Help Linear Dependent - Shortcut

Show that {x^{2} , x^{2}+x+1, x^{2}+x, x-1} is LD

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

(x^{2}+x) - 1/2(x^{2}+x+1) -1/2(x-1) -1/2(x^{2}) =0 ---> LD

Are there any formula or shortcut for this?

How about {x^{4}+x^{2}+1, x^{4}-x^{2}+1, x^{4}-x^{2}-1}?

Re: Help Linear Dependent - Shortcut

Hello, angelme!

We want nonzero numbers so that:

. .

. .

. .

Two polynomials are equal if their corresponding coefficients are equal:

. . . .

Solve the system: .

Re: Help Linear Dependent - Shortcut

Thank you Soraban!!!

So...for {x^{4}+x^{2}+1, x^{4}-x^{2}+1, x^{4}-x^{2}-1}

a(x^{4}+x^{2}+1) + b(x^{4}-x^{2}+1) + c(x^{4}-x^{2}-1) =0

(a+b+c)x^{4 }+ (a-b-c)x^{2 }+ (a+b-c) = 0

then I get

a=0

b=0

c=0

Does it mean it is LI?

Re: Help Linear Dependent - Shortcut

someone please help correct my work above

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

Re: Help Linear Dependent - Shortcut

Quote:

Originally Posted by

**angelme** Thank you Soraban!!!

So...for {x^{4}+x^{2}+1, x^{4}-x^{2}+1, x^{4}-x^{2}-1}

a(x^{4}+x^{2}+1) + b(x^{4}-x^{2}+1) + c(x^{4}-x^{2}-1) =0

(a+b+c)x^{4 }+ (a-b-c)x^{2 }+ (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.