Prove the identity matrix is unique

[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.