Theorem 6.11
Let \(V\) be an inner product space, and let \(T\) and \(U\) be linear operators on \(V\) whose adjoint exist. Then
  1. \(T + U\) has an adjoint, and \((T+U)^{*} = T^* + U^*\).
  2. \(cT\) has an adjoint, and \((cT)^* = \bar{c}T^*\) for any \(c \in \mathbf{F}\).
  3. \(TU\) has an adjoint, and \((TU)^* = U^* T^*\).
  4. \(T^*\) has an adjoint, and \(T^{**} = T\).
  5. \(I\) has an adjoint, and \(I^* = I\).


Proof:

For (a)

$$ \begin{align*} \langle (T + U)x, y \rangle &= \langle T(x) + U(x), y \rangle \\ &= \langle T(x), y \rangle + \langle U(x), y \rangle \\ &= \langle x, T^*(y) \rangle + \langle x, U^*(y) \rangle \\ &= \langle x, T^*(y) + U^*(y) \rangle \\ &= \langle x, (T^* + U^*)(y) \rangle \\ \end{align*} $$

Therefore, \((T+U)^*\) exists and it equals to \(T^*+U^*\).


Theorem 6.11 (Corollary)
Let \(A\) and \(B\) be \(n \times n\) matrices. Then
  1. \((A + B)^* = A^* + B^*\).
  2. \((cA)^* = \bar{c}A^*\) for all \(c \in \mathbf{F}\).
  3. \((AB)^* = B^*A^*\).
  4. \(A^{**} = A\).
  5. \(I^* = I\).


Proof:

[TODO]


References