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I have a homework question that is giving me a hard-time on. I am using many ideas but they seem to not work out. If someone can give me pointers, it would be much appreciated. Thank you in advance.

The question is:

In the vector space of real-valued functions $F=\{f\mid f:\Bbb R\to\Bbb R\}$, determine if the following set $S$ is linearly independent. $$S=\left\lbrace \sin 2x, \cos 2x, 2 \right\rbrace$$

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  • $\begingroup$ Do you know what the Wronskian is? $\endgroup$
    – user137731
    Commented Oct 18, 2015 at 20:58
  • $\begingroup$ Yes but we have not learned it in Linear Algebra course, so I'm unsure to use it. Do you know if there is another way? If so, could you please explain the steps or the process of solving this problem. $\endgroup$
    – user281623
    Commented Oct 18, 2015 at 21:04

2 Answers 2

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Start from a linear relation $\;a\sin 2x+b\cos 2x +c\cdot 2=0$, and set successively $\;x=0,\enspace x=\dfrac\pi4, \enspace x=\dfrac\pi6$, say.

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Hint:

Let's denote the $3$ functions $f(x) = \sin(2x), \, g(x) = \cos(2x), \, h(x) = 2$. If we can find $3$ values $x_1,x_2,x_3 \in \mathbb R$ such that: $$\begin{vmatrix} f(x_1) & g(x_1) & h(x_1)\\ f(x_2) & g(x_2) & h(x_2) \\ f(x_3) & g(x_3) & h(x_3) \end{vmatrix} \neq 0,$$ then the $3$ functions are linearly independent.

Indeed, consider $x_1 = 0, x_2 = \frac{\pi}4 , x_3 = \frac{\pi}2.$ Then $\begin{vmatrix}0 & 1 & 2\\1 & 0 & 2\\0 & -1 & 2\end{vmatrix}=-4\neq 0.$

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