Question

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

Answer 1

Answer on Question #81145 – Math – Linear Algebra

Question

Consider the linear operator $T \colon \mathbb{C}^3 \to \mathbb{C}^3$, defined by $T(z_1, z_2, z_3) = (z_1 - iz_2, iz_1 + 2z_2 + iz_3, -iz_2 + z_3)$.

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

ii) Check whether $T$ is unitary.

Solution

i) Given the rules

$$\begin{array}{l} z_1 \rightarrow z_1 - iz_2 = w_1 \\ z_2 \rightarrow iz_1 + 2z_2 + iz_3 = w_2 \\ z_3 \rightarrow -iz_2 + z_3 = w_3 \\ w = Tz \\ \end{array}$$

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

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

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

In our case

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

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 selfadjoint.

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

$$\begin{array}{l} T \cdot T^* = \begin{bmatrix} 1 & -i & 0 \\ i & 2 & i \\ 0 & -i & 1 \end{bmatrix} \cdot \begin{bmatrix} 1 & -i & 0 \\ i & 2 & i \\ 0 & -i & 1 \end{bmatrix} = \\ = \begin{bmatrix} 1(1) - i(i) + 0(0) & 1(-i) - i(2) + 0(-i) & 1(0) - i(i) + 0(1) \\ i(1) + 2(i) + i(0) & i(-i) + 2(2) + i(-i) & i(0) + 2(i) + i(1) \\ 0(1) - i(i) + 1(0) & 0(-i) - i(2) + 1(-i) & 0(0) - i(i) + 1(1) \end{bmatrix} = \\ = \begin{bmatrix} 2 & -3i & 1 \\ 3i & 6 & 3i \\ 1 & -3i & 2 \end{bmatrix} \neq I_3 \\ \end{array}$$

Therefore, $T$ is not unitary.

Answer:

i) $T$ is selfadjoint.

ii) $T$ is not unitary.

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