# Why does a non square matrix lack a multiplicative identity

Precisely, why is multiplicative identity defined to be $$IA=AI=A$$ why both sides should work why not something like $$AI=A$$ ? Is there an underlying advantage?

• If matrices are over a field, then indeed the condition $$\forall A \quad AI = A$$ gives a unique solution $I=\sum_j e_j\otimes e_j$. However, if the underlying structure is not a field, I would suppose that only one equality is not sufficient (I'll try to sketch a counterexample). Jul 16, 2019 at 9:01
• If $A$ is non square, if $AI$ is defined, the product $IA$ is not, that's all. Jul 16, 2019 at 9:06
• @Bernard , yes but that itself is my question I'm asking why is the rule defined like it has been ? Jul 16, 2019 at 9:11
• Note that $IA=A=AI$ should be written $I_mA=A=AI_n$ when $A$ is an $m\times n$ matrix. The identity matrices on the left and the right are not the same. Jul 16, 2019 at 9:56
• For $m\times n$ matrices one can define a "left identity" via $IA=A$ and a "right identity" via $AI=A$. But these are different, and neither of them is itself an $m\times n$ matrix.
– user856
Jul 16, 2019 at 9:59

The use of category theory may make this clearer. Suppose we have a category of objects $$\, V_1, V_2, V_3, \dots\,$$ An $$\,n \times m\,$$ matrix $$\,A\,$$ is an arrow from $$\,V_m\,$$ to $$\, V_n.\,$$ Matrix multiplication is only defined between compatible matrices. That is, If $$\,B\,$$ is an arrow from $$\,V_n\,$$ to $$\,V_k\,$$ then the matrix product $$\,B A\,$$ is an arrow from $$\,V_m\,$$ to $$\,V_k.\,$$ Each object $$\,V_n$$ has a unique identity arrow denoted by $$\,I_n\,$$ which is the $$\,n \times n\,$$ identity matrix. This gives us the identity $$\, A = AI_m = I_nA.\,$$ The two identity arrows are not the same unless $$\,n=m.\,$$