Identity Matrix

[diogomgf]

For any \(\displaystyle A \in M_{m*n}\) , \(\displaystyle I_{m} \cdot A = A \).
This is part of the definition of the identity matrix, and proving it is straightforward.

How do I formally prove the fact that there is no other matrix that fits this criteria?

 

[Steven G]

What have you tried? Assume that there are two identities, I and I' and conclude that they are equal. It actually is a simple proof.

 

[diogomgf]

What have you tried?

Not sure where to start on this one.

 

[Steven G]

Not sure where to start on this one.

I told you where to start. Assume that there are two identities.

 

[diogomgf]

\(\displaystyle I_n \cdot I_n = I'_n \cdot I_n \equiv I_n = I'_n \)
Just by the definition of the identity matrix... I can't really fathom where this is going.

 

[HallsofIvy]

"For any A∈M_m∗n, I_m⋅A=A.
This is part of the definition of the identity matrix."

Yes, it is part of the definition! The other part is that AI_m= A.
It has to work both ways. That is important because multiplication of matrices is not commutative.

Suppose both I_m and I'_m are identity matrices.
Then since I is an identity and I_m⋅A=A, I_mI'_m= I'_m.

And since I_m' is an identity and AI'_m= A,
I_mI'_m= I_m.

That is, I_mI'_m is equal to both I_m and I'_m. Since matrix multiplication is "well defined", I_m= I'_m.

 

[diogomgf]

Yes, it is part of the definition! The other part is that AI_m= A.
It has to work both ways. That is important because multiplication of matrices is not commutative.

That is a tangent to the original post. If you look close, you can even see that I said \(\displaystyle A_{mxn} \).
\(\displaystyle A \cdot I_m\) is not defined.

 

[Steven G]

Prof Halls, is this proof valid?
Suppose I and I' are two identity matrices of the same size.
Now consider I*I'.
Using I' as the id we get I*I' =I and using I as the id we get I*I'=I'
So I = I'

 

[diogomgf]

Prof Halls, is this proof valid?
Suppose I and I' are two identity matrices of the same size.
Now consider I*I'.
Using I' as the id we get I*I' =I and using I as the id we get I*I'=I'
So I = I'

Still, I can't understand where this is going...

 

[Dr.Peterson]

I think there are some details that need to be put together. Here are some key posts from this thread:

For any \(\displaystyle A \in M_{m*n}\) , \(\displaystyle I_{m} \cdot A = A \).
This is part of the definition of the identity matrix, and proving it is straightforward.

How do I formally prove the fact that there is no other matrix that fits this criteria?

"For any A∈M_m∗n, I_m⋅A=A.
This is part of the definition of the identity matrix."

Yes, it is part of the definition! The other part is that AI_m= A.
It has to work both ways. That is important because multiplication of matrices is not commutative.

Suppose both I_m and I'_m are identity matrices.
Then since I is an identity and I_m⋅A=A, I_mI'_m= I'_m.

And since I_m' is an identity and AI'_m= A,
I_mI'_m= I_m.

That is, I_mI'_m is equal to both I_m and I'_m. Since matrix multiplication is "well defined", I_m= I'_m.

That is a tangent to the original post. If you look close, you can even see that I said \(\displaystyle A_{mxn} \).
\(\displaystyle A \cdot I_m\) is not defined.

You're right about this last comment (if m≠n); but you also missed something in the definition of the identity matrix. The identity matrix I_m must work not only with all m×n matrixes A for one fixed n, but for any n, and on both sides. Yes, Halls slightly overstated things, ignoring n; but it is nevertheless true that when m=n, it will commute.

Therefore, if I_m is an m×m identity matrix, and I'_m is also an m×m identity matrix, then (as Jomo said)

I_m = I_mI'_m = I'_m​

because I_m is an m×m matrix, and when multiplied on the right by I'_m is left unchanged (the first "="), and likewise I'_m is an m×m matrix, and when multiplied on the left by I_m is left unchanged (the second "="). So they must be equal.

Now, if you had an I'_m that was an identity only for m×n with one particular n, and only on one side, all bets might be off. But it wouldn't be an identity matrix.

 

[diogomgf]

I think there are some details that need to be put together. Here are some key posts from this thread:



You're right about this last comment (if m≠n); but you also missed something in the definition of the identity matrix. The identity matrix I_m must work not only with all m×n matrixes A for one fixed n, but for any n, and on both sides. Yes, Halls slightly overstated things, ignoring n; but it is nevertheless true that when m=n, it will commute.

Therefore, if I_m is an m×m identity matrix, and I'_m is also an m×m identity matrix, then (as Jomo said)

I_m = I_mI'_m = I'_m​

because I_m is an m×m matrix, and when multiplied on the right by I'_m is left unchanged (the first "="), and likewise I'_m is an m×m matrix, and when multiplied on the left by I_m is left unchanged (the second "="). So they must be equal.

Now, if you had an I'_m that was an identity only for m×n with one particular n, and only on one side, all bets might be off. But it wouldn't be an identity matrix.

I think every one is giving too much emphasis on the identity matrix. I started the OP with the identity matrix with the purpose of getting it out of the way. All I want is a proof that no other matrix fits the criteria: that \(\displaystyle B_{m*m} \cdot A_{m*n} \neq A_{m*n} \)

 

[Dr.Peterson]

I think every one is giving too much emphasis on the identity matrix. I started the OP with the identity matrix with the purpose of getting it out of the way. All I want is a proof that no other matrix fits the criteria: that \(\displaystyle B_{m*m} \cdot A_{m*n} \neq A_{m*n} \)

Huh?

What are we supposed to emphasize other than what you asked about?

And hasn't that question been answered?

Maybe you need to restate what you are asking for?

 

[diogomgf]

Huh?

What are we supposed to emphasize other than what you asked about?

And hasn't that question been answered?

Maybe you need to restate what you are asking for?

The only thing that has been proven was that if there are two identity matrices I and I', then they are equal...
What I was asking is how to prove that besides the identity matrix, there isn't any other matrix B that \(\displaystyle B_{m*m} \cdot A_{m*n} = A_{m*n} \).

Sorry if I wasn't clear.

 

[Dr.Peterson]

So when you say, "How do I formally prove the fact that there is no other matrix that fits this criteria?" you aren't asking about the uniqueness of the identity matrix? That's what it means to me. Part of the problem, from the start of this discussion, is that you didn't explicitly state what the criterion is. How is it quantified?

I suppose you must mean that for a given matrix A, there is no other matrix B besides I that leaves it unchanged. But that just isn't true (which is why we wouldn't think of taking it that way). What makes the identity matrix special is that it works for everything.

As a trivial example, if you take A = 0, then B could be any matrix.

A more interesting example is that if [MATH]A = \begin{bmatrix}4 & 5\\ 0 & 0\end{bmatrix}[/MATH], then B could be anything of the form [MATH]\begin{bmatrix}1 & a\\ 0 & b\end{bmatrix}[/MATH].

Again, what makes the identity matrix unique is the whole definition as I stated it, not that it leaves any one matrix unchanged.

If this still isn't what you mean, please state your conjecture or theorem completely.

 

[diogomgf]

I suppose you must mean that for a given matrix A, there is no other matrix B besides I that leaves it unchanged. But that just isn't true(...)

A more interesting example is that if \(\displaystyle A = \begin{bmatrix}4 & 5\\ 0 & 0\end{bmatrix} \), then B could be anything of the form \(\displaystyle \begin{bmatrix}1 & a\\ 0 & b\end{bmatrix} \).

This answers my questions, thanks