Diagonalization of complex, symmetric but defect matrices

Question

Real symmetric matrices (not Hermetian $M^\dagger = M$, but symmetric $M^T = M$) can according to Autonne–Takagi factorization, as described here, be diagonalized. Thus, we can find a unitary matrix $U$ such that $A = U D U^T$, for a diagonal matrix $D$. However, as described here, the matrix
$$
A = \begin{pmatrix}
2i&1\\
1&0
\end{pmatrix}
$$
is defect, in that it only has one eigenvector $(i, 1)^T$, and thus $U$ can't be made up of eigenvectors as it usually is when diagonalizing. However, writing $U^T = (u_1, u_2, …, u_n)$, for some vectors $u_i$, we have that
$$
U A U^T = D \implies A U^T = A (u_1, …, u_n) = (Au_1, …, Au_n) = U^T D = (u_1, …, u_n) \begin{pmatrix}
d_1 &0 &0 \\
0& … &0 \\\
0&0 & d_n
\end{pmatrix}
= (d_1 u_1, …, d_nu_n).
$$
So it seems that $U$ is made up of the eigenvectors of $A$ anyway. Want went wrong?

Answer 1

It is not true that $$ UAU^T = D \implies AU^T = U^TD. $$ Note that in the case that $A$ is defective, $U$ is unitary but not orthogonal. That is, we have $U^{-1} = U^\dagger \neq U^T$.