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I get the idea of determining whether a set of vectors is dependent or independent when it involves polynomials like for example: $\{x^2-1, x^2+1, 4x, 2x-3\} in P$ in this case you can create a dependence relation:

$$λ_1(x^2-1) + λ_2(x^2+1) + λ_3(4x) + λ_4(2x-3) = 0$$

and from that you can create a matrix and row reduce it:

$$ \begin{bmatrix} -1 & 1 & 0 & -3 \\ 0 & 0 & 4 & 2\\ 1 & 1 & 0 & 0\\ \end{bmatrix} reduced \begin{bmatrix} -1 & 1 & 0 & -3 \\ 0 & 2 & 0 & -3\\ 0 & 0 & 4 & 2\\ \end{bmatrix} $$

and because each row has a pivot then I can conclude that it is independent so $λ_1 = λ_2 = λ_3 = λ_4 = 0$ (which I believe is the way to do it, but correct me if I'm wrong) *edit: I was wrong here, it's actually dependent because of the nontrivial solution.

But the question I have is how would I go upon doing this problem if instead of polynomials it involves trigs, because then in that case I can't do the matrix and row reduce to come to a conclusion.

So for this question lets say we have:

$$\{1, sin^2x, cos2x, cos^2x\} in F$$

A dependence relation:

$$λ_1(1) + λ_2(sin^2x) + λ_3(cos2x) + λ_4(cos^2x) = 0$$

From there I have no idea what to do next since I can't do matrix and row reduce, if anyone can give me some hints or show me a method for when trigs are involved that would be really appreciated.

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  • $\begingroup$ The first example is incorrect. Notice the last vector is a linear combination of $1/2v_3 + (-3/2)v_2 + 3/2v_1$. $\endgroup$ – Eoin Sep 26 '14 at 3:01
  • $\begingroup$ Hint: use $2 \sin^2 x = 1-\cos(2x), 2\cos^2 x=1+\cos(2x)$ and substitute $y=\cos(2x)$. $\endgroup$ – Graham Kemp Sep 26 '14 at 3:05
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    $\begingroup$ For that first example: linear independence is when all of the columns have pivots, not rows. So, in your example, the polynomials are not linearly independent. $\endgroup$ – Ben Grossmann Sep 26 '14 at 3:08
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if you think trigonometrics functions as functions you have

$$f(x) = λ_1(1) + λ_2(sin^2x) + λ_3(cos2x) + λ_4(cos^2x) = 0$$

for all $x \in \mathbb R$, then you can evaluate $f$ in some convenient $x$, for example $0 = f(1) = \lambda_1 + \lambda_3+\lambda_2$ doing this you can get more equations for the $\lambda_i$. but notice that $\cos (2x) = \cos(x) ^2 - \sin(x)^2$, this is a linear dependence equation.

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  • $\begingroup$ Thanks Maximiliano! I'll have to review my trig identities =p but I get how to do these questions now :) $\endgroup$ – user3412839 Sep 26 '14 at 3:37
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Consider the linear combination $$λ_1(1) + λ_2(sin^2x) + λ_3(cos2x) + λ_4(cos^2x) = 0$$ of vectors $\{1, sin^2x, cos2x, cos^2x\}.$
Note that $$\cos(x) ^2 - \sin(x)^2=\cos (2x)$$ $$\cos(x) ^2 + \sin(x)^2=1.$$ Therefore there are solutions for $λ_i, i=1,2,3,4$ wtih not all zero.
Hence the given set of vector is linearly dependent.

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  • $\begingroup$ Thanks for the help Nilan! $\endgroup$ – user3412839 Sep 26 '14 at 3:38

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