# Help Linear Dependent - Shortcut

• October 7th 2012, 09:06 AM
angelme
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?

• October 7th 2012, 09:53 AM
Soroban
Re: Help Linear Dependent - Shortcut
Hello, angelme!

Quote:

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

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

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

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

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

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

Solve the system: . $\begin{Bmatrix}a&=& t \\ b &=& t \\ c &=&\text{-}2t \\ d &=& t \end{Bmatrix}$
• October 7th 2012, 10:43 AM
angelme
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?
• October 8th 2012, 04:50 PM
angelme
Re: Help Linear Dependent - Shortcut

Question 2: {x4+x2+1, x4-x2+1, x4-x2-1}
• October 9th 2012, 07:57 AM
HallsofIvy
Re: Help Linear Dependent - Shortcut
Quote:

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 $a(x^4+ x^2+ 1)+ b(x^4+ x^2- 1)+ c(x^4- x^2- 1)= 0$ becomes $a- b- c= 0$. Taking x= 1, 3a+ b- c= 0. Taking x= 2, 21a+ 19b+ 11c= 0.