I have to find the $h$ parameter of the matrix so that it can be diagonalized by an orthogonal matrix.
\begin{pmatrix} h & h+1 & 0 \\ 2h & 1 & h-1 \\ 2h-2 & 0 & 1 \end{pmatrix}
The solution provided for this problem is that the given matrix, since it has to be diagonalized by an orthogonal matrix, must be symmetric, hence $h=1$.
The matrix that diagonalises another matrix, named $U$, is the one that has $A$'s Eigenvectors in column, so that
$D = U^{-1} AU$
and $D$ is the matrix with $A$'s Eigenvalues as its trace.
I find this incorrect but I'm not sure. I've done a lot of exercises like this and in 80% of the cases the matrix that diagonalized the matrix was orthogonal without the main matrix being symmetric.
Is it correct to assume that the matrix must be symmetric to be diagonalized by an orthogonal matrix or the solution is just partially correct?
Yes: if $U$ is an orthogonal matrix and $D = U^{-1} A U$ is a diagonal matrix, then $A = U D U^{-1} = U D U^T$ is symmetric.