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Let $V$ be a finite dimensional vector space over the complex numbers. If $u, v \in \text{End}(V)$ are unitary, then $|\det(u + v)| \leq 2^n$.

I'm not sure if my solution to this problem is correct. This is what I have. Since $u$ and $v$ are unitary I can write diagonal matrices $U$ and $V$ for $u$ and $v$, respectively, with their eigenvalues appearing in the diagonals. Thus, $U +V$ is diagonal and its eigenvalues are bounded in magnitude by 2, because $u$ and $v$ are unitary. Since the determinant $U +V$ is the product of its eigenvalues, then $|\det(U+V)| \leq 2$, so $|\det(u+v)| \leq 2$.

I'm not sure if I'm allowed to just work directly with $U+V$ or if I have to check first if they are simultaneously diagonalizable to do this.

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  • $\begingroup$ Presumably you don't just mean a vector space but actually an inner product space. Also, that's exactly the issue, when you say you "write diagonal matrices", what you mean is you pick an orthonormal basis for your inner product space and then look at the representing matrices. But there's no reason for the same orthonormal bases to work for both unitary operators. $\endgroup$
    – Smiley1000
    Commented Dec 8 at 19:56
  • $\begingroup$ However, you don't need to look at representing matrices at all for this problem, just look at the operators abstractly. $\endgroup$
    – Smiley1000
    Commented Dec 8 at 19:57
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    $\begingroup$ Related: math.stackexchange.com/questions/3396171/… $\endgroup$
    – Tri
    Commented Dec 8 at 20:00
  • $\begingroup$ Your proof certainly treats the two as simultaneously diagonalizable $\endgroup$
    – operatorerror
    Commented Dec 8 at 20:07
  • $\begingroup$ another duplicate: math.stackexchange.com/questions/3990721/… $\endgroup$
    – user8675309
    Commented Dec 9 at 0:09

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