# prove/disprove if each two in $n$ operators can be diagonalizable simultaneously then all can be diagonalizable simultaneously

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?

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