Here matrix representation of T with respect to standard basis "\\{ (1,0,0), (0,1,0), (0,0,1)\\}" over the field "C" is "A="
"\\begin{pmatrix}\n 1& i& 0 \\\\ \ni & -2&0\\\\\n0&-i&1\n\\end{pmatrix}"
Now , "T^*=" Transpose conjugate of matrix representation of "T"
"=A^H"
"=" "\\begin{pmatrix}\n 1&-i&0 \\\\\n -i&-2&0\\\\\n0&i&1\n\n\\end{pmatrix}"
Since ,"A\\neq A^H"
Hence, "T" is not a self adjoint operator.
To check unitary operators ,
Consider "A.A^H=" "\\begin{pmatrix}\n 2&-3i&-1\\\\\n3i&5&-2i\\\\\n-1&2i&2\n\\end{pmatrix}" "\\neq I"
Where "I" stands for identity matrix.
Therefore,"T" is not a unitary operator.
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