Define independency of vectors

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In summary: Anyway, you are supposed to find a polynomial that solves the equation c_1\vec{v}_1+c_2\vec{v}_2+c_3\vec{v}_3=0 for all x.
  • #1
Deimantas
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Homework Statement



It roughly translates to this: "Define linear dependency of vectors". I hope it makes any sense.

"1, x, sinx; x is defined (-∞, +∞)."

Homework Equations


The Attempt at a Solution



I believe that to solve this problem, I have to turn these vectors into a quadratic polynomial, something like this: "t^2+t+1". However I have no idea what substitution should I use in order to turn it into a polynomial. I checked the common trig substitutions and didn't find anything helpful.
 
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  • #2
Linear independency of for example 3 vectors is defined as:
If the following equation only lead to the trivial solution [itex]c_{1,2,3} = 0[/itex], then the vectors are linearly independent:
[itex] c_1\vec{v}_1+c_2\vec{v}_2+c_3\vec{v}_3 = 0[/itex]
And dependency is the oppsite. So if you are to prove dependency, find 1 nontrivial solution to the equation.
 
  • #3
But how can I solve such equation

c1(1)+c2(x)+c3(sinx)=0

where vector elements are variables(functions) instead of the usual integers?
I'm pretty sure the vectors will turn out to be independent, but I don't know how to prove it in this case...
 
  • #4
Deimantas said:
But how can I solve such equation

c1(1)+c2(x)+c3(sinx)=0

where vector elements are variables(functions) instead of the usual integers?
I'm pretty sure the vectors will turn out to be independent, but I don't know how to prove it in this case...

You need to determine if there are constants [itex] (c_1,c_2,c_3) \neq (0,0,0)[/itex] that make the equation [itex] c_1 \cdot 1 + c_2 \cdot x + c_3 \cdot \sin x = 0 [/itex] for all x. Just imagine you could find such ci and then try to work out a contradiction to some known facts.

RGV
 
  • #5
What is the exact statement of the problem? You say "Define Linear Dependency of Vectors" but that has nothing to do with "1, x, sin(x)". Are you trying to show that 1, x, and sin(x) are independent as members of the space of continuous functions? But neither of those questions has anything to do with polynomials.
 
  • #6
HallsofIvy said:
What is the exact statement of the problem? You say "Define Linear Dependency of Vectors" but that has nothing to do with "1, x, sin(x)". Are you trying to show that 1, x, and sin(x) are independent as members of the space of continuous functions? But neither of those questions has anything to do with polynomials.

Yes, I may have stated the problem in a wrong way.
Look at this exercise:

Is the system of functions sinx, cos^2(x), 1 independent? The interval of x is (0; 2π).

λ1(sinx)+λ2(cos^2(x))+λ3(1)=0 The interval of x is (0; 2π).

cos^2(x) equals 1-sin^2(x), therefore the equation is

λ1(sinx)-λ2(sin^2(x))+(λ2+λ3)=0 The interval of x is (0; 2π).

let's assume sinx = t, and (λ2+λ3)=α, then the equation is

-λ2(t^2)+λ1(t)+α=0 The interval of t is (-1; 1).

that's a polynomial, and it equals 0 only when all lambdas equal zero, therefore the system of functions is independent.


I was hoping I could do the same with 1, x, sinx somehow, that's why I mentioned polynomial. Though I don't see a way to apply the same technique in my case.
 
  • #7
Deimantas said:
Yes, I may have stated the problem in a wrong way.
Look at this exercise:

Is the system of functions sinx, cos^2(x), 1 independent? The interval of x is (0; 2π).

λ1(sinx)+λ2(cos^2(x))+λ3(1)=0 The interval of x is (0; 2π).

cos^2(x) equals 1-sin^2(x), therefore the equation is

λ1(sinx)-λ2(sin^2(x))+(λ2+λ3)=0 The interval of x is (0; 2π).

let's assume sinx = t, and (λ2+λ3)=α, then the equation is

-λ2(t^2)+λ1(t)+α=0 The interval of t is (-1; 1).

that's a polynomial, and it equals 0 only when all lambdas equal zero, therefore the system of functions is independent.


