Properties of Adjoints 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.
 A: We need the following result.

Theorem For any two linear transformations $ S $ and $ T $ on $ V $, we have $ (S \circ T)^{*} = T^{*} \circ S^{*} $.

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 $.
