I have an idea that for $n$ diagonalizable operators $A_1, A_2, ..., A_n \in \ell(V)$. if each $A_i, A_j$ can be diagonalizable simultaneously then all of them can be diagonalizable simultaneously.

if it is true then we can prove that for every permutation the answer of $A_{i_1}A_{i_2}...A_{i_n}$ is the same iff they are diagonalizable simultaneously.

can you prove or disprove that?

  • 1
    $\begingroup$ If $A_i$ and $A_j$ are simultaneously diagonalisable, then they commute: $A_iA_j=A_jA_i$. $\endgroup$ – Lord Shark the Unknown Jan 14 at 6:36
  • $\begingroup$ @LordSharktheUnknown look at the "iff" part. $\endgroup$ – Peyman mohseni kiasari Jan 14 at 6:37
  • $\begingroup$ Are we talking about bounded operators on a Hilbert space? And by diagonalizable, do you mean unitarily equivalent to a multiplication operator? $\endgroup$ – MaoWao Jan 15 at 12:40

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