1
$\begingroup$

I am trying to determine whether $\{ \sin x, \cos x\}$ is linearly dependent in vector space of all real valued functions. The definition says:

A set of vectors $\{ \vec v_1, \dots, \vec v_k \}$ in a vector space $V$ is linearly dependent if there are scalars $c_1, \dots, c_k$, at least one of which is not zero, such that

$$c_1 \vec v_1 + \dots + c_k \vec v_k=\vec 0 $$

But considering the above example, does it mean that if $\{ \sin x, \cos x\}$ is linearly dependent, then

a) there exist scalars $c,d$ (at least one of which is not zero) such that $c\sin x+d\cos x=0$ for every $x$

or

b) for every $x$, there exist scalars $c,d$ (at least one of which is not zero) such that $c\sin x+d\cos x=0$

I can't tell this from the definition. Which one is true?

$\endgroup$
  • $\begingroup$ a) and b) are the same thing. You probably meant to put "for every $x$" after the "such that" in a). Then a) will be the correct meaning. The key point here is the word "functions". The functions $\sin x$ or $\cos x$ (or, to be more precise, $x \mapsto \sin x$ and $x \mapsto \cos x$) are linearly dependent if and only if there exist scalars $c$ and $d$ such that $c \left(x \mapsto \sin x\right) + d \left(x \mapsto \cos x\right) = 0$. Now, $c \left(x \mapsto \sin x\right) + d \left(x \mapsto \cos x\right) = \left(x \mapsto c \sin x + d \cos x\right)$, so the statement says that the ... $\endgroup$ – darij grinberg Aug 9 '15 at 16:56
  • $\begingroup$ I'd use a third way to state this: there exist scalars $c$, $d$ (at least one of which is not zero), such that $c\sin x+d\cos x=0$ for every $x$. $\endgroup$ – Aretino Aug 9 '15 at 16:56
  • $\begingroup$ ... function $x \mapsto c \sin x + d \cos x$ is zero, i.e., that $c \sin x + d \cos x = 0$ for all $x$. $\endgroup$ – darij grinberg Aug 9 '15 at 16:57
  • $\begingroup$ By the way, the reason why research mathematicians rarely have to deal with this confusion is that they tend not to abbreviate the function $x \mapsto T_x$ (where $T_x$ is some expression depending on $x$) as $T_x$. For instance, you don't say "the function $x^2-3$"; you either say "the function $x \mapsto x^2-3$" or say "the function sending every $x$ to $x^2-3$" or say "the function $f$ where $f\left(x\right) = x^2-3$". (In the case of $x \mapsto \sin x$, you can also just say "the function $\sin$".) I have seen "the function $x^2-3$" mostly in school maths, and it is a bad notation. $\endgroup$ – darij grinberg Aug 9 '15 at 17:00
  • $\begingroup$ @darijgrinberg I have edited the a), that is what I meant. Thank you for clarification. $\endgroup$ – user137035 Aug 9 '15 at 17:10
2
$\begingroup$

Call $f$ the function $f(x) = \cos(x)$ and $g$ the function $g(x) = \sin(x)$. Then, you need to prove that $$ c f + d g = 0 \Longrightarrow c = d = 0, $$ where in the identity $cf + dg = 0$, $0$ is actually the origin of your vector space, i.e. the constant function $0$. In other words, $cf + d g = 0$ means $$ c f(x) + d g(x) = 0, \quad \forall x \in \mathbb{R}, $$ hence you need to prove that $$ [c \cos(x) + d \sin(x) = 0, \quad \forall x \in \mathbb{R}] \Longrightarrow c = d = 0. $$

To prove this, just consider $x = 0$ and $x = \pi/2$.

$\endgroup$
1
$\begingroup$

$\sin x$ and $\cos x$ are real numbers when $x$ is real. The vectors in the vector space of real valued functions you are talking about are $\sin$ and $\cos$. These both vectors are linearly dependants iff by definition there are two real $a,b$ with at least one which is not zero such that $$a\sin +b\cos =0$$ and the above equality holds in the vector space of the real valued functions (it is the identically zero function). It means that for all $x\in\mathbb{R}$, we have $$a\sin x +b\cos x=0$$ and this last equality holds in $\mathbb{R}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy