91Ó°ÊÓ

Recall that a linear operator T on an inner product space V is said to be normal if TT* = T*T. The adjoint of T, denoted by T*, is the unique linear operator on V such that for every pair of vectors v and w in V, we have: \( \langle Tv,w\rangle = \langle v,T^*w\rangle \) Now, we need to prove that if T is normal, then ||T(v)|| = ||T*(v)|| for every v in V.

Suppose T is a normal operator on V. We want to show that for every v in V, ||T(v)|| = ||T*(v)||. For any vector v in V, we have: \( ||T(v)||^2 = \langle T(v),T(v)\rangle \) \( = \langle v,T^*(T(v))\rangle \) (using adjoint property) Since T is normal, TT* = T*T, so we can replace T*(T(v)) with T(T*(v)): \( ||T(v)||^2 = \langle v,T(T^*v)\rangle \) \( = \langle T^*(v),T*(v)\rangle \) (using adjoint property again) \( = ||T^*(v)||^2 \) Taking the square root of both sides, we get: \( ||T(v)|| = ||T^*(v)|| \) So, if T is normal, then ||T(v)|| = ||T*(v)|| for every v in V.

Now, we need to prove the converse in the case of complex inner product spaces: if ||T(v)|| = ||T*(v)|| for every v in V, then T is normal. Suppose ||T(v)|| = ||T*(v)|| for every v in V. Let v and w be arbitrary vectors in V. We want to show that TT* = T*T. Consider the following inner products: \( \langle (TT^* - T^*T)(v), w \rangle \) Expanding the expression, we have: \( \langle TT^*v, w \rangle - \langle T^*Tv, w \rangle \) Now, using the adjoint property, we rewrite the inner products: \( \langle T^*v, T^*w \rangle - \langle Tv, Tw \rangle \) Since ||T(v)|| = ||T*(v)|| for every vector v, we can replace one of the terms: \( \langle T^*v, T^*w \rangle - \langle T^*v, T^*w \rangle = 0 \) As this identity holds for all v and w in V, we conclude that TT* = T*T. Thus, T is normal when ||T(v)|| = ||T*(v)|| for every vector v in V. In summary, we have proven that if T is normal on V, then ||T(v)|| = ||T*(v)|| for every v in V. Moreover, the converse holds in complex inner product spaces: if ||T(v)|| = ||T*(v)|| for every v in V, then T is normal.