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I need to discuss the linear independence of the following given vectors: \begin{align} \sin(x), \sin(x+1), \sin(x+2)\end{align} there are many similar questions on math.SE but most of which I have looked into deal with integrals and all I have to work with is the elementary definition of linear independence:

The given vectors are considered as linearly independent if and only if \begin{align}\lambda_1 \sin(x)+\lambda_2\sin(x+1)+\lambda_3\sin(x+2)=0 \implies \lambda_1=\lambda_2=\lambda_3=0 \end{align} The way I have approached this now was to use the trigonometric identities and then simplify the result, merging the constants together. I am not sure if this is a correct workaround but here is my attempt: \begin{align}\lambda_1 \sin(x) + \lambda_2 (\sin(1)\cos(x)+\cos(1)\sin(x))+\lambda_3(\sin(2)\cos(x)+\cos(2)\sin(x)) \\=(\underbrace{\lambda_1+\lambda_2\cos(1)+\lambda_3\cos(2)}_{:=k_1})\sin(x)+ (\underbrace{\lambda_2 \sin(1)+\lambda_3\sin(2)}_{:=k_2})\cos(x) \end{align} such that I can write: \begin{align} k_1 \sin(x) + k_2 \cos(x)=0 \implies k_1=k_2=0 \end{align} Due to the fact that $ \cos(x), \sin(x)$ are linearly independent. My problem is that I seem to have 'erased' one scalar variable with my substitution, would I need to use back substitution now and see if I can deduce $\lambda_1=\lambda_2=\lambda_3$ from there, or is my approach wrong from the beginning?


If I can bother to ask, intuition wise the given set do not look like vectors at all to me, since I am very new to this subject I may have wait with further understanding this problem, but would one of the above vectors just look like the regular function, depending on at which point $x$ they are evaluated?

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A vector is what we call a member of a vector space, whatever it might be. – Git Gud Oct 27 '13 at 21:36
"or is my approach wrong from the beginning" No, the approach is right. Maybe you should think about which result you will be able to prove. – Daniel Fischer Oct 27 '13 at 21:37
in an earlier exercise I did already have to show that $\cos(x), \sin(x)$ are linearly independent, my goal was to reach that point again which I have managed to do by merging the constants together as demonstrated. Is this already sufficient or are there some steps left? – Spaced Oct 27 '13 at 21:40
A different path you might think about... when we discuss function space the equations you write hold for all $x$. Thus, we can evaluate at special values of $x$ to cipher data. For example, you might think about $x=\pi$ and $x = 1-\pi$ etc.. – James S. Cook Oct 27 '13 at 21:40
Did you try to developp $sin(x+1+1)$ instead of $sin(x+2)$? – Katsu Oct 27 '13 at 21:42
up vote 1 down vote accepted

Expand $\sin(x+1)$ and $\sin(x+2)$ using the identity for $\sin(\alpha+\beta)$. You will find that the three "vectors" $\sin(x)$, $\sin(x+1)$ , and $\sin(x+2)$ are all linear combinations of the two "vectors" $\sin(x)$ and $\cos(x)$. What can you conclude from that? Try to work your conclusion into a proof using the definition of linear independence or linear dependence.

EDIT: I'm sorry, it looks you already did most of what I suggested above. You get a system of two linear equations that must be satisfied by the three variables $\lambda_1$, $\lambda_2$, and $\lambda_3$. What usually happens when you have fewer equations then variables? Can you find a solution in which at least of one the $\lambda$'s is nonzero?

Whoever asked you the question should have specified what the vector space is. I have to guess. I'll guess it's the set of all continuous functions from the real numbers to the real numbers. A "vector" is such a function. "Vectors" are added to each other and mutliplied by scalars in the obvious manner (if this is not obvious to you, hopefully you have a text, and that text should have examples like this one).

As Git Gud wrote, to a mathematician, a "vector" is simply an element of a vector space. Wikipedia has an article on vector spaces that begins with a long explanation, probably aimed at non-mathematicians, explaining why one might be interested in such a concept, before they get to the definition. Unfortunately, the precise definition includes a long list of rules. Fortunately, all of the rules are intuitively compelling, and it is hard to imagine (at least for me) why one would want a "vector space" not to satisfy any of the rules.

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