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Let $V_n:=\{P\in\mathbb C[X] | deg(P) \leq n\}$ with standard basis

Also $f_n: V_n\to V_n, \ f_n(P)(X):=P(X+1)-P(X)$

Ex: $P(X)=2X \Rightarrow f_n(P)(X)=2(X+1)-2X=2$

Let $n=4$. The tranf. mat. for $f_4$ is:

$A=\begin{pmatrix}0&1&1&1&1\\0&0&2&3&4\\0&0&0&3&6\\0&0&0&0&4\\0&0&0&0&0\end{pmatrix}$

I want to find a polynomial $P_4$ such that $f_4(P_4)(X)=X^3$

I thought that I could do this:

In the standard basis we can write $X^3$ as $b=(0,0,0,1,0)$

So that we get: $f_4(P_4)(X)=X^3 \Leftrightarrow AP_4=b$. Then, solving $(A|b)$ using Gauss.

But! $det(A)=0$ so it can't be solved. $A$ is not bijective but injective, so there does not exist an inverse function.

So, having such a situation where my linear map is sur/inj. how do I solve it?

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    In your question, you say "$A$ is not bijective but injective". Don't you mean that $A$ is also not injective? Note that a square matrix with determinant 0 is neither injective nor surjective. – 2017-02-03

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The fact that $\det(A) = 0$ doesn't mean that the equation $Ax = b$ can't be solved! If $\det(A) \neq 0$ then the equation $Ax = b$ has a unique solution $x \in \mathbb{F}^n$ for each $b \in \mathbb{F}^n$ (this correpsonds to (the linear map defined by) $A$ being injective and surjective). However, even if $\det(A) = 0$ the equation $Ax = b$ might have solutions. In fact that are two options:

  1. If $b$ belongs to the column space of $A$ then $Ax = b$ will have a solution. All the solutions of $Ax = b$ will be of the form $x_0 + x_1$ where $Ax_0 = b$ is some particular solution and $x_1 \in \ker(A)$ (so $Ax_1 = 0$). In particular, if $\mathbb{F} = \mathbb{R}$, there will be infinitely many solutions.
  2. If $b$ doesn't belong to the column space of $A$ then $Ax = b$ won't have a solution.

In your case, if you row-reduce $(A|b)$ you'll see that you get infinitely many solutions to the equation. They will be of the form

$$ p(X) = \frac{1}{4} X^2(X-1)^2 + c $$

where $c \in \mathbb{R}$.

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    oh jesus... How stuff makes way more sense after you selpt. :p Thanks. I'll go through several examples to make this more clear. But it does make sense. Thanks for the clarification. – 2017-02-04