Prove the identity matrix is unique

[WWGD]

I think you mean $A(I_1 - I_2) = 0$.

Ok, but just how are the two different? I'm not aware of any particular meaning of$[I_1- I_2].$.

 

[Mark44]

Ok, but just how are the two different?

I'm not aware of any particular meaning of $[I1−I2]$..

You wrote $A(I_1 - A_2)$ but I thought you meant $A(I_1 - I_2)$.
If $A(I_1 - I_2) = 0$, then using determinants you can deduce that $I_1 = I_2$, which was the whole point in being able to say that the identity matrix must be unique.

 

[WWGD]

You wrote $A(I_1 - A_2)$ but I thought you meant $A(I_1 - I_2)$.
If $A(I_1 - I_2) = 0$, then using determinants you can deduce that $I_1 = I_2$, which was the whole point in being able to say that the identity matrix must be unique.

True, if we assume $A$ is invertible. Otherwise, its right kernel isn't trivial, i.e. , it's not just $\{0\}$, so in that case of $A$ being singular, it doesn't follow that $I_1=I_2$. But if $A$ is nonsingular, then you're right.

 

[FactChecker]

True, if we assume $A$ is invertible. Otherwise, its right kernel isn't trivial, i.e. , it's not just $\{0\}$, so in that case of $A$ being singular, it doesn't follow that $I_1=I_2$. But if $A$ is nonsingular, then you're right.

From the beginning of this thread, it must be stated that the identities, $I_1$ and $I_2$, work as identities for every possible $A$. Otherwise, the conclusion that $I_1=I_2$ may be false.

 

[WWGD]

From the beginning of this thread, it must be stated that the identities, $I_1$ and $I_2$, work as identities for every possible $A$. Otherwise, the conclusion that $I_1=I_2$ may be false.

Ok, I guess I lost track of the " Initial Conditions". Using the Determinant alone ( assuming $A$ is square:

$Det(A(I_1-I_2))= DetADet( I_1-I_2))=0$ implies either of the determinants is $0$.

Though $Det( I_1-I_2)=0$ doesnt imply $I_1=I_2$.

But I admit I may have somewhat lost track of were the discussion veered.

 

[PeterDonis]

I completely missed that we were talking about rectangular matrices!

For the OP question it should not matter; uniqueness of the identity should be provable for any group (or at least any group that doesn't have some other weirdness like zero divisors). @FactChecker seems to me to have come up with the simplest derivation.

 

[PeroK]

For the OP question it should not matter; uniqueness of the identity should be provable for any group (or at least any group that doesn't have some other weirdness like zero divisors). @FactChecker seems to me to have come up with the simplest derivation.

The set of $m \times n$ matrices (where $m \ne n$) do not form a multiplicative group. The usual matrix multiplication is not even well defined.

 

[WWGD]

The set of $m \times n$ matrices (where $m \ne n$) do not form a multiplicative group. The usual matrix multiplication is not even well defined.

This refers to multiplicative groups.

 

[PeterDonis]

The set of $m \times n$ matrices (where $m \ne n$) do not form a multiplicative group. The usual matrix multiplication is not even well defined.

And in such cases there is no "identity" at all. So the OP question isn't even applicable.

 

[PeroK]

And in such cases there is no "identity" at all. So the OP question isn't even applicable.

An $n \times n$ matrix is a linear transformation on the set of $n \times m$ matrices (under left multiplication). The question related to the identity transformation in this context.