I am having problems with this exercise. Can someone help me?
a) Determine the matrix representation for $φ: \mathbb{R}[t]degree≤3 → \mathbb{R}^2 f(t) → \begin{pmatrix} f(1) − f(0)\\ f(2) − f(1) \end{pmatrix})$ with respect to the basis $(1, t, t^2, t^3)$ and the standard basis for $\mathbb{R}^2$.
b) Determine all polynomials such that $f(2) = f(1) + 1 = f(0) + 2$.
So for a) I found the mapping of the polynomials
So I have $1 \rightarrow \begin{pmatrix}0\\ 0\end{pmatrix}$
$t \rightarrow \begin{pmatrix}1\\ 1\end{pmatrix}$
$t^2 \rightarrow \begin{pmatrix}1\\ 3\end{pmatrix}$
$t^3 \rightarrow \begin{pmatrix}1\\ 7\end{pmatrix}$
So I have the matrix $A=\begin{pmatrix} 0& 1 & 1&1 \\ 0& 1& 3 & 7 \end{pmatrix}$ .
So far everything right?
For the second I am having problems.
What I wanted to do is to try solving this equations system
$\begin{pmatrix} 0& 1 & 1&1 \\ 0& 1& 3 & 7 \end{pmatrix} \begin{pmatrix} f(1)-f(0)\\ f(2)-f(1) \end{pmatrix}=\begin{pmatrix} 1 \\1 \end{pmatrix}$
But this leads me to the solution $X=\left(\begin{matrix} x_1 \\ 1+2*x_4 \\ -3*x_4 \\ x_4 \end{matrix}\right)$
For which if we think $x_1=1 x_2=t...$ we obtain $f(t)=t+(1+2)t-3t^2+t^3$
But if I do some tests with this polynomial, they show to me that the answer is wrong. Where is my mistake?
Your solution for part a is correct.
For part b, we are given the equations $$ f(2) = f(1) + 1, \quad f(1) + 1 = f(0) + 2. $$ From there, note that $$ f(1) + 1 = f(0) + 2 \iff f(1) - f(0) = 1\\ \phantom{1 + ,.}f(2) = f(1) + 1 \iff f(2) - f(1) = 1 $$ Thus, it is equivalent to find the polynomials $f$ satisfying $\pmatrix{f(1) - f(0)\\ f(2) - f(1)} = \pmatrix{1\\1}$ . That is, we wish to solve $$ \varphi(f) = \pmatrix{1\\1}. $$ Perhaps now you can see how your matrix from part a can be used.
Taking a closer look at your solution, it seems that you have correctly solved the equation $$ \pmatrix{0&1&1&1\\0&1&3&7}\mathbf x = \pmatrix{1\\1}, $$ which is indeed what you are meant to do. Indeed, the solution is $$ \mathbf x = \pmatrix{x_1\\x_2\\x_3\\x_4} = \pmatrix{x_1\\1 + 2x_4\\-3x_4\\x_4}, \quad x_1,x_4 \in \Bbb R. $$ From there, you interpreted this solution incorrectly. Letting $\mathcal B$ denote your polynomial basis and $\mathcal S_2$ the standard basis of $\Bbb R^2$, the above matrix equation can be written as $$ [\varphi]_{\mathcal S_2 \leftarrow \mathcal B}[f]_{\mathcal B} = \pmatrix{1\\1}, $$ where we note that $[\varphi]_{\mathcal S_2 \leftarrow \mathcal B}[f]_{\mathcal B} = [T(f)]_{\mathcal S_2}$. Thus, the solutions $\mathbf x$ to the matrix equation can be interpreted as coordinate-vectors of the polynomials $f$ satisfying the required condition. So, a vector $f$ will satisfy the desired condition if and only if $$ [f]_{\mathcal B} = \pmatrix{x_1\\1 + 2x_4\\-3x_4\\x_4}, \quad x_1,x_4 \in \Bbb R \iff\\ f(t) = (x_1)t + (1 + 2x_4)t + (-3x_4)t^2 + (x_4)t^3, \quad x_1,x_4 \in \Bbb R. $$