Jordan Normal Form of $T(X)=AX$ 
Let $T:M^{F}_{n \times n} \to M^{F}_{n \times n} $ be a linear transformation defined by $T(X)=AX$, where $F$ is a field.
The matrix $A$ and the transformation $T$ have the same minimal polynomial.
Find the Jordan Normal Form of $T$ when $A=\left( \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} \right)$ and $F=R$.

Here is my question:
The minimal polynomial of A is $(t-1)^2$. So is $T$'s.
According to the answer I have, the characteristic polynomial of $T$ is $(t-1)^4$ - Why?
A is a $2 \times 2$ matrix. How can it create a $4x4$ Jordan Normal Form (since the characteristic polynomial has a degree of $4$)?
My question is why the characteristic polynomial of T is $(t-1)^4$?
Thanks,
Alan
 A: Hint:
$$
T(X)=
\begin{bmatrix}
1&1\\
0&1
\end{bmatrix}
\begin{bmatrix}
x&y\\
z&t
\end{bmatrix}=
\begin{bmatrix}
x+z&y+t\\
z&t
\end{bmatrix}=
(x+z)\begin{bmatrix}
1&0\\
0&0
\end{bmatrix}+(y+t)\begin{bmatrix}
0&1\\
0&0
\end{bmatrix}
+z\begin{bmatrix}
0&0\\
1&0
\end{bmatrix}
+t\begin{bmatrix}
0&0\\
0&1
\end{bmatrix}
$$
So, the transformation $T$ is represented, in the canonical basis of $M_2(\mathbb{R})$, by the matrix:
$$
T=
\begin{bmatrix}
1&0&1&0\\
0&1&0&1\\
0&0&1&0\\
0&0&0&1
\end{bmatrix}
$$
tha has characteristic polynomial $(\lambda-1)^4$

In the canonical basis we have:
$$
X=x\begin{bmatrix}
1&0\\
0&0
\end{bmatrix}
+y\begin{bmatrix}
0&1\\
0&0
\end{bmatrix}
+z\begin{bmatrix}
0&0\\
1&0
\end{bmatrix}
+t\begin{bmatrix}
0&0\\
0&1
\end{bmatrix}=
\begin{bmatrix}
x\\y\\z\\t
\end{bmatrix}
$$
$$
T(X)=(x+z)\begin{bmatrix}
1&0\\
0&0
\end{bmatrix}+(y+t)\begin{bmatrix}
0&1\\
0&0
\end{bmatrix}
+z\begin{bmatrix}
0&0\\
1&0
\end{bmatrix}
+t\begin{bmatrix}
0&0\\
0&1
\end{bmatrix}=
\begin{bmatrix}
x+z\\y+t\\z\\t
\end{bmatrix}
$$
so the transformation acts as:
$$
T\left(\begin{bmatrix}
x\\y\\z\\t
\end{bmatrix} \right)=\begin{bmatrix}
x+z\\y+t\\z\\t
\end{bmatrix}
$$
A: You know it has to have the same roots and degree 4 so all you can do is modify the exponents. Edit: assumed algebraically closed.
A: If you take the standard basis for $M_n(\mathbb{F})$ consistings of matrices $e_{ij}$ with $1$ in the $i$-th row and $j$-th column and zero elsewhere in a specific order and represent $T$ by a matrix with respect to this basis, you will find that $T$ is represented by a $n^2 \times n^2$ block diagonal matrix consisting of $n$ copies of $A$. Thus, the characteristic polynomial of $T$ will be the characteristic polynomial of $A$ to the power of $n$. This also implies that the Jordan canonical form of $T$ over an algebraically closed field will consist of $n$ identical blocks with each block being the Jordan canonical form of $A$.
I leave it to you to find the correct ordering of $e_{ij}$ for this to work.
