Thread: 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}?

Hello, angelme!

We want nonzero numbers so that:

. .

. .

. .


Two polynomials are equal if their corresponding coefficients are equal:
. . . .

Solve the system: .

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?

someone please help correct my work above

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

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 becomes . Taking x= 1, 3a+ b- c= 0. Taking x= 2, 21a+ 19b+ 11c= 0.