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Q: Let $m, n$ be positive intergers. Let $A$ and $B$ real $n\times n$ matrices. Assume that $B$ is symmetric and positive definite. If $A$ commutes with $B^{m}$, prove that $A$ commutes with $B$.

So I don't really know where to start with this other than the decomposition theorem for symmetric positive operators, so that I can write $B$ as diagonal with positive real entries. Perhaps taking the $m^{th}$ root could be useful which we can do since $B$ has positive entries, but I'm not sure.

Many thanks!

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    $\begingroup$ It is enough to prove that there exists a polynomial $p \in \mathbb{R}\left[x\right]$ such that $B = p\left(B^m\right)$. To prove this, we can WLOG assume that $B$ is diagonal (since $B$ is diagonalizable). Now it is clear that we can set $p$ to be any polynomial which sends the $m$-th powers of the eigenvalues of $B$ to said eigenvalues. Do you know why such a polynomial exists? (You still need to use positive definiteness -- make sure that you do so, or else the proof is wrong.) $\endgroup$ – darij grinberg Dec 9 '14 at 4:29
  • $\begingroup$ I've been trying, but no progress. Would I try to use the minimal polynomial for $B$ or $B^{m}$ somehow? Or should I be trying to just write it out from scratch? $\endgroup$ – TheManWhoNeverSleeps Dec 9 '14 at 5:01
  • $\begingroup$ This last step is not really a matrix problem. Do you know about Lagrange interpolation? $\endgroup$ – darij grinberg Dec 9 '14 at 5:07
  • $\begingroup$ Reading about it now on wikipedia. So I would use Lagrange interpolation to get a polynomial $p(x)$ such that $p(\lambda^{m})=\lambda$, but how does positivity come into this? $\endgroup$ – TheManWhoNeverSleeps Dec 9 '14 at 5:14
  • $\begingroup$ Yes, that's the question :). What is the main condition for Lagrange interpolation to work? $\endgroup$ – darij grinberg Dec 9 '14 at 5:16
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Since, $B$ is symmetric and positive definite, it is Unitarily diagonalizable, that is $\exists \, U \in U_n(\mathbb{R})$, such that $U^{*}BU = D[d_1,\cdots,d_n]$ is a diagonal matrix, with $d_i > 0$, for all $i = 1(1)n$.

Denote, $U^{*}AU = W = (w_{ij})_{n \times n}$, then,

$AB^m = B^mA \implies U^{*}AU(U^{*}BU)^m = (U^{*}BU)^mU^{*}AU \implies WD^m = D^mW$

So, $d_i^mw_{ij} = w_{ij}d_j^m$ for all $i,j$

Which implies, $d_iw_{ij} = w_{ij}d_j$ for all $i,j$ (since, $d_i,d_j >0$ and $d_i^m = d_j^m \implies d_i = d_j$)

That is $WD = DW$, which in turn implies $AB = BA$.

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    $\begingroup$ Yeah that works. I like the answer given in the comments above as as well, using the Lagrange interpolation. $\endgroup$ – TheManWhoNeverSleeps Dec 11 '14 at 23:00
  • $\begingroup$ @TheManWhoNeverSleeps yes that is nice too ! :) $\endgroup$ – r9m Dec 11 '14 at 23:02

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