# Thread: how to show that?

1. ## how to show that?

Let $\displaystyle q1=1+x+x^2$
$\displaystyle q2=1+x+2x^2$
$\displaystyle q3=3x^2$

the q$\displaystyle i$form a basis for $\displaystyle P2$?
$\displaystyle P2$ is polynomial

2. Originally Posted by iloveyou7
Let $\displaystyle q1=1+x+x^2$
$\displaystyle q2=1+x+2x^2$
$\displaystyle q3=3x^2$

the q$\displaystyle i$form a basis for $\displaystyle P2$?
$\displaystyle P2$ is polynomial

Assuming you're talking about real polynomials you've to show that $\displaystyle c_1q_1+c_2q_2+c_3q_3=0 \Longleftrightarrow c_1=c_2=c_3=0\,,\,\,c_i \in \mathbb{R}$.

For this it'd be convenient to remember that by definition of equality in the ring (or vector space) of real polynomials, we have that $\displaystyle a_nx^n+a_{n-1}x^{n-1}+...+a_0=0 \Longleftrightarrow a_i=0\,\,\,\forall i$

Tonio

3. Tonio is using a fundamental theorem about "bases".

A basis $\displaystyle \{v_n\}$ of vector space V has three properties:
1) They are independent.
2) They span the space.
3) The number of vectors in the basis is equal to the dimension of the space.

And if any two of those is true, the third is also true.

Tonio is assuming that you know that $\displaystyle P^2$ has dimension 3 so, since you have 3 vectors in your set (3) above is satisfied. It is only necessary to show that either (1) or (2) is satisfied and he is suggesting (1).

4. use coordinated vectors, express the first one as $\displaystyle (1,1,1)$ (note that the order $\displaystyle a+bx+cx^2$ was given in order to express those polynomials into coordinated vectors), then put each vector below of the other to get a $\displaystyle 3\times3$ matrix, compute its determinant, and it suffices to show it's not zero to have linear independence.