= Solution
{c}
A <self-adjoint operator> satisfies $\langle Tx,y\rangle=\langle x,Ty\rangle$ 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 \b[$A_0$ 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 $2\times2$ block is a nontrivial <Jordan block>, so the matrix is not <diagonalizable>. Hence \b[$A_1$ 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 \b[$A_2$ 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.
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