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I have this homework question that I have no idea how to do:

Show that $\{t, \sin(t), \cos(2t), \sin(t)\cos(t) \}$ is a linearly independent set of functions defined on $\mathbb{R}$. Start by assuming that $$c_1 t + c_2 \sin(t) + c_3 \cos(2t) + c_4 \sin(t)\cos(t) = 0$$ for all $t$. Choose specific values of $t$ ($t=0, 1, 2\dots$) until you get a system with enough equations to determine that all the $c_i$'s must be $0$.

My only guess is to set up a matrix based on this polynomial somehow, then row reduce it to find the pivot columns. Subbing in different $t$'s would allow for some value to be attached to each $c$. Would that be the matrix? Shouldn't it be formed by the $c$ values?

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I would choose specific values from the unit circle for $t$... – The Chaz 2.0 Nov 1 '11 at 21:46
What does choosing values for t do for you though? Is the matrix formed by the t terms? – Randy Nov 1 '11 at 21:50
@Gortaur, I wouldn't say this question is a duplicate of that one, but I would certainly recommend that Randy have a look at the older question, as much of what is there is directly relevant to Randy's needs. – Gerry Myerson Nov 1 '11 at 21:55
Evaluate at four points. If you are not unlucky, you will get $4$ linearly independent vectors in $\mathbb{R}^4$. Maybe try (i) $t=0$; (ii) $t=2\pi$; (iii) $t=\pi/2$; (iv) $t=\pi/4$. Look at the question referenced by @Gortaur. There are ways other than evaluating. – André Nicolas Nov 1 '11 at 22:00

Let's say you plug-in $t=t_1,t_2,t_3,t_4$ then you'll have

$$\begin{array}{cc} c_1 t_1 + c_2\sin(t_1)+c_3 \cos(2t_1)+c_4\sin(t_1)\cos(t_1) & =0 \\ c_1 t_2 + c_2\sin(t_2)+c_3 \cos(2t_2)+c_4\sin(t_2)\cos(t_2) & =0 \\ c_1 t_3 + c_2\sin(t_3)+c_3 \cos(2t_3)+c_4\sin(t_3)\cos(t_3) & =0 \\ c_1 t_4 + c_2\sin(t_4)+c_3 \cos(2t_4)+c_4\sin(t_4)\cos(t_4) & =0 \end{array}$$

and so the corresponding matrix equation is

$$ \begin{pmatrix} t_1 & \sin(t_1) & \cos(2t_1) &\sin(t_1)\cos(t_1) \\ t_2 & \sin(t_2) & \cos(2t_2) &\sin(t_2)\cos(t_2) \\ t_3 & \sin(t_3) & \cos(2t_3) &\sin(t_3)\cos(t_3) \\ t_4 & \sin(t_4) & \cos(2t_4) &\sin(t_4)\cos(t_4) \end{pmatrix} \begin{pmatrix} c_1\\ c_2\\ c_3\\ c_4 \end{pmatrix} = \begin{pmatrix} 0\\ 0\\ 0\\ 0 \end{pmatrix}$$

By choosing the "right" values for $t_1,t_2,t_3,t_4$ you'll have a matrix which is invertible (so $c_1=c_2=c_3=c_4=0$ is the only solution). You can check this is case by making sure the determinant is non-zero. [I believe the values $t_1=0$, $t_2=1$, $t_3=2$, $t_4=3$ work, but it is not pretty.]

An alternate approach is to differentiate your equation 3 times and then plug-in a single value (like $t=0$). In this case, the coefficient matrix you'll end up with is called a Wronskian. If it's determinant is non-zero, then the matrix is invertible so the corresponding system only has the trivial solution and hence your functions are linearly independent.

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For instance, if you let $t=0$ then $t=0$, $\sin(0)=0$, and $\sin(0)\cos(0)=0$. But, $\cos(2\cdot 0)=\cos(0)=1$ and your equation becomes $c_3=0$. As The Chaz suggested, you might want to try values from the unit circle where you know the various function values of $\sin$ and $\cos$.

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At $t = 0$ we have $t = 0, \sin(t) = 0,\,\cos(2t) = 1$ and $\sin(t)\cos(t) = 0$. So if I choose $c_3 = 0$ and allow $c_1, c_2, c_4$ to take any other value, $$c_1 t + c_2 \sin(t) + c_3 \cos(2t) + c_4 \sin(t)\cos(t) = 0$$ is satisfied at $t = 0$. I think that implies that the functions are linearly dependent at $t = 0$. This would hold true for $t = k\pi$.

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