Question

Consider the linear operator T : C
4 → C
4
, defined by
T (z1,z2,z3,z4) = (−iz2,iz1,−iz4,z3).
i) Compute T
∗
and check whether T is self-adjoint.
ii) Check whether T is unitary.

Answer 1

Answer on Question 66694 - Math - Linear Algebra

Consider the linear operator $T: \mathbb{C}^4 \to \mathbb{C}^4$, defined by $T(z_1, z_2, z_3, z_4) = (-iz_2, iz_1, -iz_4, z_3)$.

i) Compute $T^{*}$ and check whether $T$ is selfadjoint.

ii) Check whether $T$ is unitary.

Solution

i) For linear operator $T$ the matrix representation is

$$\left[ \begin{array}{cccc} 0 & -i & 0 & 0 \\ i & 0 & 0 & 0 \\ 0 & 0 & 0 & -i \\ 0 & 0 & 1 & 0 \end{array} \right].$$

We recall, that an operator $T^{*}$ is called adjoint for the linear operator $T$ if for all $x, y \in \mathbb{C}^4$ $(Tx, y) = (x, T^{*}y)$. The matrix representation for $T^{*}$ can be found as

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

where $A^T$ denotes the transpose and $\overline{A}$ denotes the matrix with complex conjugated entries.

In our case

$$T^{*} = \left[ \begin{array}{cccc} 0 & -i & 0 & 0 \\ i & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & i & 0 \end{array} \right]$$

and the adjoint operator $T^{*}(z_{1},z_{2},z_{3},z_{4}) = (-iz_{2},iz_{1},z_{4},iz_{3})$. Since $T \neq T^{*}$, then $T$ is not selfadjoint.

ii) We recall, that a unitary operator is a bounded linear operator on a Hilbert space that satisfies $U^{*}U = UU^{*} = I$, where $U^{*}$ is the adjoint of $U$.

In our case

$$T \cdot T^{*} = \left[ \begin{array}{cccc} 0 & -i & 0 & 0 \\ i & 0 & 0 & 0 \\ 0 & 0 & 0 & -i \\ 0 & 0 & 1 & 0 \end{array} \right] \cdot \left[ \begin{array}{cccc} 0 & -i & 0 & 0 \\ i & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & i & 0 \end{array} \right] = \left[ \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right] = I$$

and

$$T^{*} \cdot T = \left[ \begin{array}{cccc} 0 & -i & 0 & 0 \\ i & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & i & 0 \end{array} \right] \cdot \left[ \begin{array}{cccc} 0 & -i & 0 & 0 \\ i & 0 & 0 & 0 \\ 0 & 0 & 0 & -i \\ 0 & 0 & 1 & 0 \end{array} \right] = \left[ \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right] = I$$

Therefore, $T$ is unitary.

Answer: i) $T^{*}(z_{1},z_{2},z_{3},z_{4}) = (-iz_{2},iz_{1},z_{4},iz_{3})$, $T$ is not selfadjoint; ii) $T$ is unitary.

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