# What is the necessary condition for matrix commutation?

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1. Mar 5, 2015

### fairy._.queen

Hi all!

I was wondering what the necessary condition is for two arbitrary matrices, say A and B, to commute: AB = BA.

I know of several sufficient conditions (e.g. that A, B be diagonal, that they are symmetric and their product is symmetric etc), but I can't think of a necessary one.

2. Mar 5, 2015

3. Mar 6, 2015

### fairy._.queen

Ok, so it seems the condition (quite sensible actually!) is that they must both be square and simultaneously triangularisable. Thanks a lot!

4. Mar 6, 2015

### fairy._.queen

What happens, though, when the matrix scalar field is not algebraically closed? I'm happy with the fact that, in this case, if the two matrices are diagonalisable and commute then they are simultaneously diagonalisable, but what is a necessary condition for arbitrary, say, real matrices to commute (even when they can't be diagonalised)?

Thanks!

5. Mar 7, 2015

### BruceW

you could just pretend that the scalar field is the complex numbers, and see if you can make the matrices simultaneously triangular under the complex numbers. If they are not simultaneously triangular under the complex numbers, they will not commute. And if they are simultaneously triangularizable under the complex numbers, then they do commute.

6. Mar 9, 2015

### fairy._.queen

Ok, it makes sense. Thanks a lot!

7. Mar 30, 2015

### Stephen Tashi

This remark is from a bound set of notes I found in a used book store: "If two matrices are both polynomials in the same matrix $C$, then they commute. As we shall see later, this is essentially the only way in which we can have commuting matrices."