I was hoping I could do the same with 1, x, sinx somehow, that's why I mentioned polynomial. Though I don't see a way to apply the same technique in my case.

I already told you in my first response what you need to do. Looking for polynomials would not be helpful at all, and I don't know why you would want to do it.

RGV
 
  • #8
Ray Vickson said:
I already told you in my first response what you need to do. Looking for polynomials would not be helpful at all, and I don't know why you would want to do it.

RGV

I somehow didn't manage to notice your first post, my apologies. I'll try to work it out, thanks.
 
  • #9
Ray Vickson said:
You need to determine if there are constants [itex] (c_1,c_2,c_3) \neq (0,0,0)[/itex] that make the equation [itex] c_1 \cdot 1 + c_2 \cdot x + c_3 \cdot \sin x = 0 [/itex] for all x. Just imagine you could find such ci and then try to work out a contradiction to some known facts.

RGV

Well, I couldn't find any c's, other than 0's, that would make that equation work for all x, there are only some unique cases like "x=3π/2,c1=1,c2=0,c3=1". That tells me the functions are linearly independent. But I don't know how to put it in my textbook so that it would satisfy the teacher. How should I have solved the c1(1)+c2(x)+c3(sinx)=0 equation in a reasonable way?
 
  • #10
Deimantas said:
Well, I couldn't find any c's, other than 0's, that would make that equation work for all x, there are only some unique cases like "x=3π/2,c1=1,c2=0,c3=1". That tells me the functions are linearly independent. But I don't know how to put it in my textbook so that it would satisfy the teacher. How should I have solved the c1(1)+c2(x)+c3(sinx)=0 equation in a reasonable way?

IF there are constants c_i, not all zero, that make the above equation into an identity in x, what must happen? Say c_3 ≠ 0; that means we can divide by c_3 to get
[tex] \sin(x) = -\frac{c_1}{c_3} - \frac{c_2}{c_3} x \text{ for all } x \in R, [/tex] and that would mean what? Since we get a ridiculous conclusion, we cannot have c_3 non-zero. Thus, we need c_3 = 0. Now look at c_1 + c_2 x = 0 for all x. What does that say about c_1 and c_2?

Note: the equation may hold for some x, even for a lot of different values of x, but that is not the point. The crucial issue is whether or not it holds for ALL x.

RGV
 

What is the definition of independency of vectors?

The independency of vectors refers to the property of two or more vectors being linearly independent, meaning that one vector cannot be written as a linear combination of the other vectors. In other words, the vectors are not dependent on each other and can exist independently.

How is independency of vectors determined?

The independency of vectors can be determined by using the linear dependence test, which involves setting up a system of equations and solving for the coefficients of the vectors. If the only solution is the trivial solution (all coefficients equal to 0), then the vectors are independent. If there are other solutions, then the vectors are dependent.

What is the importance of independency of vectors in mathematics and science?

The concept of independency of vectors is essential in various branches of mathematics and science, such as linear algebra, physics, and engineering. It allows for a deeper understanding of vector spaces and their properties, and it is the basis for many mathematical and scientific models and equations.

Can more than two vectors be independent?

Yes, more than two vectors can be independent. In fact, a set of n vectors is linearly independent if and only if the only solution to the system of equations formed by these vectors is the trivial solution. This means that each vector is necessary for the span of the set, and no vector can be removed without changing the span.

What is the difference between linearly independent and dependent vectors?

The main difference between linearly independent and dependent vectors is that linearly independent vectors cannot be written as a linear combination of each other, while linearly dependent vectors can. In other words, linearly independent vectors are not reliant on each other, while linearly dependent vectors are. Additionally, linearly independent vectors form a basis for the vector space they span, while linearly dependent vectors do not.

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