<s>[INST] Matrix that moves the graph of a polynomial. Consider the vector space of all polynomials of degree $\leq2$. Find the matrix which describes: $$p(t)\mapsto p(t-1)$$
So $t^2-2t+1$ for example would change to $(t-1)^2-2(t-1)+1$.
(The graph moves one unit to the right.)
First I set up this equation for the example above:
$$
\begin{pmatrix}
\cdot  & \cdot & \cdot\\
\cdot  & \cdot & \cdot\\
\cdot  & \cdot & \cdot
\end{pmatrix}
\begin{pmatrix}
t^2\\
-2t\\
1\\
\end{pmatrix}
=\begin{pmatrix}
(t-1)^2\\
-2(t-1)\\
1\\
\end{pmatrix}=
\begin{pmatrix}
t^2-2x+1\\
-2t+2\\
1\\
\end{pmatrix}
$$
I figured out that the matrix has to be
\begin{pmatrix}
1  & 1 & 1\\
0 & 1 & 2\\
0  & 0 & 1\\
\end{pmatrix}
Which looks very nice, so I thought I was done. Just to be sure, I took another example:
$$t^2+t+1 ~~~ \longmapsto ~~~(t-1)^2+(t-1)+1$$
But in this case, the matrix seems to be
\begin{pmatrix}
1  & -2 & 1\\
0 & 1 & -1\\
0  & 0 & 1\\
\end{pmatrix}
Finally, I multiplied out the first example to $t^2-4t+4$, took the representing vector and the matrix for that transformation had only entries on the diagonal (obviously, but still) - so which one is it/what am I doing wrong?
 [/INST] What you seem to be describing is similar to the problem of finding the matrix of a linear transformation. You can find a matrix, $A_T$ which applies a linear map $T$, from a coordinate vector of one basis to a coordinate vector of another basis. 
In your example, if you just use the standard basis, it means instead of multiplying $A_T$ by $(a, bt, ct^2)^T$, you multiply by $(a, b, c)^T$. Then the answer you get is a coordinate vector, $(d, e, f)^T$, which represents the polynomial $d +  et + ft^2$.
Computing $A_T$ is not too difficult. We first find what the transformation $T$ does to the standard basis vectors, $\{1, t, t^2\}$:
\begin{align*}
T(1) &= 1\\
T(t) &= t - 1\\
T(t^2) &= (t-1)^2 = t^2 - 2t + 1\\
\end{align*}
Now, we write these as a linear combination of standard basis vectors:
\begin{align*}
1 &= 1 + 0t + 0t^2\\
t - 1 &= -1 + 1t + 0t^2\\
t^2 - 2t + 1 &= 1 -2t + 1t^2\\
\end{align*}
Then the matrix $A_T$ which applies the map $T$ from $\{1, t, t^2\}$ to $\{1, t, t^2\}$ has the above coordinate vectors as columns:
\begin{align*}
A_T =  
\begin{bmatrix}
1 & -1 & 1\\
0 & 1 & -2\\
0 & 0 & 1\\
\end{bmatrix}
\end{align*}
So, if you want to find $T(p)$, first find the coordinate vector of $p$ with respect to the standard basis, which is easy. For $p = 1 - 2t + t^2$, we have:
\begin{align*}
\begin{pmatrix}
1 \\
-2\\
1\\
\end{pmatrix}
\end{align*}
Then we multiply this by $A_T$ to get the coordinate vector with respect to the standard basis:
\begin{align*} 
\begin{bmatrix}
1 & -1 & 1\\
0 & 1 & -2\\
0 & 0 & 1\\
\end{bmatrix}
\begin{pmatrix}
1 \\
-2\\
1\\
\end{pmatrix}
=
\begin{pmatrix}
4 \\
-4\\
1\\
\end{pmatrix}
\end{align*}
So $T(1 - 2t + t^2) = 4 -4t + t^2$.
</s>