Help Linear Dependent - Shortcut

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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?

How about {x⁴+x²+1, x⁴-x²+1, x⁴-x²-1}?
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 {x⁴+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

someone please help correct my work above

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 {x⁴+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.