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

[nuuskur]

Homework Statement

Given the system of vectors \cos x, \cos (x+2), \sin (x-5). 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:
k_1\cos x + k_2\cos (x+2) + k_3\sin (x-5) = 0. 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 x\in\mathbb{R}, 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.

 

[Mark44]

Homework Statement

Given the system of vectors \cos x, \cos (x+2), \sin (x-5). 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:
k_1\cos x + k_2\cos (x+2) + k_3\sin (x-5) = 0. 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.

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 x\in\mathbb{R}, 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.

 

[Ray Vickson]

Homework Statement

Given the system of vectors \cos x, \cos (x+2), \sin (x-5). 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:
k_1\cos x + k_2\cos (x+2) + k_3\sin (x-5) = 0. 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 x\in\mathbb{R}, 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 $\sin(x)$.

 

[nuuskur]

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.

 

[Mark44]

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.

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.

 

[Ray Vickson]

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?

 

[nuuskur]

tells me that a non-trivial solution exists and therefore the system is linearly dependant.Thanks, I get it now.