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?
 A: A polynomial of degree $\le 2$ is the linear combination of $x^0, x^1, x^2$.  It may be represented using $\begin{bmatrix} A \\ B \\ C\end{bmatrix}$ to represent $Ax^2 + Bx + C$.
Suppose you want to shift $P_1$ to the right $k$ units to produce $P_2$:
$$\begin{align}P_2(x) &= P_1(x - k) \\
                     &= A(x - k)^2 + B(x - k) + C \\
                     &= Ax^2 - 2Akx + Ak^2 + Bx - Bk + C \\
                     &= (A)x^2 + (B - 2Ak)x + (Ak^2 - Bk + C)
                     \end{align}$$
This leaves us to find some matrix $M$:
$$\begin{align}P_2 &= M(k)\times P_1 \\
               \begin{bmatrix} A \\ B - 2Ak \\ Ak^2 - Bk + C \end{bmatrix} &=
               \begin{bmatrix} \text{Your} \\ \text{Matrix} \\ \text{Here} \end{bmatrix}
               \begin{bmatrix} A \\ B \\ C \end{bmatrix}
\end{align}$$
So consider the top row of your matrix:
$$\begin{align}
               \begin{bmatrix} A \\ B - 2Ak \\ Ak^2 - Bk + C \end{bmatrix}  &=
               \begin{bmatrix} m_1 & m_2 & m_3 \\ &  \cdots  & \\  & \cdots &  \end{bmatrix}
               \begin{bmatrix} A \\ B \\ C \end{bmatrix}
\end{align}$$
This results in the equation:
  $$A = m_1 A + m_2 B + m_3 C$$
From this we see that $m_1 = 1, m_2 = 0$, and $m_3 = 0$.  The second row of the matrix gives us another result:
$$\begin{align}
               \begin{bmatrix} A \\ B - 2Ak \\ Ak^2 - Bk + C \end{bmatrix}  &=
               \begin{bmatrix} &  \cdots  & \\ m_4 & m_5 & m_6 \\  & \cdots &  \end{bmatrix}
               \begin{bmatrix} A \\ B \\ C \end{bmatrix}
\end{align}$$
Giving the equation $B - 2Ak = m_4 A + m_5 B + m_6 C$.  We can see from this that $m_4 = -2k, m_5 = 1$, and $m_6 = 0$.  The last row of the matrix gives 
  $$Ak^2 - Bk + C = m_7 A + m_8 B + m_9 C \tag{Last Row}$$
Find $m_7, m_8, m_9$ the same we way found the previous six matrix entries, and your final matrix becomes:
$$             \begin{bmatrix} A \\ B - 2Ak \\ Ak^2 - Bk + C \end{bmatrix} =
               \begin{bmatrix} 1 & 0 & 0 \\ -2k & 1 & 0 \\ m_7 & m_8 & m_9 \end{bmatrix}
               \begin{bmatrix} A \\ B \\ C \end{bmatrix}
$$
One way to check if your result is right, since shift a matrix to the right by $k$ twice is the same as shifting it to the right $2k$ once, you should check to see if:
$$\underbrace{M(k)^2}_{\text{shift right twice}} = \underbrace{M(2k)}_{\text{shift right once}}$$
A: You need to carefully revise what is meant by the matrix of a linear transformation.  The basic idea is that we replace the transformation $T$ by matrix multiplication:
$$T({\bf x})=A{\bf x}\ .$$
However in many cases this has to be modified: in your problem you would be looking at
$$p(t-1)=Ap(t)$$
where $p$ is a polynomial and $A$ is a $3\times3$ matrix, which does not make sense.  We need to replace the vectors $\bf x$ and $A\bf x$ by their coordinate vectors with respect to a specified basis.  The proper definition is this:

the matrix of $T:V\to W$ with respect to the basis $B$ for $V$ and the basis $C$ for $W$ is $A$

means

$[T({\bf x})]_C=A\,[{\bf x}]_B$ for all ${\bf x}$ in $V$,

where $[{\bf x}]_B$ means the coordinate vector of ${\bf v}$ with respect to $B$.
Note a consequence of this: you should never refer just to the matrix of a linear transformation, but to the matrix of a linear transformation with respect to certain bases.  As you have not specified the bases, I will assume that you want to use the standard basis $\{1,t,t^2\}$.
There are various ways to find the matrix, the following is often convenient.  Consider a basis $B=\{{\bf b}_1,\ldots,{\bf b}_n\}$, and substitute ${\bf x}={\bf b}_1$ in the definition.  We have
$$[T({\bf b}_1)]_C=A[{\bf b}_1]_B=A\pmatrix{1\cr0\cr\vdots\cr}
  =\{\hbox{$1$st column of $A$}\}\ .$$
That is, we have the following "recipe" for the first column of $A$: take the first basis vector from $B$; calculate $T$ of this vector; find the coordinates of the result with respect to $C$.  In your case you have
$${\bf b}_1=1\ ,\quad T(1)=1=1(1)+0(t)+0(t^2)\ ,\quad
  [T(1)]_C=\pmatrix{1\cr0\cr0\cr}\ .$$
So we get
$$A=\pmatrix{1&?&?\cr0&?&?\cr0&?&?\cr}\ .$$
We find the second column in the same way, starting with the second basis vector:
$${\bf b}_2=t\ ,\quad T(t)=t-1=-1(1)+1(t)+0(t^2)\ ,\quad
  [T(t)]_C=\pmatrix{-1\cr1\cr0\cr}$$
and so
$$A=\pmatrix{1&-1&?\cr0&1&?\cr0&0&?\cr}\ .$$
See if you can similarly do the working for the third column to show that the final answer is
$$A=\pmatrix{1&-1&1\cr0&1&-2\cr0&0&1\cr}\ .$$
To help understand what the matrix means, multiply $A$ by any randomly chosen vector, for example,
$$\pmatrix{1&-1&1\cr0&1&-2\cr0&0&1\cr}\pmatrix{7\cr-2\cr11\cr}
  =\pmatrix{20\cr-24\cr11\cr}\ .$$
Translating back to $T$ by using the definition, this means that
$$T(7(1)+(-2)(t)+11(t^2))=20(1)+(-24)(t)+11(t^2)\ ,$$
that is,
$$7-2(t-1)+11(t-1)^2=20-24t+11t^2\ ,$$
and you can easily confirm that this is correct.
Final note: observe that we will always have the matrix $A$ times a vector consisting of numbers.  So the line where you have a matrix times the vector $(t^2,-2t,1)$ is certainly wrong.
A: \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}
$$
\begin{bmatrix} a \\ b \\ c \end{bmatrix} \mapsto \begin{bmatrix} a \\ b-2a \\ c+a-b \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0  \\
-2 & 1 & 0 \\ 1 & -1 & 1 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \end{bmatrix}
$$
A: 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 $T(1 - 2t + t^2) = 4 -4t + t^2$.
