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Let $A, B \in \mathbb{R}^{n \times n}$ two positive definite matrices such that $AB = BA$, that is $A$ commutes with $B$. It is easy to prove that $A^{1/2}$ commutes with $A$, indeed $AA^{1/2} = A^{3/2}=A^{1/2}A$, but I am wondering whether it is true that $A^{1/2}B = BA^{1/2}$, in other words that $AB = BA$ implies $A^{1/2}B = BA^{1/2}$. The best clue I have - and it isn't much - is that I cannot prove it.

gosbi
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1 Answers1

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Let $P$ be a polynomial such that $P(\lambda) = \sqrt{\lambda}$ for every eigenvalue $\lambda$ of $A$. Then $A^{1/2} = P(A)$. A polynomial in $A$ commutes with every matrix that commutes with $A$.

Robert Israel
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  • thank you for your answer. Could you suggest a reference for the proof of this fact (a polynomial in A commutes with any matrix that commutes with A)? I had never come across this result before. – gosbi Jan 23 '15 at 18:20
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    It's obvious. If $A$ commutes with $B$, then so does $A^n$ for all positive integers $n$, and then so does any linear combination of these. – Robert Israel Jan 23 '15 at 21:40
  • @RobertIsrael In what sense is the square root a linear combination of these? – HerpDerpington Mar 21 '24 at 21:01
  • a polynomial is a linear combination of $A^n$ terms – J. W. Tanner Mar 21 '24 at 21:22