Prove the identity matrix is unique

[PeroK]

Please everyone accept my apologies for my posts on this thread. I completely missed that we were talking about rectangular matrices!

 

[FactChecker]

Please everyone accept my apologies for my posts on this thread. I completely missed that we were talking about rectangular matrices!

Your point about the non-square matrices makes me wonder about my simple "proof" (post #50) which seems to say that two matrices with different dimensions are equal. I will have to think about that.

 

[PhDeezNutz]

Maybe I'm simplifying this too much but can't this done by a "simple proof by contradiction" (and with index notation)?

Let $A$ be a $\left(n \times n \right)$ (a square) matrix
Let $B$ be a $\left(n \times p \right)$ matrix
Let $AB = B$ which is a $\left(n \times p \right)$

Therefore $B_{ij} = \left(AB\right)_{ij} = A_{ik} B_{kj}$.

Now suppose $A_{ik} \neq \delta_{ik}$ then what would that mean about $B_{ij} \neq B_{ij}$. Therefore $A_{ik}$ must equal $\delta_{ik}$.

Did I dun goof?

 

[Demystifier]

An identity matrix $I$ is a matrix having the property $IA=A$ for every $A$. (Here I assume that matrices are squared and have a fixed dimension, so "every" means every squared matrix with that fixed dimension.)To prove that $I$ is unique, the idea is to assume the opposite and derive a contradiction. So let as assume that it's not unique, in which case we have two different matrices $I_1$ and $I_2$ having the properties $I_1A=A$ and $I_2A=A$ for every $A$, which implies
$$(I_1-I_2)A=A-A=0$$
for every $A$. Since it must be true for every $A$, it follows in particular that it must be true for every invertible $A$. But for invertible $A$ we can multiply this from the right with $A^{-1}$, which gives
$$I_1-I_2=0$$
i.e. $I_1=I_2$, which contradicts the initial assumption that $I_1$ and $I_2$ were different. Hence the initial assumption was wrong, which proves that $I$ is unique. Q.E.D.

 

[martinbn]

An identity matrix $I$ is a matrix having the property $IA=A$ for every $A$. (Here I assume that matrices are squared and have a fixed dimension, so "every" means every squared matrix with that fixed dimension.)To prove that $I$ is unique, the idea is to assume the opposite and derive a contradiction. So let as assume that it's not unique, in which case we have two different matrices $I_1$ and $I_2$ having the properties $I_1A=A$ and $I_2A=A$ for every $A$, which implies
$$(I_1-I_2)A=A-A=0$$
for every $A$. Since it must be true for every $A$, it follows in particular that it must be true for every invertible $A$. But for invertible $A$ we can multiply this from the right with $A^{-1}$, which gives
$$I_1-I_2=0$$
i.e. $I_1=I_2$, which contradicts the initial assumption that $I_1$ and $I_2$ were different. Hence the initial assumption was wrong, which proves that $I$ is unique. Q.E.D.

How is $A^{-1}$ defined? The definition is: a matrix, if it exists, such that $AA^{-1}=A^{-1}A=I$. But which identity if we don't know there is only one!

 

[Demystifier]

How is $A^{-1}$ defined? The definition is: a matrix, if it exists, such that $AA^{-1}=A^{-1}A=I$. But which identity if we don't know there is only one!

Good point, here is a correct proof. The identity matrix is actually defined by two properties $IA=A$ and $AI=A$, for every $A$. Now suppose there are two such $I$'s, namely
$$I_1A=AI_1=A$$
and
$$I_2A=AI_2=A$$
for every $A$. Considering the cases $A=I_2$ in the first line and $A=I_1$ in the second, we get the equalities
$$I_1I_2=I_2I_1=I_2$$
and
$$I_2I_1=I_1I_2=I_1$$
which implies $I_1=I_2$, Q.E.D.

 

[martinbn]

Good point, here is a correct proof. The identity matrix is actually defined by two properties $IA=A$ and $AI=A$, for every $A$. Now suppose there are two such $I$'s, namely
$$I_1A=AI_1=A$$
and
$$I_2A=AI_2=A$$
for every $A$. Considering the cases $A=I_2$ in the first line and $A=I_1$ in the second, we get the equalities
$$I_1I_2=I_2I_1=I_2$$
and
$$I_2I_1=I_1I_2=I_1$$
which implies $I_1=I_2$, Q.E.D.

Yes, that is in @FactChecker post #44.

 

[nuuskur]

No additional structure is required. Given a set $A$ and a multiplication $\cdot : A\times A\to A$. If there exists $e\in A$ such that $ea=a=ae$ for every $a\in A$, then for any $e'\in A$ with this property, one has $e=ee'=e'$. In short, the identity is unique.

 

[WWGD]

You may also argue, if $AI_1=AI_2$, then $A[I_1-A_2]=0$. In the most general sense, $I_1-I_2$ is in the right kernel of $A$. If $A$ is an invertible matrix, this right kernel must be trivial. As Fresh_42 points, out , beyond that setting of invertible matrices, it depends on what type of object $A$ is.

 

[Mark44]

You may also argue, if $AI_1=AI_2$, then $A[I_1-A_2]=0$.

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

 

[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.