# Proving Linearly Independence from System of Equation

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 - ## 2 Answers

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 $$0$$.

• 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? – Consider Non-Trivial Cases Jul 3 '20 at 15:29
• 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. – José Carlos Santos Jul 3 '20 at 15:32

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 $$c_0=c_1=c_2=c_3=0$$.

• All $x$ except the solutions of equation $(1)$? – Consider Non-Trivial Cases Jul 3 '20 at 14:20
• @Andrew I'm sorry, I don't know what you mean. – K.defaoite Jul 3 '20 at 14:41
• $c_0+c_1x+c_2x^2+c_3x^3=0$ 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. – Consider Non-Trivial Cases Jul 3 '20 at 15:14
• @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 $c_0=...=c_3=0.$ – K.defaoite Jul 3 '20 at 15:19