1. ## Matrices

I cannot get my head around this question. I would very much appreciate any help please.
If we
are given invertible matrices A, B and P so that A = PB, we can say that A is ‘left equivalent to B’. Prove that ‘left equivalence’ is an ‘equivalence relation’.

2. ## Re: Matrices

Originally Posted by mlg
I cannot get my head around this question. I would very much appreciate any help please.
If we
are given invertible matrices A, B and P so that A = PB, we can say that A is ‘left equivalent to B’. Prove that ‘left equivalence’ is an ‘equivalence relation’.
You need to give a complete definition of "left equivalent".
Is the matrix $P$ fixed? If not how does it function?

3. ## Re: Matrices

Thanks.
This type of maths is new to me.
I need to study it in more detail.

4. ## Re: Matrices

Originally Posted by mlg
Thanks.
This type of maths is new to me.
I need to study it in more detail.
BUT why can you not type in the exact definition from your text/lecture material?

5. ## Re: Matrices

Thanks again.
It mentions 'prove' in the question.
Would this suggest that a definition is not sufficient?

6. ## Re: Matrices

Originally Posted by mlg
Thanks again.
It mentions 'prove' in the question.
Would this suggest that a definition is not sufficient?
Plato is asking you to state what your text/course material is giving you for a definition of "equivalence relation."

-Dan

7. ## Re: Matrices

Thanks Dan.
My text material is giving me the following:
Matrix similarity is an equivalence relation and to show it we need to check in turn each the following criteria for equivalence: Reflexive, Symmetric and Transitive.

8. ## Re: Matrices

Hi,
I think you are still missing Plato's point. You need the exact definition of "left equivalence". Almost surely, the definition is:
Given invertible matrices A and B of the same size, A is left equivalent to B if and only if there is an invertible matrix P with A = PB.

So now you can prove that left equivalence is an equivalence relation on the set of all invertible $n\times n$ matrices for a given constant size $n$.

Reflexive: Let A be an invertible $n\times n$ matrix. Then A = IA (I is the identity $n\times n$ matrix). So this shows there is an invertible matrix P with A = PA, namely P = I.

You should now be able to prove the symmetric and transitive properties.

Edit: I should have added this thought. For a "meaningful" equivalence relation, let m and n be fixed positive integers and $M(n,m)$ the collection of all $n\times m$ matrices with entries from the reals (or any other field). Then for $A,\,B\in M(n,m)$, A is left equivalent to B if and only if there is an invertible $n\times n$ matrix P with A = PB.

9. ## Re: Matrices

Thank you johng.
I will now study this in more detail.

10. ## Re: Matrices

Thanks again johng for your time and effort.
As I said before, this type of maths is new to me. So it will take me a while to study it in more depth.

11. ## Re: Matrices

By the way, you say "If we are given invertible matrices A, B and P so that A = PB, we can say that A is ‘left equivalent to B". That doesn't look right to me because there is no "P" in the conclusion and you appear to be saying that A and B being equivalent depends upon some given matrix "P". I suspect what you really mean is "Two matrices, A and B, are left equivalent if and only if there exist an invertible matrix, P, such that A= PB.

You need to show:
1) Reflexive. Can you find an invertible matrix, P such that A= PA?

2) Symmetric. Suppose there exist an invertible matrix, P, such that A= PB. Can you find an invertible matrix, Q, such that B= QA?

3) Transitive. Suppose there exist an invertible matrix, P, such that A= PB and an invertible matrix Q such that B= QC. Can you find an invertible matrix, R, such that A= RC?

12. ## Re: Matrices

Thank you HallsofIvy.
I'll now work on your suggestions.

13. ## Re: Matrices

I'll give you a big hint for proving symmetry of left-equivalent.

We will assume A is left-equivalent to B, and to show symmetry we need to prove that B is left-equivalent to A.

Since A is left-equivalent to B, we know there is an invertible matrix P with A = PB.

Since P is invertible, P-1 exists, so multiply both sides of the equation A = PB by it.

14. ## Re: Matrices

Thanks Deveno.
I'll go ahead and try it.