# Thread: [SOLVED] Prove polynomials are a generating system (vector spaces)

1. ## [SOLVED] Prove polynomials are a generating system (vector spaces)

Let V be the vector space of the polynomials of degree $\displaystyle \leq$ 2 over $\displaystyle \mathbb{R}$.

Prove that the polynomials:

$\displaystyle p_1 = 2T^2 + T + 1$
$\displaystyle p_2 = 4T^2 + T$
$\displaystyle p_3 = -2T^2 + 2T + 1$

span, or are a generating system of, V.

If I solve the system of linear equations- how do I prove that they span V?
Thanks.

2. Originally Posted by bmp05
Let V be the vector space of the polynomials of degree $\displaystyle \leq$ 2 over $\displaystyle \mathbb{R}$.

Prove that the polynomials:

$\displaystyle p_1 = 2T^2 + T + 1$
$\displaystyle p_2 = 4T^2 + T$
$\displaystyle p_3 = -2T^2 + 2T + 1$
.
You just need to show these polynomial are linearly independent.
Say that $\displaystyle a_1p_1 + a_2p_2+a_3p_3 = \bold{0}$ where $\displaystyle \bold{0}$ is the zero-polynomial.
This means that,
$\displaystyle 2a_1+4a_2-2a_3=0$
$\displaystyle a_1 + \ \ \ a_2 + 2a_3 = 0$
$\displaystyle a_1 + \ \ \ \ \ \ \ \ \ + a_3 = 0$

Argue that $\displaystyle a_1=a_2=a_3=0$ is the only solution.

3. Write the polynomials as vectors in $\displaystyle \mathbb{R}^{3}$ .

$\displaystyle A= \left[\begin{matrix} 1 & 0 & 1\\ 1 & 1 & 2 \\ 2 & 4 & -2 \end{matrix}\right]$

Row operations will show that this matrix A has 3 pivot positions. So it therefore spans $\displaystyle \mathbb{R}^{3}$ and by the isomorphism between $\displaystyle \mathbb{R}^{3} \mbox{ and } \mathbb{P}_{3}$ also $\displaystyle \mathbb{P}_{3}$ .