Let $V = \bigl\{f:\mathbb{R} \to \mathbb{R} \mid \exists a_0,\dots,a_4\in \mathbb{R},\, f(x)=\sum_{i=0}^{4}a_ix^i \bigr\}$, we define the linear map $\phi:V \rightarrow V$ as follows: $$ \phi(f)(x)=f''(x)+xf'(x)-f(x+1). $$ Find $\phi^{-1}(\{g\})$ where $g(x)=2x^3+1$.
I'm unsure of my solution to this problem and I need someone to help me. In a previous exercise I determined the transformation matrix as follows, but the problem is that the matrix is obviously not invertible. $$ \begin{pmatrix} -1&-1&1&-1&-1 \\ 0&0&-2&3&-4 \\ 0&0&1&-3&6 \\ 0&0&0&2&-1 \\ 0&0&0&0&3 \end{pmatrix} $$ So I was wondering if the answer is simply "$\phi^{-1}(g)(x)$ doesn't exist. "
It looks like you're using the standard monomial basis for $V$: $$ \beta = \bigl\{ 1, x, x^2, x^3, x^4 \bigr\}, $$ so the columns of the matrix express the coefficients of the image of these basis vectors when expressed back in terms of the basis. Explicitly, for $0 \leq j \leq 4$, column $j$ expresses $\phi(x^j)$ as linear combination of the $x^i$: $$ \phi(x^j) = \sum_{i=0}^4 a_{ij} x^i. $$
We're interested in the preimage of $g$ under this transformation, denoted $$ \phi^{-1}(g) = \bigl\{f \in V \mid \phi(f) = g \bigr\}. $$ This is a set of polynomials, each of which maps to $g$ under the transformation $\phi$.
Let's express $g$ in coordinates with respect to the basis: $$ g = 2x^3 + 1 = \color{red}{1} + \color{orange}{0}x + \color{green}{0}x^2 + \color{blue}{2}x^3 + \color{purple}{0}x^4 \quad\implies\quad \bigl(g\bigr)_\beta = \begin{pmatrix} \color{red}{1} \\ \color{orange}{0} \\ \color{green}{0} \\ \color{blue}{2} \\ \color{purple}{0} \end{pmatrix}. $$ So we want to solve the linear system $$ \begin{pmatrix} -1&-1&1&-1&-1 \\ 0&0&-2&3&-4 \\ 0&0&1&-3&6 \\ 0&0&0&2&-4 \\ 0&0&0&0&3 \end{pmatrix} \begin{pmatrix} a_0 \\ a_1 \\ a_2 \\ a_3 \\ a_4 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 2 \\ 0 \end{pmatrix}. $$ Note that $a_{3, 4} = -4$, which differs from your calculation, although it doesn't change the conclusion. We can express this matrix-vector equation as the augmented matrix $$ \left( \begin{matrix} -1&-1&1&-1&-1 \\ 0&0&-2&3&-4 \\ 0&0&1&-3&6 \\ 0&0&0&2&-4 \\ 0&0&0&0&3 \end{matrix} \;\; \middle| \;\; \begin{matrix} 1 \\ 0 \\ 0 \\ 2 \\ 0 \end{matrix} \right). $$ Unfortunately, a few row operations show that the system is inconsistent, i.e. that there are no solutions. Hence, $$ \phi^{-1}(g) = \varnothing. $$ This preimage set always exists. It just happens to be empty.