Let ( V , ⟨ , ⟩ ) be a finite dimensional inner product space over C, and let T : V → V be a linear operator. Show that the following two conditions are equivalent:

Question

Let ( V , ⟨ , ⟩ ) be a finite dimensional inner product space over C, and let T : V → V be a linear operator. Show that the following two conditions are equivalent:

1)  is invertible and   

2)    for all     .

(These are the two equivalent conditions defining a unitary operator.)

Answer 1

((i)-->(ii)) We assume that the Condition (i) holds. That is T is invertible and T^(-1) = T^(*). For any x and y in V, we have

<T(x), T(y)> = <x, T^*(T(y))> = <x, T^(-1)(T(y))>=<x, y>, where T^(-1)(T(y))=y. So the Condition (ii) is true.

((ii)-->(i)) On the other hand, we assume that for any x, y in V, <T(x), T(y)>=<x,y> holds. That is, <x, T^*(T(y))>=<x,y> for any x, y in V. In particular, let y be any vector in V, then 0 = <x, T^*(T(y)) - y>=<x, (T^(*)oT - I)(y)> for each x in V, whereT^(*)oT is the composition of T^(*) and T, and I is the identity map that sends every vector to itself. Since <x, v>=0 for each v if and only if v=0, then we have (T^(*)oT - I)(y)=0; this is true for each y in V, so T^(*)oT - I must be the zero map on V. Therefore, we have T^(*)oT- I =0 or T^(*)oT=I. This implies that T has the inverse T^(*) and so T is invertible and T^(*)=T^(-1). So the Condition (i) is true.

Now we can conclude that the two conditions are equivalent.