# Show (1,a,a^2) (1,b,b^2) (1,c,c^2) are linearly independant

• Apr 16th 2011, 08:24 PM
genericguy
Show (1,a,a^2) (1,b,b^2) (1,c,c^2) are linearly independant
Hi there,
I'm attempting to solve the following question: Show {1,a,a^2),(1,b,b^2),(1,c,c^2)} are linearly independant if a, b and c are distinct.

I'm sure I'm so close!
What I've come up with so far is putting the three vectors in rows in a matrix, and finding the determinant with row expansion, which gives a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2)

I know that they are linearly independant when this determinant doesn't equal zero, and it's simple to show it DOES equal zero if a=b, b=c or c=a. For example, when a=b, b^3-bc^2+bc^2-b^3+c(b^2-b^2)=0

However, I don't think showing this is sufficient proof that when they are distinct, the determinant will never equal zero.

Could you point me in the right direction?
Thanks!
• Apr 16th 2011, 08:39 PM
alexmahone
You can factorise the determinant as:
det=(a-b)(b-c)(c-a)

The result follows immediately.
• Apr 16th 2011, 08:48 PM
genericguy
Wow, I was trying so hard to factorise it but missed that completely. :|
Thanks!
• Apr 17th 2011, 07:34 PM
Drexel28
Quote:

Originally Posted by genericguy
Hi there,
I'm attempting to solve the following question: Show {1,a,a^2),(1,b,b^2),(1,c,c^2)} are linearly independant if a, b and c are distinct.

I'm sure I'm so close!
What I've come up with so far is putting the three vectors in rows in a matrix, and finding the determinant with row expansion, which gives a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2)

I know that they are linearly independant when this determinant doesn't equal zero, and it's simple to show it DOES equal zero if a=b, b=c or c=a. For example, when a=b, b^3-bc^2+bc^2-b^3+c(b^2-b^2)=0

However, I don't think showing this is sufficient proof that when they are distinct, the determinant will never equal zero.

Could you point me in the right direction?
Thanks!

Just as a point of interest, in general (1,x_1,s,x_1^n),...,(1,x_n,...,x_n^n) are linearly independent if x_j != x_k k,j in [n] This is the Vandermonde matrix.