Linear Independence of \cos x, \cos (x+2), \sin (x-5)

In summary: This is because, as I said, you can write the third one as a linear combination of the first two, and the second one can then be written as a linear combination of the first one and the third one. So, all three are dependent.In summary, the system of vectors given by cos x, cos (x+2), and sin (x-5) is linearly dependent. This can be shown by using the addition formulas for sin and cos to rewrite cos (x+2) and sin (x-5) in terms of cos x and sin x. Taking the derivatives of the equation k1cos x + k2cos(x+2) + k3sin
  • #1
nuuskur
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Homework Statement


Given the system of vectors [itex]\cos x, \cos (x+2), \sin (x-5)[/itex]. Determine whether the system is linearly independent.

Homework Equations

The Attempt at a Solution


If it were linearly dependent, there would exist a non-trivial linear combination, such that:
[itex]k_1\cos x + k_2\cos (x+2) + k_3\sin (x-5) = 0[/itex]. Do not suggest the Wronski determinant, please - we haven't covered it in our Algebra, yet. There's got to be a simpler way to justify the system's (in)dependence.

If I fix any two constants as 0, then the third one will automatically be zero, as the equality has to be true for every [itex]x\in\mathbb{R}[/itex], therefore leaving us only with a trivial solution.

There is also a Lemma in the textbook that says whenever a sub-system turns out to be linearly dependent (provided the sub-system consists of at least two vectors, as according to another Lemma, the solution is automatically trivial if the sub-system consists of only one vector) then from there it follows the whole system is linearly dependent.

I'm quite sure the opposite isn't true - that if any sub-system is not linearly dependent then the whole system is linearly independent.

I'm not sure what to do here.
 
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  • #2
nuuskur said:

Homework Statement


Given the system of vectors [itex]\cos x, \cos (x+2), \sin (x-5)[/itex]. Determine whether the system is linearly independent.

Homework Equations

The Attempt at a Solution


If it were linearly dependent, there would exist a non-trivial linear combination, such that:
[itex]k_1\cos x + k_2\cos (x+2) + k_3\sin (x-5) = 0[/itex]. Do not suggest the Wronski determinant, please - we haven't covered it in our Algebra, yet. There's got to be a simpler way to justify the system's (in)dependence.
Since the equation above is identically true for all values of x, taking the derivative of both sides will give you another equation that is also identically true. Then you'll have two equations in three unknowns (##k_1, k_2, k_3##). To get the third equation, take the derivative one more time.
nuuskur said:
If I fix any two constants as 0, then the third one will automatically be zero, as the equality has to be true for every [itex]x\in\mathbb{R}[/itex], therefore leaving us only with a trivial solution.

There is also a Lemma in the textbook that says whenever a sub-system turns out to be linearly dependent (provided the sub-system consists of at least two vectors, as according to another Lemma, the solution is automatically trivial if the sub-system consists of only one vector) then from there it follows the whole system is linearly dependent.

I'm quite sure the opposite isn't true - that if any sub-system is not linearly dependent then the whole system is linearly independent.

I'm not sure what to do here.
 
  • #3
nuuskur said:

Homework Statement


Given the system of vectors [itex]\cos x, \cos (x+2), \sin (x-5)[/itex]. Determine whether the system is linearly independent.

Homework Equations

The Attempt at a Solution


If it were linearly dependent, there would exist a non-trivial linear combination, such that:
[itex]k_1\cos x + k_2\cos (x+2) + k_3\sin (x-5) = 0[/itex]. Do not suggest the Wronski determinant, please - we haven't covered it in our Algebra, yet. There's got to be a simpler way to justify the system's (in)dependence.

If I fix any two constants as 0, then the third one will automatically be zero, as the equality has to be true for every [itex]x\in\mathbb{R}[/itex], therefore leaving us only with a trivial solution.

There is also a Lemma in the textbook that says whenever a sub-system turns out to be linearly dependent (provided the sub-system consists of at least two vectors, as according to another Lemma, the solution is automatically trivial if the sub-system consists of only one vector) then from there it follows the whole system is linearly dependent.

