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

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- Jul 23rd 2010, 05:27 AMadam_leedsLinearly 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 AMHallsofIvy
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 AMadam_leeds
- Aug 4th 2010, 07:00 AMadam_leeds