Prove the identity matrix is unique

[FactChecker]

Possibly that the matrices commute, I thought about that but I'm not sure how it helps.

Which matrix/matrices does $I_1I_2$ equal?

 

[Steve4Physics]

Hi @askmathquestions. It may be worth noting the following (if not already clear).

Suppose $A$ is any $m \times n$ matrix where $m \ne n$.

There are two (each unique) different identity matrices, $I_{m \times m}$ and $I_{n \times n}$ such that
$I_{m \times m} A = A$ and
$A I_{n \times n} = A$

Of course if $m=n$, then only a single identity matrix exists.

Proof of the uniqueness should not use inverses for various reasons. Not all matrices are invertible but all matrices can be (left or right) multiplied by the approriate identity matrix.. But more fundamentally, inverses are defined in terms of a unique identity matrix; so a proof of identity uniqueness - using inverses - is a circular argument.

 

[askmathquestions]

Hi @askmathquestions. It may be worth noting the following (if not already clear).

Suppose $A$ is any $m \times n$ matrix where $m \ne n$.

There are two (each unique) different identity matrices, $I_{m \times m}$ and $I_{n \times n}$ such that
$I_{m \times m} A = A$ and
$A I_{n \times n} = A$

Of course if $m=n$, then only a single identity matrix exists.

Proof of the uniqueness should not use inverses for various reasons. Not all matrices are invertible but all matrices can be (left or right) multiplied by the approriate identity matrix.. But more fundamentally, inverses are defined in terms of a unique identity matrix; so a proof of identity uniqueness - using inverses - is a circular argument.

Thanks for pointing that out, so there are two different, each unique identities, one for left multiplication and one for right multiplication.

I guess I'm interested in left multiplication, that, for a given vector (or square matrix) $x$, we have $Ax$ for some appropriately sized square matrix $A$.
I'm not sure though why you required $m \neq n$, isn't the identity unique when acting on square matrices too?

 

[fresh_42]

Thanks for pointing that out, so there are two different, each unique identities, one for left multiplication and one for right multiplication.

May happen, usually doesn't. I guess if you ask for an example in a thread, it will take a while before someone has an answer, if at all, except for the case of unequal dimensions.

 

[askmathquestions]

You need to solve $X\cdot A=0.$ These are $n^2$ variables and $n\cdot p$ linear equations. It is immediately clear that

1. there is possibly more than one solution, i.e. $X$ isn't unique if $n<p.$
2. there is possibly no solution, i.e. $I_1$ and $I_2$ do not exist if $n>p.$
3. there is only a unique solution guaranteed if $n=p.$

The $n\cdot p$ many linear equations have variables $x_{ij}$ and parameters $a_{ij}.$
I still do not know where $A$ is allowed to be from, but identity means, that it has to hold for all entries from whatever this be-from is. I assume we can plugin any real number. In that case, plugin $a_{11}=1,$ and $a_{ij}=0$ elsewhere, then proceed with $a_{12}=1$ and $a_{ij}=0$ elsewhere etc. This gives you tons of equations that all have to be true.

Well, I'm interested in an argument that works for square matrices acting on vectors, as well as square matrices acting on other square matrices, I'm interested in both because both vectors and square matrices have many conventional properties and applications, which I generalized by saying $A$ is an $n$ x $p$ matrix in the original problem. So if you require $n = p$, does that mean your domain of identities can no longer include vectors? i.e. there does not exist a unique identity for vectors?

 

[askmathquestions]

No additional assumptions are required. The identity element must be unique.

Considering the case where we have only the zero matrix and hence no identity is unnecessarily muddying the waters.

If vectors were invertible matrices, then taking the inverse would be sufficient for most purposes and I would be more satisfied with that, I realized that specifically because it wouldn't matter "which" identity you ended up with after obtaining $AA^{-1}$, it would simplify to $I_1 = I_2$ in either of any case.

Still no one has addressed the $x$ $y$ example I brought up, which seems to suggest vectors are invertible matrices.

Though in practice, it is a better proof if the proof can be extended to non-vectors and non-square matrices too.

There also might be some confusion in all the back and forth between what is $A$ $I$ and $x$.

In my original problem, I took $A$ to be any $n$ x $p$ matrix that is not identically the zero matrix, so that this would cover both vectors and square matrices. I'm mainly interested in left-multiplication by an identity $IA = A.$

 

[fresh_42]

If vectors were invertible matrices, ...

They are not, simply because vector times vector isn't a vector anymore.

