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My friend asked the following question related to matrices:

Assume A is

$$ \begin{pmatrix} \frac{1}{2} & 0 & \frac{1}{2} \\ 1 & 0 & 0 \\ 0 & \frac{1}{2} & \frac{1}{2} \\ \end{pmatrix} $$

and B is

$$ \begin{pmatrix} \frac{1}{4} & \frac{1}{4} & \frac{1}{2} \\ \frac{1}{2} & 0 & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{4} & \frac{1}{4} \\ \end{pmatrix} $$

AB = BA =$$ \begin{pmatrix} \frac{3}{8} & \frac{1}{4} & \frac{3}{8} \\ \frac{1}{4} & \frac{1}{4} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{8} & \frac{3}{8} \\ \end{pmatrix} $$

Why is this true? From the knowledge that I have, $A, B$, and the product are not identity matrices. $A=B^{n}$ is not true.

saru
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ssd21345
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    Having $AB=I$ or $A=B^n$ will guarantee that $AB=BA$ but they are not the only ways you can have $AB=BA$. Another example is two diagonal matrices will commute. – Ross Millikan Jun 01 '20 at 14:43
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    If you verified that $AB=BA$ via computations, and this does not satisfy you, are you looking for a characterization of matrices that commute? – Fimpellizzeri Jun 01 '20 at 14:44
  • @Fimpellzieri. Yes, is there any other way to prove these particular Matrices? – ssd21345 Jun 01 '20 at 14:48
  • You might be interested in simultaneously diagonalizable – Cloud JR K Jun 01 '20 at 14:49
  • @JMoravitz Thanks, I did read it and trying to calculate eigenvectors to prove them simultaneously diagonalizable but after I calculate it doesn't really know what to do with the eigenvectors , sorry didn't mention it before-hand. – ssd21345 Jun 01 '20 at 14:49
  • To be honest, slapping most of the commute questions with "When is matrix multiplication commutative?" isn't useful. (I spent ~1 hr to search math StackExchange before asking that) Since most of the answers of that question are about simultaneously diagonalizable, while you can find that there are more possibilities of matrices being commuting at here, hope someone in the future read this question find it's more useful. – ssd21345 Jun 01 '20 at 16:26

2 Answers2

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It is not true that $A = B^n$ for an integer $n$; however, it is true that $B = A^2$. This is enough to see that the matrices will commute.

Ben Grossmann
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If matrices $A$ and $B$ commute and $A$ has distinct eigenvalues, then $B$ must be a polynomial function of $A$. In this case, as Omnomnomnom pointed out, $B$ happens to be $A^2$.

On the other hand, $A$ is not a function of $B$ in this case, because $B$ has a double eigenvalue ($-1/4$) and $A$ has distinct eigenvalues.

Robert Israel
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  • Perhaps it is worth clarifying that "not a function of" in this case means that there is no polynomial (or analytic function) $p$ for which $A = p(B)$. – Ben Grossmann Jun 01 '20 at 15:22