Thread: Linearly independence and dependence of polynomials

Hi

How do we work out if the vectors in P2 are linearly dependent or independent.



Thanks

The most direct way is to use the definition of "linear indepdence" and "linear dependence":
If there exist numbers, , , ..., , not all 0, such that , where that "0" is the 0 vector, then the vectors , , ..., are dependent, if not, then they are independent.

So you start by setting up the linear combination which, here, is where that "0" is the 0 function, f(x)= 0 for all x.

Now you can "combine like terms" and use the fact that if a polynomial is 0 for all x then all coefficients must be 0 to get three equations to solve for a, b, and c. Since all three equations will be "= 0", the have the obvious solution a= b= c= 0. If that is only solution, then the vectors are independent. It that solution is not unique, if there exist other, non-zero, solutions, then they are dependent.

(You don't have to actually find a, b, and c. You can use the fact, if you aready know it, that a system of equations has a unique solution if and only if the determinant of the coefficient matrix is non-zero.)

Originally Posted by HallsofIvy

The most direct way is to use the definition of "linear indepdence" and "linear dependence":
If there exist numbers, , , ..., , not all 0, such that , where that "0" is the 0 vector, then the vectors , , ..., are dependent, if not, then they are independent.

So you start by setting up the linear combination which, here, is where that "0" is the 0 function, f(x)= 0 for all x.

Now you can "combine like terms" and use the fact that if a polynomial is 0 for all x then all coefficients must be 0 to get three equations to solve for a, b, and c. Since all three equations will be "= 0", the have the obvious solution a= b= c= 0. If that is only solution, then the vectors are independent. It that solution is not unique, if there exist other, non-zero, solutions, then they are dependent.

(You don't have to actually find a, b, and c. You can use the fact, if you aready know it, that a system of equations has a unique solution if and only if the determinant of the coefficient matrix is non-zero.)

Thank you so the determinant of that is -32, therefore it doesnt have a unique solution therefore its linearly dependent.

Originally Posted by adam_leeds

Thank you so the determinant of that is -32, therefore it doesnt have a unique solution therefore its linearly dependent.

Is this right?

I did the determinant of

2 -1 4
3 6 2
2 10 -4