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I am super confused on how to get started on this problem. A starting hint would be great. I am given that $A \in M_n(\mathbb{C})$, and that for $B,C \in M_n(\mathbb{C})$, $AB=BA, CA=AC$ and that $A$ has distinct eigenvalues. The problem is to show that $BC=CB$. I am really stuck on this problem. A helpful hint would be much appreciated.

9301293
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1 Answers1

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Hint : When two matrices (or linear maps) commute, the eigenspaces of one are left invariant by the other.

Hint n°2 (for solution n°2) : Consider $\tilde{A} = PAP^{-1}$ the diagonalized form of $A$, and in the same way $\tilde{B} = PBP^{-1}$. Now write the coefficients of $\tilde{A}\tilde{B}$, as well as the coefficients of $\tilde{B}\tilde{A}$, in term of those of $\tilde{A}$ and $\tilde{B}$ ; comparing them should give you a nicer form for $\tilde{B}$ (that is, it will restrict the coefficients of $\tilde{B}$). Do the same for $\tilde{C} = PCP^{-1}$.

mennecybean
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  • I do not think that is a theorem or topic that we have covered. There was a hint in the problem that said "consider the digonilazation of $A$". – 9301293 May 13 '15 at 08:17
  • Is there another way to go about this problem? – 9301293 May 13 '15 at 08:20
  • Yes : writing $A$, $B$, $C$ in a basis which diagonalizes $A$, the condition $AB=BA$ yields conditions on the coefficients of $B$. – mennecybean May 13 '15 at 08:30
  • Would you mind typing up an answer to this, I am really not understanding it... – 9301293 May 13 '15 at 08:37
  • Using the second (diagonalization) method ? I tried to make the hint more explicit in the edit : I hope there's enough for you to conclude :) – mennecybean May 13 '15 at 08:53
  • Segipp, I collected some equivalent results, satisfied by the "distinct eigenvalues" condition but allowing slightly more generality, at http://math.stackexchange.com/questions/92480/given-a-matrix-is-there-always-another-matrix-which-commutes-with-it/92832#92832 – Will Jagy May 13 '15 at 19:14