Linear Independence of trigonometric functions

  • #26
S.G. Janssens
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If the objective is the same, then the system is still linearly dependent. For any sub-system's linear dependence implies the entire system if linearly dependent.
Sure, but the OP's question was to determine the dimension of the span of these functions, not just to determine whether or not they are independent.
 
  • #27
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Then we need to determine the maximal linearly independent sub-system owing to the result that ##r## linearly independent vectors span an ##r##-dimension subspace.
 
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  • #28
So, just to be sure, can you confirm that in the correct version of the problem we have the functions
$$
\sin{x}\cos{x}, \sin{2x}, \cos{2x}, \sin^2{x}, \cos^2{x}
$$
Yes, this is the correct form. I'll review the rest of your message later today.

If the objective is the same, then the system is still linearly dependent. For any sub-system's linear dependence implies the entire system is linearly dependent.
Is it? If it is I have no idea how to find the dimension (I see you talk about it, but I can't comprehend it right now).
 
  • #29
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Is it? If it is I have no idea how to find the dimension (I see you talk about it, but I can't comprehend it right now).
Yes, a system of vectors is linearly dependent if any vector in the system is a linear combination of the remaining vectors.
 
  • #30
Yes, a system of vectors is linearly dependent if any vector in the system is a linear combination of the remaining vectors.
Although I know this theorem (that's what we were taught it is), I still "don't know" that the remaining functions are linearly dependent. In this case, the answer should be one dimension.
What made me doubt you, was that Krylov mentioned I can prove that all the factors are equal to zero using a method ("evaluating the condition in some well chosen points...") I haven't use before. Wouldn't that make the said functions linearly independent?

By the way, is there a tool I can use to input functions and see if they're or not linearly independent? I could use that right now to see where exactly I'm heading.
 
  • #31
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For this particular problem you will not be able to prove that all the constants are equal to the zero element, because we have both told you there exists a vector that is a linear combination of the remaining vectors.

Overall, yes - if you prove that zero vector is produced only by a trivial linear combination (it means all constants are equal to the zero element), then you will have proved that the vectors are linearly independent.


Let us consider the vectors [itex]\sin x\cos x, \sin 2x,\cos 2x, \sin ^2x,\cos ^2x[/itex]
What is the rank (number of linearly independent vectors) of this system?
We have determined that [itex]\cos 2x = \cos ^2x -\sin ^2x + 0\ldots [/itex]
The rank can therefore be, at most, four. The vector [itex]\cos 2x[/itex] offers us no new information, so we "remove it from play".

We are left to consider [itex]\sin x\cos x, \sin 2x, \sin ^2x,\cos ^2x[/itex]
[itex]\sin x\cos x = \frac{1}{2}\sin 2x + 0\ldots [/itex]. Bam! Another one!


Consider now
[itex]\sin 2x, \sin ^2x,\cos ^2x[/itex]
and continue analogously.

Note that if a system consists of only one vector, then it is linearly independent if and only if that vector is a zero vector.

The dimension of the spanned subspace is 3.

In a general case. If we are given [itex]n[/itex] vectors [itex]a_k\in V_F, k\in \{1,2,\ldots ,n\}[/itex] we want to know the dimension of the subspace spanned by these vectors, we must determine the number of linearly independent vectors in that system of vectors. It's a recursive method:
1) If the number of L.I vectors is [itex]n[/itex], then dimension of spanned subspace is also [itex]n[/itex]
2) Assume [itex]r[/itex]-th vector is a linear combination of the remaining vectors, that means [itex]a_r = \sum_{j=1}^{r-1} a_j + \sum_{j=r+1}^n a_j[/itex]. "Remove the vector [itex]a_r[/itex] from play. Left with [itex]n-1[/itex] vectors. If 1), then result, else step2).

Things to note: we can never get rid of all vectors in the system, because ultimately we will come down to there being one vector left, but then it is linearly independent if and only if it's a zero vector. Hang on, that means the Entire system is linearly dependent all the way.

What dimension subspace does a zero vector span?

If you can wrap your head around all that then I have nothing left to teach you. You are now a Jedi!
 
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  • #32
I'm confused, because you both suggested ways to solve it different than those we used in class. Anyway, I'm going to read the topic again and try to understand all of your information.
 

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