Though in practice, it is a better proof if the proof can be extended to non-vectors and non-square matrices too.

That's why I asked you about the domains right from the beginning. This seems to make no sense. Vectors are elements of a vector space. A vector space has certain properties. A matrix represents a function between vector spaces. Such a space of linear functions has certain properties. You compare apples with oranges.

 

[askmathquestions]

They are not, simply because vector times vector isn't a vector anymore.That's why I asked you about the domains right from the beginning. This seems to make no sense. Vectors are elements of a vector space. A vector space has certain properties. A matrix represents a function between vector spaces. Such a space of linear functions has certain properties. You compare apples with oranges.

Okay, it sounds like you're saying the uniqueness of the identity matrix acting on square matrices and the uniqueness of the identity matrix acting on vectors are two different problems.

I don't quite know if that's true, because let's say you have a matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ and a vector $\begin{bmatrix} x \\ y \end{bmatrix}$. Well, the left-hand multiplicative identity for both of these is the matrix $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
Recall I assumed that $I_1$ and $I_2$ were two $n$ x $n$ matrices, does this help?

 

[fresh_42]

Okay, it sounds like you're saying the uniqueness of the identity matrix for square matrices and the uniqueness of the identity matrix for vectors are two different problems.

No. I am saying that you do not properly distinguish between vectors and matrices. You say that $A$ is a $n$ times $p$ matrix, then you call it a vector, i.e. $p=1.$ But these are fundamentally different multiplications:
$I_1$ and $I_2$ are functions from a $n$-dimensional vector space into a $n$-dimensional vector space.
p>1
$A$ is a function from a $p$-dimensional vector space into a $n$-dimensional vector space.
Hence, $I_1\cdot A$ is a multiplication of functions.
p=1
$I_1\cdot A$ is the function $I_1$ evaluated on the vector $A.$

If you only want to solve
$$
\begin{bmatrix}a&b\\c&d\end{bmatrix}\cdot \begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}x\\y\end{bmatrix}
$$
for all possible $x,y$, then solve
$$
\begin{bmatrix}a&b\\c&d\end{bmatrix}\cdot \begin{bmatrix}1\\0\end{bmatrix}=\begin{bmatrix}1\\0\end{bmatrix} \text{ and }
\begin{bmatrix}a&b\\c&d\end{bmatrix}\cdot \begin{bmatrix}0\\1\end{bmatrix}=\begin{bmatrix}0\\1\end{bmatrix}
$$
and calculate $a,b,c,d.$

If you want to prove $I_1A=I_2A=A \Longrightarrow I_1=I_2$ by general properties then specify the linear function spaces. Such a specification determines which conclusions are allowed.

I don't quite know if that's true, because let's say you have a matrix $[[a,b],[c,d]]$ and a vector $[[x],[y]]$. Well, the multiplicative identity for both of these is the matrix $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$.
Recall I assumed that $I_1$ and $I_2$ were two $n$ x $n$ matrices, does this help?

Yes, see above. And the general case is accordingly.

 

[askmathquestions]

No. I am saying that you do not properly distinguish between vectors and matrices. You say that $A$ is a $n$ times $p$ matrix, then you call it a vector, i.e. $p=1.$ But these are fundamentally different multiplications:
$I_1$ and $I_2$ are functions from a $n$-dimensional vector space into a $n$-dimensional vector space.
p>1
$A$ is a function from a $p$-dimensional vector space into a $n$-dimensional vector space.
Hence, $I_1\cdot A$ is a multiplication of functions.
p=1
$I_1\cdot A$ is the function $I_1$ evaluated on the vector $A.$

If you only want to solve
$$
\begin{bmatrix}a&b\\c&d\end{bmatrix}\cdot \begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}x\\y\end{bmatrix}
$$
for all possible $x,y$, then solve
$$
\begin{bmatrix}a&b\\c&d\end{bmatrix}\cdot \begin{bmatrix}1\\0\end{bmatrix}=\begin{bmatrix}1\\0\end{bmatrix} \text{ and }
\begin{bmatrix}a&b\\c&d\end{bmatrix}\cdot \begin{bmatrix}0\\1\end{bmatrix}=\begin{bmatrix}0\\1\end{bmatrix}
$$
and calculate $a,b,c,d.$

If you want to prove $I_1A=I_2A=A \Longrightarrow I_1=I_2$ by general properties then specify the linear function spaces. Such a specification determines which conclusions are allowed.Yes, see above. And the general case is accordingly.

