Jordan Normal Form of $T(X)=AX$

Question

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

Answer 1

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 $(\lambda-1)^4$

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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: $$ T\left(\begin{bmatrix} x\\y\\z\\t \end{bmatrix} \right)=\begin{bmatrix} x+z\\y+t\\z\\t \end{bmatrix} $$

Answer 2

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.

Answer 3

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.