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I am trying to understand the proof of Linearly Independence of the basis set $\{1, x, x^2, x^3\}$. It is written that -

Substituting $3$ other values of $x$ into the above equation yields a system of $3$ linear equations in the remaining $3$ unknowns $c_1, c_2$ and $c_3$.

The equation we are considering right now is - $$c_0 \cdot (1) +c_1x+c_2x^2+c_3x^3=0 \cdots (1)$$

First we plug in $x= 0$ to get the value of $c_0$ in equation $(1)$, we get -

$$ c_0 \cdot (1) +c_1\cdot 0 +c_2\cdot 0 +c_3\cdot 0=0\implies c_0 =0$$ For arbitrary 3 non-zero values $ x_1, x_2, x_3 \in \mathbb{R}$ where $ x_1, x_2, x_3$, are not solutions of equation $(1)$ and for $c_0=0$, we will get $3$ equations from equation $(1)$-

$$c_1x_1+c_2x_1^2+c_3x_1^3=0$$ $$c_1x_2+c_2x_2^2+c_3x_2^3=0$$ $$c_1x_3+c_2x_3^2+c_3x_3^3=0$$

Now, how can it be shown that the only solution of the above system of equations is the trivial solution $c_1 = c_2 = c_3 = 0$ and therefore the set $\{1, x, x_2, x_3\}$ is linearly independent?

Thanks.

The source of the problem and background is given below -

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The matrix of that system is$$\begin{bmatrix}x_1&x_1^{\,2}&x_1^{\,3}\\x_2&x_2^{\,2}&x_2^{\,3}\\x_3&x_3^{\,2}&x_3^{\,3}\end{bmatrix}.$$It is a Vandermonde matrix and its determinant is $(x_1-x_2)(x_1-x_3)(x_2-x_3)$. So, if the numbers $x_1$, $x_2$, and $x_3$ are distinct, the only solution of the system is $(0,0,0)$, since then the determinant is not $0$.

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  • $\begingroup$ Sorry to disturb you, but I am having a little problem to understand the line of your argument.... i have assumed $x_1, x_2, x_3$ are distinct, so why we need Vandermonde matrix with non-zero determinant which implies $x_1, x_2, x_3$ are distinct? $\endgroup$ – Consider Non-Trivial Cases Jul 3 '20 at 15:29
  • $\begingroup$ It's the other way around: since those numbers are distinct, then the determinant is not $0$, and therefore the system has one and only one solution. $\endgroup$ – José Carlos Santos Jul 3 '20 at 15:32
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An alternative approach: one might notice that $c_0+c_1x+c_2x^2+c_3x^3$ is a cubic polynomial and has at most three roots because of the fundamental theorem of algebra. Thus the only way this polynomial is $=0 ~ \forall x$ is if $c_0=c_1=c_2=c_3=0$.

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  • $\begingroup$ All $x$ except the solutions of equation $(1)$? $\endgroup$ – Consider Non-Trivial Cases Jul 3 '20 at 14:20
  • $\begingroup$ @Andrew I'm sorry, I don't know what you mean. $\endgroup$ – K.defaoite Jul 3 '20 at 14:41
  • $\begingroup$ $c_0+c_1x+c_2x^2+c_3x^3=0$ is the equation $(1)$, if $x$ is a solution then the polynomial could be $0$ without $c_0=c_1=c_2=c_3=0$. So, your answer seems to be faulty. $\endgroup$ – Consider Non-Trivial Cases Jul 3 '20 at 15:14
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    $\begingroup$ @Andrew I know. But linear independence requires $c_0+c_1 x +c_2 x^2 + c_3 x^3 =0$ for all $x$. I.e, the polynomial has an infinite number of roots. But the fundamental theorem of algebra tells us that any polynomial has only finitely many roots, therefore the only way (1) $=0 ~ \forall x$ is if $c_0=...=c_3=0.$ $\endgroup$ – K.defaoite Jul 3 '20 at 15:19

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