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What we did in linear algebra 1. is to prove that two Polynomials are linear independent. I understand the general concepts, but I had a problem with the specific proof.

We have $S=\{ x^2 - 4, x + 6 \}$. Our assertion is that they are linear independent.

What the prof did, was: She said that since it must be true for all $x$, then it must as well be true for our "favorite" $x$.

We choose arbitrary e.g. $-6$

and then we get that

$$\alpha_1( 36 - 4) + \alpha_20 = 0$$

And then she just jumps to the concluson that $\alpha_1$ must be $0$ (analog for $\alpha_2$). BUT I mean if we look at the equation we see that in this case $\alpha_2$ could be anything, which means that it is not a trivial linear combination anymore. Isn´t that a problem? Shouldn´t we be able to show that both $\alpha$ -s have to be $0$ at the time? If not, why? Because it´s just a system of linear equations and then it does not matter whether in between it´s a non-tivial or a trivial linear combination we´re solving? But then again: shouldn´t these be equivalent transformations?

Thank you in advance

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    We are looking for $\alpha_1,\ \alpha_2$ such that $\alpha_1(x^2-4) + \alpha_2(x+6) = 0$ for all $x$. If we determined that $\alpha_1=0$ that leaves us with $\alpha_2(x+6) = 0$ for all $x$, not just $x=-6$. What must the value of $\alpha_2$ be then? – 2017-01-01
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    Yep, but then again if we talk about wether an equation is linear indipendent or not, it`s not about a special case (which you creat by saying that α1 = 0). I do know how to solve this mathematically, but what I`m curious about is the interpretations and the questions I asked above. If any of them are unclear please ask and I`ll try to put it in another way. – 2017-01-01
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    Basically what I`m saying is that finding a trivial linear combination is not enough to prove that it`s linear indipendent (by definition). And by fixing α1 = 0 you`re just finding one of the many trivial solutions, but you don`t prove that there is no non-trivial one. – 2017-01-01
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    But this is showing that the trivial solution is the only solution. We are not creating a special case by fixing $\alpha_1=0$, we are showing that for $\alpha_1,\ \alpha_2$ to satisfy the equation we must have $\alpha_1=0$. We use that conclusion to show that we must also have $\alpha_2=0$. – 2017-01-01

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Suppose there are two constants $a_1, a_2$, at least one of which is nonzero, with $$a_1(x^2-4)+a_2(x+6)=0$$ If this is true, then plugging in $x=-6$ we get $$a_1(32)+a_2(0)=0$$ from which it follows that $a_1=0$. On the other hand, plugging in $x=2$ we get $$a_1(0) + a_2(8) = 0$$ from which it follows that $a_2=0$. But we began with the assumption that at least one of $a_1, a_2$ was nonzero, so we have reached a contradiction.

Notice that in the above we are not setting $a_1=a_2=0$; we are assuming the existence of a nontrivial linear combination and deducing that $a_1=a_2=0$.