I'm currently going through Harvard's Abstract Algebra using Michael Artin's book, and have no real way of verifying my proofs, and was hoping to make sure that my proof was right.
The question reads:
Let $V$ be the vector space of functions on the interval $[0, 1]$. Prove that the functions $x^{3}$, $\sin(x)$, and $\cos(x)$ are linearly independent.
My proof goes as follows:
For these to be linearly dependent there must exist an $a_{i} \neq0$, where $ i = 1, 2, 3$ such that
$$a_{1}x^{3} + a_{2}\sin(x) + a_{3}\cos(x) = 0. $$
So, we'll do this in 3 cases:
Case 1: $x = 0$
In this case, $x^{3} = 0$, $\sin(x) = 0$ but $\cos(x) = 1$.
So, we have
$$0\times a_{1} + 0\times a_{2} + 1\times a_{3} = 0.$$
So, $a_{1}$ and $a_{2}$ could be anything but $a_{3}$ must be 0.
Case 2: $x \in (0,1)$
In this case, $x^{3} \neq 0$, $\sin(x) \neq 0$ and $\cos(x) \neq 0$.
So, for this to be true, $a_{1}$, $a_{2}$ and $a_{3}$ all must be $0$.
Case 3: $x = 1$
In this case, $x^{3} = 1$, $\sin(x) = .8…$ and $\cos(x) = .5…$.
So, we have
$$1\times a_{1} +.8\times a_{2} + .5\times a_{3} = 0.$$
So, $a_{3}$ could be any value, while $a_{1}$ and $a_{2}$ must be $0$.
So, if $a_{1} \neq 0$ then we have a problem in Case 3. If $a_{2} \neq 0$ we have a problem in Case 3. If $a_{3} \neq 0$ we have a problem in Case 1. So, we know that all of the $a$ values must be $0$ and we complete the proof.
Best Answer
If you make no assumptions about $a_i$s and then show they must be zero, you have proved linear independence.
From the evaluation at $x=0$ (case 1) you correctly conclude that $a_3=0$. At this point, I suggest simplifying your expression to $a_1x^3+a_2\sin(x)$.
In your evaluation at $x\in(0,1)$ (case 2) you have a logical error of assuming that a sum of numbers being zero requires each number to be zero. $a+b=0$ does not imply $a$ and $b=0$
There is also an error here in the structure of your proof, because if you had shown that all of the $a_i$s are zero, you would be done already.
You have a similar logical error for the evaluation at $x=1$ (case 3).
It seems you are thinking of the outcome of the values of $a_i$s as depending on what $x$ is, hence the "cases" that are treated independently. But the $a_i$s are always the same, for all $x$. Your idea of evaluating at particular points is a good one, and they build on each other. The evaluation at $x=0$ shows that $a_3=0$. Evaluation at two other particular values of $x$, say, $x=\frac12$ and $x=1$, would give you two equations in two unknowns for $a_1$ and $a_2$, and you could show the only solution is $a_1=a_2=0$.
Alternatively, you could take the derivative of your expression to get $3a_1x^2+a_2\cos(x)=0$, and then evaluation at $x=0$ will give you $a_2=0$ directly, after which you'll be left with $a_1x^3\equiv 0$ from which it isn't hard to show $a_1=0$ either by evaluation at $x=1$ or taking the derivative $3$ times.