Thread: Linearly independence and dependence of polynomials

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