Linear transformation - basis/standard basis/linear independendence

Let $V_4 = \{p(x) \in \mathbb{R}[X], \text{ such that } \deg{p(x)} \leq 4\}.$

Consider the linear map $f: V_4 \longrightarrow \mathbb{R}^2$ given by $$p(x) \longmapsto \begin{bmatrix} p(1)\\\ p(2)\end{bmatrix}.$$

1. Show that $\{(x-1)(x-2), x(x-1)(x-2), x^2(x-1)(x-2)\}$ is a basis of the nullspace $U$ of $f$.

2. Show that $B = \{1, x, (x-1)(x-2), x(x-1)(x-2), x^2(x-1)(x-2)\}$ is a basis of $V_4$

3. Compute the matrix of $f$ with respect to the basis $B$ of $V_4$ and $C =$ standard basis $\{(1,0), (0,1)\}$ of $\mathbb{R}^2$. I know the answer is: $$\begin{bmatrix} 1 & 1 & 0 & 0 & 0 \\\ 1 & 2 & 0 & 0 & 0 \end{bmatrix}$$ but I don't know how to reach this answer. It has something to do with computing $f(1), f(x)$ etc, which equal $(1,1)$ and $(1,2)$, respectively, but I don't understand how you work this out.

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I tried to fix your formatting and typed it into LaTeX. Please check that I haven't introduced any mistakes. –  t.b. Apr 16 '11 at 11:00
thanks for that! it is dy not deg thats the only thing that is wrong! thanks. any clue as to the answers? –  user9656 Apr 16 '11 at 11:14
I don't understand the notation dy. I'm pretty sure it must be $\deg$ (the degree of a polynomial). –  t.b. Apr 16 '11 at 11:15

1. The image of $f$ is two-dimensional, e.g. $f(x-2) = \begin{bmatrix} -1 \\\ 0 \end{bmatrix}$ and $f(x-1) = \begin{bmatrix} 0 \\\ 1\end{bmatrix}$, so the kernel has dimension $3$ because $\dim{\operatorname{Ker}{f}} + \dim{\operatorname{Im}{f}} = \dim V_4 = 5$. It remains to check that the given polynomials are in $U$ (obvious) and that they are linearly independent (consider their degrees, see also my comments on point 2).
2. Consider the degrees. In more detail: let me write $B = \{b_0, \ldots, b_4\}$. Since the dimension of $V_4$ is $5$ we need only show that the vectors of $B$ are linearly independent. Consider the equation $$\lambda_0 b_0 + \cdots + \lambda_{4} b_4 = 0.$$ Since the polynomial $b_4$ is the only one of degree four, we must have $\lambda_4 = 0$. Now $b_3$ is the only polynomial of degree three, and so on. Therefore $\lambda_0 = \lambda_1 = \cdots = \lambda_4 = 0$.
3. The $j$th column of the matrix contains the entries of $f(b_j)$ with respect to the basis $C$.