Matrix representation and characterstic polynomial of linear transformation from the 0 vector space to a finitely-dimensional space Relevant definition:

In this definition, what happens if $V = \{0\}$, the 0 vector space? The basis of the zero vector space is the empty set, so the definition would not make much sense. Do we just say that matrix representation does not exist?
And what if $V = W = \{0\}$?
And indeed, let's say we have $T: V \rightarrow W$ with $V = W = \{0\}$.
The characteristic polynomial of a linear operator is defined as

So the characteristic polynomial of $T$ as above would depend on how we define the matrix representation of $T$, which I do not know how it is defined.
So how would one define the characteristic polynomial here?
Thanks for your time!
 A: You can extend all the definitions so that they work when $\dim V = 0$ or $\dim W = 0$ although it's not very interesting. Representing a linear map by a matrix should give you a one-to-one correspondence between linear maps and matrices. A linear map $T \colon V \rightarrow W$ where $\dim V = n$ and $\dim W = m$ is represented by an $m \times n$ matrix. But what is a "$0 \times n$" or "$m \times 0$" matrix? Well, it is just an "empty matrix". Formally, an $m \times n$ matrix $A$ is a function $A \colon \{ 1, \dots, m \} \times \{ 1, \dots, n \} \rightarrow \mathbb{F}$. Then:


*

*If $V = \{ 0_V \}$ then $\dim V = 0$ and the only basis for $V$ is the empty set which indeed has zero elements. There is only one linear map $T \colon V \rightarrow W$ which is the map $T(0_V) = 0_W$. You can say that such a map is indeed represented by a "$m \times 0$" matrix which is a function $A \colon \{1, \dots, m\} \times \emptyset \rightarrow \mathbb{F}$. Since $\{1, \dots, m\} \times \emptyset = \emptyset$, there is only one function $A \colon \emptyset \rightarrow \mathbb{F}$ which is the empty function.

*If $W = \{ 0_W \}$ then $\dim W = 0$ and the only basis for $W$ is the empty set which indeed has zero elements. There is only one linear map $T \colon V \rightarrow W$ which given by $T(v) = 0_W$ for all $v \in V$. You can say that such a map is indeed represented by a $0 \times n$ matrix which is a function $A \colon \emptyset \times \{ 1, \dots, n \} \rightarrow \mathbb{F}$. Since again $\emptyset \times \{ 1, \dots, n \} = \emptyset$, there is only one function $A \colon \emptyset \rightarrow \mathbb{F}$ which is the empty function. 


When $\dim V = \dim W = 0$, again, there is only one $0 \times 0$ matrix $A$ which corresponds to the empty function $A \colon \emptyset \times \emptyset \rightarrow \mathbb{F}$. You might denote such an "empty" matrix by $[]$. The determinant of a linear map $T \colon V \rightarrow V$ when $V = 0$ is defined (or calculated, according to your definitions) to be $1$. This is consistent with the fact that $T$ is the identity mapping of $V$ to itself. The determinant of $[] = I_0$ is also one. Finally, this implies that the characteristic polynomial of the unique operator $T \colon V \rightarrow V$ when $\dim V = 0$ is $1$. This is consistent with the fact that the characteristic polynomial of an operator on an $n$-dimensional space is a polynomial of degree $n$ whose leading coefficient is $(-1)^n$.
