# Linearly independence and dependence of polynomials

• Jul 23rd 2010, 05:27 AM
Linearly independence and dependence of polynomials
Hi

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

$\displaystyle p1= 2-x + 4x^2; p2= 3 + 6x +2x^2; p3= 2+10x-4x^2$

Thanks
• Jul 23rd 2010, 05:49 AM
HallsofIvy
The most direct way is to use the definition of "linear indepdence" and "linear dependence":
If there exist numbers, $\displaystyle a_1$, $\displaystyle a_2$, ..., $\displaystyle a_n$, not all 0, such that $\displaystyle a_1v_1+ a_2v-2+ \cdot\cdot\cdot+ a_nv_n= 0$, where that "0" is the 0 vector, then the vectors $\displaystyle v_1$, $\displaystyle v_2$, ..., $\displaystyle v_n$ are dependent, if not, then they are independent.

So you start by setting up the linear combination $\displaystyle a_1v_1+ a_2v-2+ \cdot\cdot\cdot+ a_nv_n$ which, here, is $\displaystyle a_1p_1+ a_2p_2+ a_3p_3= a_1(2- x+ 4x^2)+ a_2(3+ 6x+ 2x^2)+ a_3(2+ 10x- 4x^2)= 0$ 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.)
• Jul 23rd 2010, 05:59 AM
Quote:

Originally Posted by HallsofIvy
The most direct way is to use the definition of "linear indepdence" and "linear dependence":
If there exist numbers, $\displaystyle a_1$, $\displaystyle a_2$, ..., $\displaystyle a_n$, not all 0, such that $\displaystyle a_1v_1+ a_2v-2+ \cdot\cdot\cdot+ a_nv_n= 0$, where that "0" is the 0 vector, then the vectors $\displaystyle v_1$, $\displaystyle v_2$, ..., $\displaystyle v_n$ are dependent, if not, then they are independent.

So you start by setting up the linear combination $\displaystyle a_1v_1+ a_2v-2+ \cdot\cdot\cdot+ a_nv_n$ which, here, is $\displaystyle a_1p_1+ a_2p_2+ a_3p_3= a_1(2- x+ 4x^2)+ a_2(3+ 6x+ 2x^2)+ a_3(2+ 10x- 4x^2)= 0$ 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.
• Aug 4th 2010, 07:00 AM