Finding a basis for a polynomial vector space

Question

I came across the following question:

Let $V$ be the real vector space of polynomials with degree $\le 2$. For $\alpha,\beta \in \mathbb{R}$ let $W_\alpha = \{f \in V |f(\alpha) = 0\}$ and $D_\beta = \{f \in V |f'(\beta) = 0\}$. $f'$ describes the derivative of $f$.

Find a basis for $W_\alpha$ and $D_\beta$.

Unfortunatly I am not quite getting my head wrapped around vector spaces with polynomials yet. So after failing the question I had a look at the solution and there it says:

By solving the system of linear equations:

$W_\alpha: \{X-\alpha, X^2-\alpha^2\}$

$D_\beta: \{1, X^2-2\beta X\}$

More explanation is not given.

I know that the elements in $W_\alpha$ in the form of $f(x) = c_2X^2+c_1X+c_0$ are zero at $\alpha$, so $f(\alpha) = c_2\alpha^2+c_1\alpha+c_0 = 0$.

Respectively for the elements in $D_\beta$: $f'(\beta) = 2c_2\beta+c_1 = 0$.

However, it is not clear to me how I get from that information to the basis of $W_\alpha$ and $D_\beta$ using systems of linear equations.

Answer 1

Note that for $W$ form the condition

• $f(\alpha) = c_2\alpha^2+c_1\alpha+c_0 = 0\implies c_0=-c_2\alpha^2-c_1\alpha$

we obtain

• $f(x) = c_2X^2+c_1X-c_2\alpha^2-c_1\alpha=c_2(X^2-\alpha^2)+c_1(X-\alpha)$

can you proceed for $D$?