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.

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Does the $n=1$ case imply $T = T^*$? How can this be true for any arbitrary linear transformation? – Muphrid Jan 11 '13 at 23:14
@Muphrid: no, it implies $T^{\ast} = T^{\ast}$. – Qiaochu Yuan Jan 11 '13 at 23:57
@Miami: if $V$ is a complex inner product space, $f$ needs to have real coefficients. – Qiaochu Yuan Jan 11 '13 at 23:58
@MiamiMath: Are you assuming $V$ to be a finite-dimensional inner-product space over $\mathbb{R}$? – Haskell Curry Jan 12 '13 at 3:20

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 $f \in \mathbb{R}[X]$.

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 $p = q$.
Thanks, this is a great help. I probably should have specified that we're considering $V$ as a real inner product space. Very thorough solution. – Mathmo Jan 12 '13 at 0:17
Can we show that the minimum polynomials of $T$ and $T^{*}$ are the same? – Mathmo Jan 12 '13 at 0:23