Question

Consider the linear operator T:C^3 be defined by
T(z1,z2,z3)= (z1+iz2, iz1-2z2+iz3,-iz2+z3).
Compute T^* and check whether T is self-adjoint.
Check whether T is unitary.

Answer 1

Given the rules

$$z_1\to z_1+iz_2=w_1$$ $$z_2\to iz_1-2z_2+iz_3=w_2$$ $$z_3\to -iz_2+z_3=w_3$$

For linear operator 𝑇 the matrix representation is

$$T=\begin{bmatrix} 1 & i & 0 \\ i & -2 & i \\ 0 & -i & 1 \end{bmatrix}$$

We recall, that an operator T^* is called adjoint for the linear operator  T if

$$for \ all\ x, y \in \mathbb{C}^3 \ (Tx, y)=(x, T^*y).$$

The matrix representation for T^* can be found as

$$T^*=(\overline{T})^T=\overline{(T^T)}$$

where T^T denotes the transponse

\overline{T} \ denotes the matrix with complex conjugated entries.

In our case

$$T^*=\begin{bmatrix} 1 & -i & 0 \\ -i & -2 & i \\ 0 & -i & 1 \end{bmatrix} \mathrlap{\,/}{=} \ T$$

The adjoint operator 

$$T^*(z_1, z_2, z_3)=(z_1-iz_2, -iz_1-2z_2+iz_3, -iz_2+z_3).$$

Therefore, T is not selfadjoint.

We recall, that a unitary operator is a bounded linear operator on a Hilbert space that satisfies 

$$U^*U=UU^*=I_3,$$

where U^* is the adjoint of U.

In our case

$$T^*T=\begin{bmatrix} 1 & -i & 0 \\ -i & -2 & i \\ 0 & -i & 1 \end{bmatrix}*\begin{bmatrix} 1 & i & 0 \\ i & -2 & i \\ 0 & -i & 1 \end{bmatrix}=\begin{bmatrix} 2 & 3i & 1 \\ -3i & 6 & -i \\ 1 & i & 2 \end{bmatrix} \mathrlap{\,/}{=} \ I_3$$

Therefore, T is not unitary.