The Unapologetic Mathematician

The Category of Matrices II

As we consider the category $\mathbf{Mat}(\mathbb{F})$ of matrices over the field $\mathbb{F}$, we find a monoidal structure.

We define the monoidal product $\boxtimes$ on objects by multiplication — $m\boxtimes n=mn$ — and on morphisms by using the Kronecker product. That is, if we have an $m_1\times n_1$ matrix $\left(s_{i_1}^{j_1}\right)\in\hom(n_1,m_1)$ and an $m_2\times n_2$ matrix $\left(t_{i_2}^{j_2}\right)\in\hom(n_2,m_2)$, then we get the Kronecker product

$\left(s_{i_1}^{j_1}\right)\boxtimes\left(t_{i_2}^{j_2}\right)=\left(s_{i_1}^{j_1}t_{i_2}^{j_2}\right)$

Here we have to be careful about what we’re saying. In accordance with our convention, the pair of indices $(i_1,i_2)$ (with $1\leq i_1\leq m_1$ and $1\leq i_2\leq m_2$) should be considered as the single index $(i_1-1)m_2+i_2$. It’s clear that this quantity then runs between ${1}$ and $m_1m_2$. A similar interpretation goes for the index pairs $(j_1,j_2)$.

Of course, we need some relations for this to be a monoidal structure. Strict associativity is straightforward:

$\left(\left(r_{i_1}^{j_1}\right)\boxtimes\left(s_{i_2}^{j_2}\right)\right)\boxtimes\left(t_{i_3}^{j_3}\right)=\left((r_{i_1}^{j_1}s_{i_2}^{j_2})t_{i_3}^{j_3}\right)=\left(r_{i_1}^{j_1}(s_{i_2}^{j_2}t_{i_3}^{j_3})\right)=\left(r_{i_1}^{j_1}\right)\boxtimes\left(\left(s_{i_2}^{j_2}\right)\boxtimes\left(t_{i_3}^{j_3}\right)\right)$

For our identity object, we naturally use ${1}$, with its identity morphism $\left(1\right)$. Note that the first of these is the object the natural number ${1}$, while the second is the $1\times1$ matrix whose single entry is the field element ${1}$. Then we can calculate the Kronecker product to find

$\left(t_i^j\right)\boxtimes\left(1\right)=\left(t_i^j\right)=\left(1\right)\boxtimes\left(t_i^j\right)$

and so strict associativity holds as well.

The category of matrices also has duals. In fact, each object is self-dual! That is, we set $n^*=n$. We then need our arrows $\eta_n:1\rightarrow n\boxtimes n$ and $\epsilon_n:n\boxtimes n\rightarrow1$.

The morphism $\eta_n$ will be a $1\times n^2$ matrix. Specifically, we’ll use $\eta_n=\left(\delta^{i,j}\right)$, with $i$ and $j$ both running between ${1}$ and $n$. Again, we interpret an index pair as described above. The symbol $\delta^{i,j}$ is another form of the Kronecker delta, which takes the value ${1}$ when its indices agree and ${0}$ when they don’t.

Similarly, $\epsilon_n$ will be an $n^2\times1$ matrix: $\epsilon_n=\left(\delta_{i,j}\right)$, using yet another form of the Kronecker delta.

Now we have compatibility relations. Since the monoidal structure is strict, these are simpler than usual:

$(\epsilon_n\otimes1_n)\circ(1_n\otimes\eta_n)=1_n$
$(1_{n^*}\otimes\epsilon_n)\circ(\eta_n\otimes1_{n^*})=1_{n^*}$

But now all the basic matrices in sight are various Kronecker deltas! The first equation reads

$\left(\delta_a^b\delta^{c,d}\right)\left(\delta_{b,c}\delta_d^e\right)=\delta_a^e$

which is true. You should be able to verify the second one similarly.

The upshot is that we’ve got the structure of a monoidal category with duals on $\mathbf{Mat}(\mathbb{F})$.