Hey I have some problems with this problem
Consider all polynomials with $\leq \deg 3$ and the following map:
$f: \mathbb{R}[t] \rightarrow \mathbb{R}[t]$
$f(t) \rightarrow f(t+1)-f(t)$
a) Determine the mapping matrix $A=M_B^B$ with the base $B=(1,t,t^2,t^3)$
b) Find a matrix $S \in Gl_n(\mathbb{R})$ with $A \cdot S=\begin{pmatrix} 1 & 0 &0 &0 \\ 0& 1 &0 &0 \\ 0& 0& 1&0 \\ 0& 0& 0& 0 \end{pmatrix}$
So for a) I found the mapping of the polynomials in the Basis $B$.
So I have $1 \rightarrow 0$
$t \rightarrow 1$
$t^2 \rightarrow 2t+1$
$t^3 \rightarrow 3t^2+3t+1$
So I have the matrix $A=\begin{pmatrix} 0 & 1 &1 &1 \\ 0& 0 &2 &3 \\ 0& 0& 0&3 \\ 0& 0& 0& 0 \end{pmatrix}$ .
So far everything right?
For b) I thought to do $S=A^{-1} \cdot \begin{pmatrix} 1 & 0 &0 &0 \\ 0& 1 &0 &0 \\ 0& 0& 1&0 \\ 0& 0& 0& 0 \end{pmatrix}$ but the problem is that $A$ is not invertible. So what I have done is solving this system
$\begin{vmatrix} 0& 1& 1&1 \\ 0& 0& 2& 3\\ 0& 0& 0& 3\\ 0& 0& 0& 0 \end{vmatrix}\left.\begin{matrix} 1& 0& 0&0 \\ 0& 1& 0& 0\\ 0& 0& 1 & 0\\ 0& 0& 0&0 \end{matrix}\right|$
$\begin{vmatrix} 0& 1& 1&0 \\ 0& 0& 2& 0\\ 0& 0& 0& 3\\ 0& 0& 0& 0 \end{vmatrix}\left.\begin{matrix} 1& 0& -1/3&0 \\ 0& 1& -1& 0\\ 0& 0& 1 & 0\\ 0& 0& 0&0 \end{matrix}\right|$
$\begin{vmatrix} 0& 1& 0&0 \\ 0& 0& 1& 0\\ 0& 0& 0& 1\\ 0& 0& 0& 0 \end{vmatrix}\left.\begin{matrix} 1& -1/2& 1/6&0 \\ 0& 1/2& -1/2& 0\\ 0& 0& 1/3 & 0\\ 0& 0& 0&0 \end{matrix}\right|$
$\begin{vmatrix} 1& 0& 0&0 \\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 0 \end{vmatrix}\left.\begin{matrix} -1/2& 1/6&0&1 \\ 1/2& -1/2& 0&0\\ 0& 1/3 & 0&0\\ 0& 0&0 &0 \end{matrix}\right|$
The problem is that when I put this matrices on matrixcalc.org I don't become the right solution. Where Is my error?
In the final step, you can only swap rows, not columns.
So you can get the second, third and fourth rows of S by bringing the row of zeroes to the top.
The first row of S makes no difference to A×S, so choose a row that makes S invertible.