 Solution

A self-adjoint operator satisfies $⟨Tx,y⟩=⟨x,Ty⟩$ for every $x,y$. The finite-dimensional finite-dimensional spectral theorem gives an orthonormal basis of eigenvectors and real eigenvalues; eigenvectors for distinct eigenvalues are orthogonal. Consequently a matrix represents a self-adjoint transformation in some, possibly nonorthonormal, basis exactly when it is diagonalizable with real eigenvalues.

For $A_0$, the eigenvalues are $2,1,0$, all distinct and real. Eigenvectors are $e_1$, $e_2$, and $(0,-1,1{)}^T$. Thus $A_0$ can represent a self-adjoint transformation after choosing a suitable basis.

For $A_1$, the eigenvalue one has algebraic multiplicity two but its eigenspace has dimension one. The lower $2\times 2$ block is a nontrivial Jordan block, so the matrix is not diagonalizable. Hence $A_1$ cannot represent a self-adjoint transformation in any basis.

For $A_2$, the eigenspace for two is spanned by $e_1$ and $(0,1,1{)}^T$, and the eigenvalue-one eigenspace is spanned by $e_2$. These three vectors form an eigenbasis. Therefore $A_2$ can represent a self-adjoint transformation. To see sufficiency on the given inner product space, first define the real diagonal transformation in an orthonormal basis and then apply the inverse of a diagonalizing basis change to obtain the requested matrix. Symmetry of the displayed matrix itself would be required only for an orthonormal representing basis.