Are These Vectors Linearly Independent in R*R?

In summary, The conversation is discussing whether sets of vectors in a vector space R*R of functions from R to R are linearly independent. For set b, the definition of linear independence is used to show that the equation c1*sin(t) + c2*sin2(t) + c3*sin3(t) = 0 has only one solution for the constants c1, c2, and c3. For set c, the functions 1, sin2(t), and cos2(t) are shown to be linearly dependent through the equation c1*1 + c2*sin2(t) + c3*cos2(t) = 0 having a solution where not all constants are zero.
  • #1
ak123456
50
0

Homework Statement


determine whether the following are linear independent sets of vectors in the vector space R*R of all functions from R to R
a)fn:=1+t+...+t^n for n=1,...4
b) sin,sin^2 ,sin^3
c)1,sin^2,cos^2

Homework Equations





The Attempt at a Solution


can i do like this for (b)(c)
sin^3=sin(sin^2)
sin^2+cos^2=1
so they are linearly dependent sets
 
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  • #2
ak123456 said:
sin^3=sin(sin^2)
That doesn't look like a linear combination.
 
  • #3
wywong said:
That doesn't look like a linear combination.

why? I think sin^3=(sin) * (sin^2) is right
 
  • #4
ak123456 said:
why? I think sin^3=(sin) * (sin^2) is right
It's right but completely irrelevant.

For b, use the definition of linear independence to show that the equation
c1*sin(t) + c2*sin2(t) + c3*sin3(t) = 0 has exactly one solution for the constants c1, c2, and c3. (If there is more than one solution, the functions are linearly dependent.)

For c, the functions 1, sin2(t) and cos2(t) are linearly dependent, as you said. The first function, 1, is a linear combination of the other two.

Another way to look at this is that the equation
c1*1 + c2*sin2(t) + c3*cos2(t) = 0 has a solution where not all of the constants are zero. One such solution is c1 = -1, c2 = 1, c3 = 1. There are lots of solutions.
 
  • #5
Mark44 said:
It's right but completely irrelevant.

For b, use the definition of linear independence to show that the equation
c1*sin(t) + c2*sin2(t) + c3*sin3(t) = 0 has exactly one solution for the constants c1, c2, and c3. (If there is more than one solution, the functions are linearly dependent.)

For c, the functions 1, sin2(t) and cos2(t) are linearly dependent, as you said. The first function, 1, is a linear combination of the other two.

Another way to look at this is that the equation
c1*1 + c2*sin2(t) + c3*cos2(t) = 0 has a solution where not all of the constants are zero. One such solution is c1 = -1, c2 = 1, c3 = 1. There are lots of solutions.

i see thanks a lot
 

1. What is a linear independent set?

A linear independent set is a collection of vectors in a vector space that are not linear combinations of each other. This means that no vector in the set can be written as a combination of the other vectors using scalar multiplication and addition.

2. How can you determine if a set is linear independent?

To determine if a set is linear independent, you can use the linear independence test. This involves setting up a system of equations using the vectors in the set and solving for the coefficients. If the only solution is when all coefficients are equal to zero, then the set is linear independent.

3. Can a set with only one vector be linear independent?

Yes, a set with only one vector can be linear independent. This is because there are no other vectors in the set for it to be a linear combination of.

4. What is the relationship between linear independence and linear dependence?

Linear independence and linear dependence are opposite concepts. A set is linear independent if none of its vectors can be written as a linear combination of the others, while a set is linear dependent if at least one vector can be written as a linear combination of the others.

5. Why is linear independence important in linear algebra?

Linear independence is important in linear algebra because it allows us to create a basis for a vector space. A basis is a set of linear independent vectors that span the entire vector space, meaning that any vector in the space can be written as a linear combination of the basis vectors. This allows for easier calculations and understanding of vector spaces.

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