I'm quite sure the opposite isn't true - that if any sub-system is not linearly dependent then the whole system is linearly independent.

I'm not sure what to do here.

Use the addition formulas for ##\sin## and ##\cos##, to re-express ##\cos(x+2)## and ##\sin(x-5)## in terms of ##\cos(x)## and ##\sin(x)##.
 
  • #4
Mark44 said:
Since the equation above is identically true for all values of x, taking the derivative of both sides will give you another equation that is also identically true. Then you'll have two equations in three unknowns (##k_1, k_2, k_3##). To get the third equation, take the derivative one more time.
Ah, so I have to make sure the system determinant is not 0 and then I can safely deduce the only solution is trivial and therefore the system is linearly independent. (Attempting to solve via Cramer's method)Edit:
Trouble: the system determinant is 0, so for every coefficient I get 0/0. According to Wronski, the system is linearly dependant if the determinant is 0, but is there another way to justifying this - we haven't gone over the Theorem in our Algebra course, yet.
 
Last edited:
  • #5
nuuskur said:
Ah, so I have to make sure the system determinant is not 0 and then I can safely deduce the only solution is trivial and therefore the system is linearly independent. (Attempting to solve via Cramer's method)
What I was thinking of was not Cramer's method, but writing the coefficients in a matrix and using row reduction to solve for the constants. If the system is dependent you'll get an infinite number of solutions, something that is suggested by Cramer's giving you 0 for the determinant.
nuuskur said:
Edit:
Trouble: the system determinant is 0, so for every coefficient I get 0/0. According to Wronski, the system is linearly dependant if the determinant is 0, but is there another way to justifying this - we haven't gone over the Theorem in our Algebra course, yet.
 
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  • #6
nuuskur said:
Ah, so I have to make sure the system determinant is not 0 and then I can safely deduce the only solution is trivial and therefore the system is linearly independent. (Attempting to solve via Cramer's method)Edit:
Trouble: the system determinant is 0, so for every coefficient I get 0/0. According to Wronski, the system is linearly dependant if the determinant is 0, but is there another way to justifying this - we haven't gone over the Theorem in our Algebra course, yet.

If you do what I suggested in #3, you will see that the three vectors u1 = cos(x), u2=cos(x+2) and u3 = sin(x-5) are linear combinations of the two vectors v1 = cos(x) and v2 = sin(x). What does that tell you about whether or not the three vectors u1, u2, u3 are linearly independent or not?
 
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  • #7
tells me that a non-trivial solution exists and therefore the system is linearly dependant.Thanks, I get it now.
 
Last edited:

What is the definition of linear independence?

Linear independence is a term used in linear algebra to describe a set of vectors that are not multiples of each other. In other words, no vector in the set can be written as a linear combination of the others.

What is the significance of linear independence in mathematics and science?

Linear independence is an important concept in mathematics and science because it allows us to determine if a set of vectors can span a vector space. It also plays a crucial role in solving systems of linear equations and understanding the behavior of linear systems.

How can we determine if a set of functions, such as \cos x, \cos (x+2), \sin (x-5), are linearly independent?

To determine if a set of functions is linearly independent, we can use the Wronskian determinant. If the Wronskian determinant is non-zero, then the set of functions is linearly independent. In this case, the Wronskian determinant for the given set of functions is non-zero, so they are linearly independent.

What is the relationship between linear independence and orthogonality?

Linearly independent vectors are also orthogonal to each other, meaning they are perpendicular in n-dimensional space. This is because if two vectors are linearly independent, they cannot be multiples of each other, and therefore cannot lie on the same line.

How can we use the concept of linear independence in practical applications?

Linear independence is used in many fields, including physics, engineering, and computer science. It is used to solve systems of linear equations, analyze the stability of systems, and determine if a set of data is redundant. In machine learning, linear independence is used to select the most important features in a dataset and reduce dimensionality.

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