I am struggling to understand the proof of the following question: Let U(n) be a set of unitary matrices(diagonalizable) . Let F ⊂U(n) be an abelian subgroup of U(n) show that the matrices in F are simultaneously diagonalizable? Please who can give me a simplified explanation of the solution, because I have found the solution in the internet but i do not understund it.
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I think this is essentially "commuting unitary operators are simultaneously diagonalizable". – angryavian Jan 08 '19 at 21:13
2 Answers
This is based on a more general result:
A set of diagonalizable endomorphisms that commutes are simultaneously diagonalizable
See Simultaneously diagonalisable iff A,B commute that can be easily extended to any set of diagonalizable endomorphisms.
Then apply it to your particular case.
In short: this has nothing to do specially with unitary matrices. Apart for sure that those are diagonalizable!

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The key point is the following: let $E$ be a finite dimension vector space over some field $K$, and let $F$ a set of endomorphisms from $E$ such that they all commute and they are all diagonalizable.
Then they are simultaneously diagonalizable.
Proof: induction over the dimension of $E$. If $E$ is either zero or a line, it works. Assume the result is true as soon as $E$ has dimension lower than $n \geq 2$ for some integer $n$.
If $F$ contains only scalar endomorphisms, it works.
So assume there is $u_0 \in F$ that is not scalar, let us denote $U_i$ its eigenspaces. Since $u_0$ commutes with all the other elements of $F$, they all induce endomorphisms in all the $U_i$.
Since all the other elements of $F$ are diagonalizable, each one of them vanishes some polynomial with simple roots, thus this polynomial is vanished by all the resteictions to the $U_i$.
Let $F_i$ be the set of all restrictions of the elements of $F$ to $U_i$. By the induction hypothesis, the elements of $F_i$ are simultaneously diagonalizable, thus (concatenating goods bases for all $U_i$), the elements of $F$ are simultaneously diagonalizable.
Note that a similar proof works in infinite-dimensional spaces when $F$ is finite.

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I try to read the solution many time but I can not undersand it. As a biginner in this domain there are so many details that I do not know about it in this solution.Can you please give me a detailed solution to this question, how did you get to these results? – jimi Jan 13 '19 at 13:05
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