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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?

Aditya Prakash
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  • 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).

     – TZakrevskiy Jul 16 '19 at 09:01
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    If $A$ is non square, if $AI$ is defined, the product $IA$ is not, that's all. – Bernard Jul 16 '19 at 09:06
  • @Bernard , yes but that itself is my question I'm asking why is the rule defined like it has been ? – Aditya Prakash Jul 16 '19 at 09:11
  • Do you mean the general rule for matrix product? – Bernard Jul 16 '19 at 09:17
  • The $2 \times 3$ matrices don't form a ring, so what is the meaning of your question ? – reuns Jul 16 '19 at 09:41
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    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. – Lutz Lehmann Jul 16 '19 at 09:56
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    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. –  Jul 16 '19 at 09:59

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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 $\,n=m.\,$

Somos
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  • I honestly don't see how the use of category theory makes this any "clearer" than the bare linear algebra facts would... –  Jul 16 '19 at 18:28