Suppose we have a linear transformation $ T $ on some real inner product space $ V $, with adjoint $ T^{*} $.
How can we go about showing that $$ (T^{n})^{*} = (T^{*})^{n} $$ for a positive integer $ n $? Also, does this means that $ [f(T)]^{*} = f(T^{*}) $ for any polynomial $f$?
Any help would be appreciated.
We need the following result.
Proof: By the definition of ‘adjoint transformation’, we have \begin{align} \forall \mathbf{v}_{1},\mathbf{v}_{2} \in V: \quad \langle \mathbf{v}_{1},(S \circ T)(\mathbf{v}_{2}) \rangle &= \langle \mathbf{v}_{1},S(T(\mathbf{v}_{2})) \rangle \\ &= \langle {S^{*}}(\mathbf{v}_{1}),T(\mathbf{v}_{2}) \rangle \\ &= \langle {T^{*}}({S^{*}}(\mathbf{v}_{1})),\mathbf{v}_{2} \rangle \\ &= \langle (T^{*} \circ S^{*})(\mathbf{v}_{1}),\mathbf{v}_{2} \rangle. \end{align} Therefore, $ (S \circ T)^{*} = T^{*} \circ S^{*} $. $ \quad \spadesuit $
To prove the identity in question, we induct on $ n \in \mathbb{N} $.
For each $ n \in \mathbb{N} $, let $ P(n) $ denote the statement $$ (T^{n})^{*} = (T^{*})^{n}. $$ The truth of $ P(1) $ is tautological. Next, suppose that $ P(k) $ is true for some $ k \in \mathbb{N} $. Then \begin{align} (T^{k + 1})^{*} &= (T^{k} \circ T)^{*} \\ &= T^{*} \circ (T^{k})^{*} \quad (\text{By the theorem.}) \\ &= T^{*} \circ (T^{*})^{k} \quad (\text{By the induction hypothesis.}) \\ &= (T^{*})^{k + 1}. \end{align} Hence, $ P(k + 1) $ is true. By mathematical induction, $ P(n) $ is true for all $ n \in \mathbb{N} $.
By linearity, we conclude that $ [f(T)]^{*} = f(T^{*}) $ for any polynomial $ f \in \mathbb{R}[X] $.
Addendum
This part was prepared in response to the OP’s question in his comment below. Let $ p $ and $ q $ be the minimal polynomials of $ T $ and $ T^{*} $ respectively. By the solution above, we have $$ p(T^{*}) = [p(T)]^{*} = \mathbf{0}^{*} = \mathbf{0}. $$ Hence, $ q $ divides $ p $. Next, we have $$ q(T) = q((T^{*})^{*}) = [q(T^{*})]^{*} = \mathbf{0}^{*} = \mathbf{0}. $$ Hence, $ p $ divides $ q $ also. Therefore, as both $ p $ and $ q $ are monic polynomials, we conclude that $ p = q $.