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I want to determine whether 3 functions are linearly independent:

\begin{align*} x_1(t) = 3 \\ x_2(t) = 3\sin^2(t) \\ x_3(t) = 4\cos^2(t) \end{align*}

Definition of Linear Independence: $c_1x_1 + c_2x_2 + c_3x_3 = 0 \implies c_1=c_2=c_3=0$ (only the trivial solution)

So we have: \begin{align} 3c_1 + 3c_2\sin^2(t) + 4c_3\cos^2(t) = 0 \end{align}

My first idea is to differentiate both sides and get:

$6c_2\sin(t)\cos(t) - 8c_3\cos(t)\sin(t) = 0$

Then we can factor to get:

$\sin(t)\cos(t)(6c_2 - 8c_3) = 0$

So $c_3= \dfrac{6}{8}c_2$ gives the equation equals zero. Thus all $c$ are not $0$ and thus $x_1, x_2, x_3$ are linearly dependent.

Is this correct? Or is there a cleaner way to do this?

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How could you take advantage of the Pythagorean identity? –  David Mitra Jun 26 '13 at 22:13
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Since $\cos^2 \varphi + \sin^2 \varphi \equiv 1$, you can directly see that $4x_2 + 3x_3$ is a constant. –  Daniel Fischer Jun 26 '13 at 22:13
    
I was thinking about that, but I how do I deal with the coefficients? $(\sqrt(3c_2)\sin(t))^2 + (\sqrt(4c_3)\sin(t))^2$ –  CodeKingPlusPlus Jun 26 '13 at 22:15
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3 Answers 3

up vote 8 down vote accepted

Yes, indeed, your answer is fine. And it would have been a particularly fine determining the linear (in)dependence of a system of equations that doesn't readily admit of another observation about the relationship between $\cos^2 t$ and $\sin^2 t$ $(\dagger)$. Indeed, you're one step away from working with the Wronskian, which is a useful tool to prove linear independence.

$(\dagger)$ Now, to the observation previously noted: You could have also used the fact that $$x_1(t) - \left[(x_2(t) +\frac 34 x_3(t)\right] = 3 - (3 \sin^2 t + 3\cos^2 t)= 3 - 3\left(\underbrace{\sin^2(t) + \cos^2(t)}_{\large = 1}\right)=0$$

and saved yourself a little bit of work: you can read off the nonzero coefficients $c_i$ to demonstrate their existence: $c_1 = 1, c_2 = -1, c_3 = -\frac 34$, or you could simply express $x_1$ as a linear combination of $x_2, x_3$, to conclude the linear dependence of the vectors. (But don't count on just any random set of vectors turning out so nicely!)

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Clean write up +1 –  Amzoti Jun 27 '13 at 0:53
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It is much more easier to use known identity $\sin^2{t}+\cos^2{t}=1$. We have $x_1(t)-x_2(t)-\frac{3}{4}x_3(t)=3-3\cos^2{t}-3\sin^2{t}=0$, so functions are linearly dependent.

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Do not worry about the downvote. It is an attack on my answers. The answer is correct. This answer solves the problem from scratch. The main question in this answer is: when the system has a unique solution? The answer is: when the Determinant of the matrix of coefficients does not equal zero. The answer is totally correct.

A related problem. Here is an approach. We differentiate the equation $$ c_1 x_1(t) + c_2 x_2(t) + c_3 x_3(t) = 0 $$ twice to get the system

$$ c_1 x_1(t) + c_2 x_2(t) + c_3 x_3(t) = 0 $$ $$ c_1x'_1(t) + c_2 x'_2(t) + c_3 x'_3(t) =0 $$ $$ c_1x''_1(t) + c_2 x''_2(t) + c_3 x''_3(t) =0 $$

The above system of equations has a solution $c_1=c_2=c_3=0$ if the determinant $D\neq 0$ or in other words the matrix of coefficients is invertible. Note that, you are solving for $c_1,c_2,c_3$ Now, just work out the details.

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Replace "The above system of equations has a solution c1=c2=c3=0 if the determinant D≠0" by "The above system of equations has no other solution than c1=c2=c3=0 if and only if the determinant D≠0". –  Did Jun 27 '13 at 14:48
    
@Downvoters: What's the downvote for? The answer is correct. –  Mhenni Benghorbal Jun 29 '13 at 22:05
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I'm not a downvoter, but I suspect the fact that your answer is irrelevant may be an explanation. –  M Turgeon Jun 29 '13 at 22:13
    
@MTurgeon: What do you mean by irrelevant? –  Mhenni Benghorbal Jun 29 '13 at 22:19
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Not wrong but irrelevant, since c1=c2=c3=0 is ALWAYS a solution, whether the determinant D is zero or not. –  Did Jun 30 '13 at 9:32
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