# Thread: Help Linear Dependent - Shortcut

1. ## Help Linear Dependent - Shortcut

Show that {x2 , x2+x+1, x2+x, x-1} is LD

I took like 20 mins to find the solution with trial and error to get

(x2+x) - 1/2(x2+x+1) -1/2(x-1) -1/2(x2) =0 ---> LD

Are there any formula or shortcut for this?

2. ## Re: Help Linear Dependent - Shortcut

Hello, angelme!

$\displaystyle \text{Show that: }\:\begin{Bmatrix}x^2 \\ x^2+x+1 \\ x^2+x \\ x-1\end{Bmatrix}\,\text{ are linearly dependent.}$

We want nonzero numbers $\displaystyle a,b,c,d$ so that:

. . $\displaystyle a(x^2) + b(x^2+x+1) + c(x^2+x) + d(x - 1) \;=\;0$

. . $\displaystyle ax^2 + bx^2 + bx + b + cx^2 + cx + dx - d \;=\;0$

. . $\displaystyle (a+b+c)x^2 + (b+c+d)x + (b-d) \;=\;0x^2 + 0x + 0$

Two polynomials are equal if their corresponding coefficients are equal:
. . . . $\displaystyle \begin{Bmatrix} a+b+c \;=\;0 \\ b+c+d \;=\;0 \\ b-d \;=\;0 \end{Bmatrix}$

Solve the system: .$\displaystyle \begin{Bmatrix}a&=& t \\ b &=& t \\ c &=&\text{-}2t \\ d &=& t \end{Bmatrix}$

3. ## Re: Help Linear Dependent - Shortcut

Thank you Soraban!!!

So...for {x
4+x2+1, x4-x2+1, x4-x2-1}

a(
x4+x2+1) + b(x4-x2+1) + c(x4-x2-1) =0
(a+b+c)
x4 + (a-b-c)x2 + (a+b-c) = 0

then I get

a=0
b=0
c=0

Does it mean it is LI?

4. ## Re: Help Linear Dependent - Shortcut

Question 2: {x4+x2+1, x4-x2+1, x4-x2-1}

5. ## Re: Help Linear Dependent - Shortcut

Originally Posted by angelme
Thank you Soraban!!!

So...for {x
4+x2+1, x4-x2+1, x4-x2-1}

a(
x4+x2+1) + b(x4-x2+1) + c(x4-x2-1) =0
(a+b+c)
x4 + (a-b-c)x2 + (a+b-c) = 0

then I get

a=0
b=0
c=0

Does it mean it is LI?
Yes.

Yet another way to do this is to simply choose three numbers for x, getting three equations for a, b, and c.

For example, taking x= 0, the $\displaystyle a(x^4+ x^2+ 1)+ b(x^4+ x^2- 1)+ c(x^4- x^2- 1)= 0$ becomes $\displaystyle a- b- c= 0$. Taking x= 1, 3a+ b- c= 0. Taking x= 2, 21a+ 19b+ 11c= 0.