Originally Posted by

**HallsofIvy**

2) A "basis" for a vector space is defined by two properties:

a) It spans the space.

b) The vectors in it are independent.

To show that this set spans the space, you must show that any polynomial in it, $\displaystyle y= ax^2+ bx+ c$ for any numbers a, b, and c, can be written as a linear combination of $\displaystyle x^2-1$, x+ 1, and $\displaystyle x^2+ x+ 1$. That is, given that $\displaystyle ax^2+ bx+ c= \alpha(x^2- 1)+ \beta(x+1)+ \gamma(x^2+ x+ 1)$, show that you can solve for $\displaystyle \alpha$, $\displaystyle \beta$, and $\displaystyle \gamma$ in terms of a, b, c.

To show that these are independent, show that the only way you can have $\displaystyle \alpha(x^2- 1)+ \beta(x+1)+ \gamma(x^2+ x+ 1)= 0$ is if $\displaystyle \alpha= \beta= \gamma= 0$.

You can do both of those by using the fact that if two polynomials are equal for all x, then "corresponding coefficients" must be equal.

Wicked only suggests showing that the functions are independent. That is because a basis has **three** properties- the two I mentioned above,

a) It spans the space.

b) The vectors in it are independent.

and

c) The number of vectors in it is the same as the dimension of the space.

Further, any **two** of those is sufficient to prove the third.

So **if** you know that the dimension of $\displaystyle P^2$ is 3, then showing that the vectors are independent is enough.

.