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Consider the space of all functions $f: \mathbb{R}\longrightarrow \mathbb{C}$. Prove that $\{1, e^{ix}, e^{-ix}\}$ are linearly independent vectors.

Grobber
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5 Answers5

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Linear independence of $\{f,g,h\}$ means that $\alpha f + \beta g + \gamma h=0$ implies $\alpha=\beta=\gamma=0$. Here $f(x)=1$, $g(x)=\mathrm e^{\mathrm ix}$ and $h(x)=\mathrm e^{-\mathrm ix}$.

To see this, let's just evaluate the functions at three different points. Say $x=0$, $x=\pi/2$ and $x=\pi$. Then we get $$\begin{align} \alpha f(0) + \beta g(0) + \gamma h(0) &= \alpha + \beta + \gamma &= 0 &&(1)\\ \alpha f(\pi/2) + \beta g(\pi/2) + \gamma h(\pi/2) &= \alpha + \mathrm i\beta-\mathrm i\gamma &= 0 &&(2)\\ \alpha f(\pi) + \beta g(\pi) + \gamma h(\pi) &= \alpha - \beta -\gamma &= 0 &&(3) \end{align}$$ Addint equation (1) and (3) already gives $\alpha=0$. If we insert this into (1) and (2) and divide (2) by $\mathrm i$, we get $$\begin{align} \beta + \gamma &= 0 &(4)\\ \beta - \gamma &= 0 &(5) \end{align}$$ Adding (4) and (5) gives $\beta=0$, subtracting those equations gives $\gamma=0$.

Thus the three functions are linearly independent.

Note that if we had found a solution, this would not have proven linear dependence, because it doesn't exclude that at other points the functions are different (imagine if we had chosen the values $0$, $2\pi$ and $4\pi$ to evaluate the functions at).

celtschk
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  • Would it be possible to elaborate on your final sentence of this answer? My understanding is that by definition if any non-trivial solution is found, the functions must be linearly dependent. – Silly Goose Jun 02 '22 at 22:12
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    @SillyGoose: If you insert $0$, $2\pi$ and $4\pi$, the resulting equations are $\alpha+\beta+\gamma=0$ in each case. Quite obviously, those equations are linearly dependent (it's three times the same equation, after all), although the functions aren't. – celtschk Jun 03 '22 at 03:28
  • I see! Thank you so much. – Silly Goose Jun 03 '22 at 15:04
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Recall the definition of linear independence for vector $u,v$ is $$C_1u + C_2 v = 0\iff C_1=C_2=0.$$ Now consider $$C_1 \mathbf{1}+C_2 e^{ix}=0,\quad \forall x\in \mathbb{K},$$ $\mathbb{K}$ can be either $\mathbb{R}$ or $\mathbb{C}$. The only solution is $C_1=C_2=0$.

So for your case, it would be $$C_1 \mathbf{1}+C_2 e^{ix} + C_3 e^{-ix}=0,\quad \forall x\in \mathbb{K}\implies C_1=C_2=C_3=0.$$

newbie
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    Perhaps I am confused, but isn't this answer just restating the problem more explicitly? Is there a suggested approach to solving the problem beyond that? – Jonas Meyer Jul 05 '13 at 15:51
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    I agree with Jonas. This answers nothing. – MJD Jul 05 '13 at 15:53
  • @JonasMeyer Maybe I'm wrong, but when I first encounted the similar question years ago, it was a mind turning from solid vector in Euclidean space to function space. – newbie Jul 05 '13 at 15:58
  • @newbie: I don't entirely understand your comment. Perhaps you are suggesting that the obstacle is in translating the problem statement to the more explicit statement about the functions, because it is difficult to get used to functions being vectors? I do agree that problems of this sort are often easy once one is comfortable working with the definitions. – Jonas Meyer Jul 05 '13 at 16:01
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    @JonasMeyer well, it states for all $x\in\mathbb{K}$, the rest is to construct a system of linear equations. If that is what you think is missing. I took a wild guess the OP may be able to figure the little puzzle himself. Since the main step would be how to formulate the question in linear independence, in my opinion. – newbie Jul 05 '13 at 16:05
  • @newbie I agree, the $\forall{x}\in\mathbb{K}$ part is what I was missing when I was trying to solve it myself. This is a good answer. – rurouniwallace Jul 05 '13 at 18:22
  • Questions like this always boil down to using the definitions. – John Jul 05 '13 at 23:49
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it's easy to see that it's equivalent to saying that $1, \sin(x), \cos(x)$ are linearly independent. (because you can get both $\sin, \cos$ as a combination.) to do this assume $$a + b \sin(x) + c\cos(x) = 0$$ as a function, so for all $x$. if so then also all its derivatives must vanish, the first one in particular, so $$b \cos(x) - c \sin(x) = 0$$ for all $x$. but if $x = \pi$ then $\sin(\pi) = 0$ and $\cos(\pi) = -1$ hence b = 0 and so c = 0, done.

mm-aops
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  • Minor point (which doesn't really affect the argument): If the scalars are only real, do you still get $\sin$, or only $i\sin$? – Jonas Meyer Jul 05 '13 at 15:52
  • you're right, you only get $i \sin$, thanks. but still, if the functions in question are linearly independent over $\mathbb{C}$ then they're linearly independent over $\mathbb{R}$. or you just go along with $i \sin$ and do the same thing – mm-aops Jul 05 '13 at 15:56
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Assume $f(x):=a\cdot 1+b\cdot e^{ix}+c\cdot e^{-ix}=0$ for all $x$. After multiplying with $e^{ix}$ and writing $y$ for $e^{ix}$ we obtain that $c+ay+by^2=0$ for all $y$ in the range of $x\mapsto e^{ix}$. This range contains at least three distinct points, and a quadratic polynomial is uniquely determined by its values at three points ...

Hagen von Eitzen
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I assume that you want linear independence over $\mathbb{C}$. Hint: The three functions have different behaviors as $x\rightarrow i\infty$.

Marci
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