The problem is:
Show that $\{ 1, \cos t, \cos^2 t, \dots, \cos^6 t \}$ is a linearly independent set of functions defined on $\mathbb{R}$.
The problem expects the student to use a computer program such as Matlab.
To solve the problem I created a matrix in Matlab with $t$ ranging from $1$ to $7$, then row reduced it and got the identity matrix as result. I.e. the columns of the matrix is linearly independent.
Is it enough to just show for $t=1\to 7$; could it not possibly break down at some other number? And is there a more elegant way of solving this without the use of "brute force"?
Here's the Matlab code if necessary:
for t = 1:7
A(t,:) = [1 cos(t) (cos(t))^2 (cos(t))^3 (cos(t))^4 (cos(t))^5 (cos(t))^6];
end
Best Answer
The computational approach that you used works well. For our particular collection of functions, or more generally for the powers of a single function $f(t)$, there is a straightforward non-computational approach. As a bonus, the argument below works just as well for $666$ as it does for $6$.
Suppose to the contrary that our collection $\{1,\cos t, \cos^2 t, \dots, \cos^6 t\}$ of functions is not linearly independent. Then there are constants $a_0, a_1, \dots, a_6$, not all $0$, such that $$a_0+a_1 \cos t +a_2 \cos^2 t +\cdots +a_6 \cos^6 t=0 \quad\text{ for all $t$}.$$
Let $P(x)=a_0+a_1x+a_2x^2+\cdots +a_6x^6$. Not all the coefficients $a_i$ are $0$, so the equation $P(x)=0$ has at most $6$ solutions. (A non-zero polynomial of degree $\le n$ has at most $n$ roots.) Thus if we can show that the function $\cos t$ can take on more than $6$ different values, we obtain the desired contradiction.
But $\cos t$ takes on infinitely many different values as $t$ ranges over the reals, or indeed over any interval of non-zero length. This completes the proof.