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?

\begin{align}
p(t) & = at^2+bt+c \\
p(t-1) & = a(t-1)^2+b(t-1)+c \\
& = at^2 + (b-2a)t+ (c+a-b).
\end{align}
$$
\begin{bmatrix} a \\ b \\ c \end{bmatrix} \mapsto \begin{bmatrix} a \\ b-2a \\ c+a-b \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0  \\
-2 & 1 & 0 \\ 1 & -1 & 1 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \end{bmatrix}
$$
