should be distinct. See Vandermonde matrix. – Seewoo Lee Aug 26 '16 at 02:58 — Seewoo Lee, Aug 26 '16 at 02:58
@ArcticChar edited my mistake – IntegrateThis Aug 26 '16 at 03:00 — IntegrateThis, Aug 26 '16 at 03:00

Caleb Stanford · Answer 1 · 2016-08-26T03:03:15.827

9

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 $f$?

edited Aug 26 '16 at 03:03

answered Aug 26 '16 at 02:58

Caleb Stanford

45,895

this seems to just evaluate to the equality I mentioned. – IntegrateThis Aug 26 '16 at 03:01
@JamesDickens The main hint was that it's a quadratic polynomial. But since you didn't pick up on it, I expanded slightly on the hint. – Caleb Stanford Aug 26 '16 at 03:03
still confused the roots depend on the coefficients and so..? – IntegrateThis Aug 26 '16 at 03:08
3

This is even simpler than using the determinant to show this. – user84413 Aug 26 '16 at 03:09
@JamesDickens How many roots does a quadratic polynomial have? Do you see now? – Caleb Stanford Aug 26 '16 at 03:09
@6005 2 roots, so.. Sorry to bother I just want to understand but I'm not sure what I am missing about your idea. – IntegrateThis Aug 26 '16 at 03:10
3

@JamesDickens All right, I'll just tell you. $a,b,c$ are three distinct roots of $f$. But $f$ can only have two distinct roots. So that means $f$ has to be the zero polynomial. – Caleb Stanford Aug 26 '16 at 03:11
@6005 Ok brilliant! – IntegrateThis Aug 26 '16 at 03:13
1

I didn't pick it up also, but this is really brilliant! :-) – YoTengoUnLCD Aug 26 '16 at 03:46

score 8 · Accepted Answer · answered Aug 26 '16 at 03:02

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.

This is perfect thank you. – IntegrateThis Aug 26 '16 at 03:03 — IntegrateThis, Aug 26 '16 at 03:03

Show that $(1,1,1)$, $(a,b,c)$, $(a^2, b^2, c^2)$ are linearly indepdenent for distinct $a,b,c$

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