Are These Vectors Linearly Independent in R*R?

ak123456
Messages
50
Reaction score
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
 
Physics news on Phys.org
ak123456 said:
sin^3=sin(sin^2)
That doesn't look like a linear combination.
 
wywong said:
That doesn't look like a linear combination.

why? I think sin^3=(sin) * (sin^2) is right
 
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.
 
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
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top