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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
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1 Answer

up vote 3 down vote accepted

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 $.

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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
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