Couldn't I just say $A \in \mathbb{C}^{n \ x \ p}$? You're saying vectors and matrices are different. For practical purposes I agree, but for this specific proof, both vectors and square matrices are a class of $n$ x $p$ matrices, so if you can prove uniqueness for the $n$ by $p$ case, then you've proven uniqueness for both vectors and other matrices.

 

[fresh_42]

Couldn't I just say $A \in \mathbb{C}^{n \ x \ p}$? You're saying vectors and matrices are different. For practical purposes I agree, but for this specific proof, both vectors and matrices are a class of $n$ x $p$ matrices, so if you can prove uniqueness for the $n$ by $p$ case, then you've proven uniqueness for both vectors and other matrices.

Yes, but the word vector is confusing then. Obviously.

You need to multiply matrices! Solve $(x)_{rs} \cdot E_{uv}=E_{uv}$ for all $u,v$ and calculate $(x)_{rs}.$

 

[askmathquestions]

You need to multiply matrices!

Right, you can multiply a vector by a matrix, or you can multiply another matrix by a matrix, in either case you're multiplying by a matrix, so I don't quite understand what the issue is. Both circumstances can be generalized as multiplying an $n$ x $p$ matrix $A$ by a square $n$ by $n$ matrix $I$.

 

[askmathquestions]

Also, I didn't notice there was a linear and abstract algebra section of this website. Would it be possible for a moderator to move this thread there?

 

[FactChecker]

It is not clear to me what the full problem statement, assumptions, and prior proven facts are.
If you are given a right and a left multiplicative identity, $I_R, I_L$, respectively, then you know that $I_L = I_L I_R = I_R$.
Does that help you?

 

[askmathquestions]

It is not clear to me what the full problem statement, assumptions, and prior proven facts are.
If you are given a right and a left multiplicative identity, $I_R, I_L$, respectively, then you know that $I_L = I_L I_R = I_R$.
Does that help you?

You're sure that this is true for a general $n$ x $p$ matrix? Another user said those identities are different. Perhaps you're assuming $A$ is a square matrix.

 

[FactChecker]

You're sure that this is true for a general $n$ x $p$ matrix? Another user said those identities are different. Perhaps you're assuming $A$ is a square matrix.

It is true by the definition of the multiplicative identity.
$I_L = I_L I_R$ by the definition of the right multiplicative identity matrix.
$I_L I_R = I_R$ by the definition of the left multiplicative identity matrix.
I thought that you specified earlier that the matrices were nxn.

 

[askmathquestions]

It is true by the definition of the multiplicative identity.
$I_L = I_L I_R$ by the definition of the right multiplicative identity matrix.
$I_L I_R = I_R$ by the definition of the left multiplicative identity matrix.
I thought that you specified earlier that the matrices were nxn.

The left-multiplicative identity is $n$ by $n$, I don't know that the right-multiplicative identity necessarily is, I wasn't assuming the same restrictions on that since I'm concerned partly about left-hand transformations on vectors.

 

[FactChecker]

The left-multiplicative identity is $n$ by $n$, I don't know that the right-multiplicative identity necessarily is, I wasn't assuming the same restrictions on that since I'm concerned partly about left-hand transformations on vectors.

IMHO, with all of the "I wasn't assuming" and "I'm concerned", it is not clear what the exact problem statement in the book was versus how much you modified the problem.

 

[askmathquestions]

IMHO, with all of the "I wasn't assuming" and "I'm concerned", it is not clear what the exact problem statement in the book was versus how much you modified the problem.

There is no problem statement in a book, I said that already, and the fact that this question isn't well-posed yet should also let you know that it's not from any official text. This is my own inquiry, and further why I asked for this thread to be moved to the linear algebra section. I explicitly said I need to figure out what assumptions need to be made and then amended my statement accordingly.

 

[FactChecker]

There is no problem statement in a book, I said that already,

Sorry, I missed that.

and the fact that this question isn't well-posed yet should also let you know that it's not from any official text. This is my own inquiry, and further why I asked for this thread to be moved to the linear algebra section. I explicitly said I need to figure out what assumptions need to be made and then amended my statement accordingly.

Ok. I see that is part of the goal of the question.

Suppose that you show that the matrix $I$ = diag(1,1,1,1,...,n) is a right multiplicative identity and that you have $I_1$ and $I_2$ as left multiplicative identities.
Then you can use the appropriate property at each step to say that $I_1 = I_1 I = I = I_2 I = I_2$.

 

[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 $A(I_1 - I_2) = 0$.