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As we all know that two $n \times n$ matrices $A, B$ need not have the relation $AB = BA.$ But when do two $n \times n$ matrices have such a property?

Yes
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    I think an answer is much longer than you would expect. Try searching necessary and sufficient conditions for matrices to commute if no one gives an answer here. I see some articles on the topic. – Eoin Nov 12 '14 at 06:20
  • @Eoin: Ah, thanks so much. I was just working through linear algebra material and this question was triggered. After googling about this topic, I felt it would be better to ask here to see if some experts may help out with. – Yes Nov 12 '14 at 06:23
  • haha yeah I thought so. I'm glad someone had a fairly short answer after looking at the articles. – Eoin Nov 12 '14 at 06:39
  • You might take a look at this question – Thomas Nov 12 '14 at 06:51

2 Answers2

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To give a completely general answer is too complicated to be of general use. The simplest nice case: if $A$ and $B$ are diagonalizable and share the same eigenspaces, then they commute. This is because sharing the same eigenspaces implies that $A$ and $B$ can be simultaneously diagonalized, i.e. $A=CDC^{-1}$ and $B=CD'C^{-1}$ where $D,D'$ are diagonal and thus commute. This is also a necessary condition among diagonalizable matrices, that is, commuting diagonalizable matrices can be simultaneously diagonalized.

Kevin Carlson
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While I am not aware of any necessary and sufficient condition for matrices to commute with $A\in K^{n\times n}$, it is possible to completely characterise the set of matrices that commute with $A\in \mathbb R^{2\times 2}$ (and other special cases) by setting up a system of equations equating the individual coefficients of the two products $AB$ and $BA$ and finding the conditions for those equations to be fulfilled.

For small dimensions, this yields quite usable results. An explicit example for the $2 \times 2$ case can be found here.