Characteristic polynomial of an endomorphism over 2x2 matrices I am given the endomorphism over the vector space of 2x2 matrices Mat$_2(\mathbb{R})$ defined by
$$ f(X) = \begin{pmatrix}2&2\\0&2\end{pmatrix}X.$$
I am intending to find the characteristic polynomial and the Jordan normal form of this endomorphism. But looking at the definition of the characteristic polynomial it doesn't seem like the vectorspace is important for the calculation?
Would this just be like calculating the characteristic polynomial of a 2x2 matrix or am I missing something?
Our definition of the characteristic polynomial for $A \in$ Mat$_n(\mathbb{K})$ is
$$ \chi_A(x) := \det(x \cdot I_n-A),$$
where $x \cdot I_n -A \in$ Mat$_n(\mathbb{K}[x])$. The characteristic polynomial of an endomorphism $f \in \text{End}_\mathbb{K}(V)$, where $V$ is a finite dimensional vector space, is defined as
$$ \chi_f(x) = \chi_{M^B_f}(x).$$
 A: Our basic situation is that $f$ is a function with domain (and co-domain) Mat$_2(\mathbb{K})$, where $\mathbb{K}$ is the field of real numbers. In symbols:
$$f:\text{Mat}_2(\mathbb{R}) \to \text{Mat}_2(\mathbb{R}),$$
where $f$ is left-multiplication by your fixed matrix. This map is an endomorphism of its domain, since matrix multiplication is linear.
There is some higher-level theory involved in really understanding, for example, what is the thing Mat$_n(\mathbb{K}[x])$. When $\mathbb{K}$ is any field, then $\mathbb{K}[x]$ is a polynomial ring, and so on. For finding the characteristic polynomial of your $f$, maybe you don't need to master all that stuff now.
You do need to understand $M^B_f$ though, which appears at the end of your post. This is the matrix of your linear transformation in a particular basis $B$. The comment of Evariste is demonstrated by the relation
$$ \chi_f(x) = \chi_{M^B_f}(x).$$
For we see $B$ on the right but not on the left ($\chi_f$ "is independent of the choice of the basis").
For your immediate problem, you were indeed missing something big, as follows. The $V$ you need is a four-dimensional vector space, because a 2 by 2 matrix has four entries. So I'd say an outline of your to-do list is

*

*Write down the standard basis for your real vector space $V\,=\,\text{Mat}_2(\mathbb{R})$.


*Recall that a linear transformation (an endomorphism) of a finite-dimensional vector space can be represented by a matrix.


*With $B$ the standard basis, figure out $M^B_f$, the matrix of your endomorphism $f$. (Will we have $M^B_f\,\in\,\text{Mat}_2(\mathbb{R})\,$?)
Note we can use the word 'vector' in two ways here. It can mean an ordinary $n$ by $1$ column of numbers, or more abstractly, it can mean any element of a given vector space. So a 2 by 2 matrix is legitimately called a vector.
