According to the spectral theorem, every unitary matrix $A\in U(n)$ is diagonalizable, i.e. there is another unitary matrix $U\in U(n)$ such that
$$A=UDU^{-1}$$
where $D$ is a diagonal matrix consisting of the eigenvalues of $A$. Now since $\mathrm{SU}(n)\subset U(n)$, clearly this is also true for every $A\in \mathrm{SU}(n)$. My question is now if we can then say that also $U\in \mathrm{SU}(n)$. I have seen this in some book, but it is not clear to me why this is the case. For example, when I take the determinat of the equation $A=UDU^{-1}$, then $\mathrm{det}(U)$ and $\mathrm{det}(U^{-1})$ always cancel, so I do not get any information from that. So, how can I see that $U$ has determinant $1$?
Yes, since
$$ A = UDU^{-1} = (B U) D ( B U)^{-1}.$$
where
$$B = \begin{pmatrix} 1/ \det U & \vec 0\\ \vec 0& I_{n-1} \end{pmatrix}$$
and $BU \in SU(n)$.