Final answer:
To prove that T is Hermitian if and only if iT is skew-Hermitian, we multiply T by i and apply the adjoint operator. We then demonstrate that T+T*, T*T, and TT* are Hermitian, and T-T* is skew-Hermitian by applying properties of adjoints. For Hermitian T, the inner product ⟨T(v),v⟩ is shown to be real.
Step-by-step explanation:
Understanding Hermitian and Skew-Hermitian Transformations
To show that T is Hermitian if and only if iT is skew-Hermitian, consider the definition of a Hermitian operator (T = T*). Multiplying both sides by i, we get iT = iT*. Taking the adjoint of both sides of this equation given that the adjoint is a linear operation and the adjoint of i is -i since i is purely imaginary, gives (iT)* = -iT. Hence, if T is Hermitian then iT is skew-Hermitian, and vice versa.
For the next part, we want to prove that T+T*, T*T, and TT* are Hermitian, while T-T* is skew-Hermitian. This comes down to showing that the adjoint of each expression is equal to the expression itself for Hermitian, or the negative of the expression for skew-Hermitian. Using properties of the adjoint operation, we can confirm each property respectively. All these combinations involve real scalars when vectors are taken as argument, hence proving their Hermitian nature.
To prove (c), we decompose T into a sum of two components, T = S1 + iS2, where S1 and S2 are Hermitian. S1 can be taken as (T + T*)/2 and S2 as (T - T*)/(2i), which are both Hermitian as shown prior. The uniqueness of S1 and S2 follows from the properties of the inner product space.
Finally, for a Hermitian T, the inner product ⟨T(v),v⟩ is real because it equals ⟨v,T*(v)⟩=⟨v,T(v)⟩, as T = T*. Since the inner product of a vector with itself must be real, the conclusion follows.