The categorical dimension of a vector space is the scalar image of its linear dimension

Let $V$ have basis $(v_1,\dots ,v_n)$. With the usual duality from [L3], the canonical pivotal structure has component $J_V:V\to V^{\ast \ast }$ given by $J_V(v)(f)=f(v)$, and ${\dim }_J(V)$ is by definition the categorical trace of $J_V$ The dimension of an object relative to a pivotal structure.

Using the same basis-and-dual-basis computation as for categorical trace, ${\dim }_J(V)={\sum }_{i=1}^n{J}_V(v_i)(v_i^{\ast })={\sum }_{i=1}^n{v}_i^{\ast }(v_i)=n\cdot 1_k\in k\mathrm{.}$

The natural number $n$ is the ordinary dimension of $V$ by Finite-dimensional vector space, and its dimension ${\dim }_{F}V$; infinite-dimensional means having no finite basis. Hence the categorical dimension is its scalar image $n\cdot 1_k$ in $\mathrm{End}(k)\cong k$. In positive characteristic this scalar can differ from $n$ as a natural number; for example, a $p$-dimensional space has categorical dimension $0\in k$ when $\mathrm{char}k=p$.