Why $K^0 = \{0\}$? I am reading "Linear Algebra" by Takeshi SAITO.
Why $n \geq 0$ instead of $n \geq 1$?
Why $K^0 = \{0\}$?
Is $K^0 = \{0\}$ a definition or not?
He wrote as follows in his book:
Let $K$ be a field, and $n \geq 0$ be a natural number.  
$$K^n =  \left\{\begin{pmatrix}
           a_{1} \\
           a_{2} \\
           \vdots \\
           a_{n}
         \end{pmatrix} \middle| a_1, \cdots, a_n \in K \right\}$$
is a $K$ vector space with addition of vectors and scalar multiplication.
$$\begin{pmatrix}
           a_{1} \\
           a_{2} \\
           \vdots \\
           a_{n}
         \end{pmatrix} +
\begin{pmatrix}
           b_{1} \\
           b_{2} \\
           \vdots \\
           b_{n}
         \end{pmatrix} = \begin{pmatrix}
           a_{1}+b_{1} \\
           a_{2}+b_{2} \\
           \vdots \\
           a_{n}+b_{n}
         \end{pmatrix}\text{,}$$
$$c \begin{pmatrix}
           a_{1} \\
           a_{2} \\
           \vdots \\
           a_{n}
         \end{pmatrix} =
\begin{pmatrix}
           c a_{1} \\
           c a_{2} \\
           \vdots \\
           c a_{n}
         \end{pmatrix}\text{.}$$
When $n = 0$, $K^0 = 0 = \{0\}$.

It can be thought of as a "useful" definition. Any subspace of $K^n$ is isomorphic to $K^m$ for some $m\leq n$. If you don't define $K^0=\{0\},$ then this isn't true for the $0$-subspace.
Another approach is to define $K^n$ as the set of functions from a set of $n$ elements to $K$. When $n=0$, the set of functions from the empty set to any set is $1.$
It's worth noting that the three occurences of $0$ in the equality $K^0=0=\{0\}$ are representing three different things.
The first zero is the natural number $0.$
The second is a trivial space, a vector space with one element.
The third $0$ is the element of that trivial space.
You might then write it as:
$$K^0=\mathbf 0=\{\vec 0\}$$
