= Solution
Set $N=M_i-I$. Direct multiplication gives
$$
N=\begin{pmatrix}0&i&i\\-i&0&0\\i&0&0\end{pmatrix},\qquad
N^2=\begin{pmatrix}0&0&0\\0&1&1\\0&-1&-1\end{pmatrix},\qquad N^3=0,
$$
with $N\neq0$. If $M_i$ were diagonalisable, then subtracting the identity in the same basis would diagonalise $N$. A diagonal matrix whose cube is zero has every diagonal entry zero and is itself zero. Similarity would then imply $N=0$, a contradiction. Hence
$$
\boxed{M_i\text{ is not diagonalisable}.}
$$
This proves failure directly; merely observing that it is not self-adjoint would not suffice, because non-self-adjoint matrices can still be diagonalisable.
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