Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
How could you take advantage of the Pythagorean identity? – David Mitra Jun 26 '13 at 22:13
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
@CodeKingPlusPlus Linear independence is invariant under multiplication of individual vectors by nonzero constants. So you can just multiply your three vectors by respectively $\frac13,\frac13,\frac14$, and the problem becomes easy. – Marc van Leeuwen Dec 26 '14 at 6:24
up vote 11 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!)

share|cite|improve this answer

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.

share|cite|improve this answer

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.

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.

share|cite|improve this answer
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
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
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
Once again, @Mhenni, please try to read what people take the pain to write to answer your queries: "zero is always a solution" means that zero is always in the set of solutions and says nothing about the existence of other (nonzero) solutions. – Did Jun 30 '13 at 13:28
I've deleted some comments here. I'm not entirely sure how this ancient thread is so active. Stop defacing your answers. This includes edits to cast off your downvoters as misinformed. – mixedmath Dec 26 '14 at 17:19

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.