Let $m \in \mathbb{N}$. I wish to prove the linear independence of the set of vectors {$\sin(2^m x), \sin(2^{m-1}x), \ldots, \sin(2x), \sin(x)$} $\subset F(\mathbb{R})$. I had attempted to prove the property by induction upon $m$: the base case, quite evidently, was trivial to prove (insofar as $\sin(x)≠0$), but I could not proceed very well with the induction hypothesis. How might I do so? Or is there perhaps a better way by which to show the property holds?

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    $\begingroup$ Here's a nice trick: They are all eigenvectors of the linear operator $\frac{d^2}{dx^2}$ with distinct eigenvalues $\endgroup$
    – lc2r43
    Aug 23, 2020 at 17:45

1 Answer 1


We can prove a more general result. Let $n \ge 1.$ We show that the set $$\{\sin(x), \sin(2x), \ldots, \sin(nx)\} \subset F(\Bbb R)$$ is linearly independent.

Indeed, suppose that $a_1, \ldots, a_n$ are reals such that $$a_1\sin(x) + \cdots + a_n\sin(nx) = 0$$ for all $x \in \Bbb R.$

Fix $k$ such that $1 \le k \le n$. Multiplying both sides with $\sin(kx),$ we get $$a_1\sin(x)\sin(kx) + \cdots + a_n\sin(nx)\sin(kx) = 0$$ for all $x\in\Bbb R$.

Now, integrate both sides from $0$ to $2\pi$ and note that $$\int_0^{2\pi}\sin(ix)\sin(kx){\mathrm d}x = 0 \iff i \neq k.$$

Thus, we get that $$a_k\int_0^{2\pi}(\sin(kx))^2{\mathrm d}x = 0$$ or $$a_k = 0.$$

As $k$ was arbitrary, we get that each $a_i$ is zero, proving linear independence.

EDIT: Note that if we consider the vector space $\mathcal{C}[0, 2\pi]$, the set of real-valued continuous functions on $[0, 2\pi],$ we can define an inner product on it as $$\langle f, g\rangle := \int_0^{2\pi}f(t)g(t){\mathrm d}t.$$ The above shows that $\{\sin(x), \ldots, \sin(nx)\}$ is an orthogonal subset of $\mathcal{C}[0, 2\pi]$ under this inner product. In particular, it is linearly independent. (Since each vector is nonzero.)

From this, it is easy to see that the functions considered as elements of $F(\Bbb R)$ must also be linearly independent.

  • $\begingroup$ Not mentioning orthogonal bases and inner products seems like a missed opportunity. $\endgroup$
    – Pedro
    Aug 23, 2020 at 18:05
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    $\begingroup$ I did wish to mention inner products but the ambient vector space is $F(\Bbb R)$ and not $\cal C[0, 2\pi]$. Though you are right, I still should mention it. I'll add a note to this effect. $\endgroup$ Aug 23, 2020 at 18:07

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