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Please offer a solution to the following problem. It was offered in class by my professor as an additional exercise to try on one's own.

Let $V$ be the inner product space, and assume that $\alpha \in End(V)$. Suppose that $A$ is the representation matrix of $\alpha$ with respect to an orthonormal basis {$v_1,...,v_n$}.

(i) Prove that $\alpha$ is self-adjoint if and only if $A = A^*$.

(ii) Prove that if $B$ is the representation matrix of $\alpha$ with respect to another orthonormal basis {$w_1,...,w_n$}, then $B=U^*AU$ for some matrix $U$ such that $U^*U=I$.

Thank you for your assistance.

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You should really do something about that $0\%$ accept rate. Since the exercise was offered for you to try on your own, it would be nice for you to tell us what you've tried. –  EuYu Nov 19 '12 at 0:51
    
The formula $\alpha v = \alpha (\sum_k x_k v_k) = \sum_k x_k \alpha v_k = \sum_k x_k \sum_i [A]_{ik} v_i = \sum_k \sum_i [A]_{ik} x_k v_i = \sum_i [Ax]_i v_i$ may be helpful. –  copper.hat Nov 19 '12 at 0:56

1 Answer 1

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For part 1, use Proposition 16.16 from Golan.

Let $V$ and $W$ be finitely-generated inner product spaces, having ONB $B=\{v_1, \ldots, v_n\}$ and $D=\{w_1, \ldots, w_n\}$, respectively. Let $\alpha: V \to W$ be a linear transformation. Then $\Phi_{BD}(\alpha)$ is the matrix $A=[a_{ij}]$, where $a_{ji}=\langle \alpha(v_i),w_j\rangle$ and $\Phi_{DB}(\alpha^*)=A^H$.

The proof of this proposition follows: For all $1 \leq i \leq n$, let $\alpha(v_i)= \displaystyle\sum_{h=1}^k a_{hj} w_h$. Then for all $1 \leq j \leq k$, we have $\langle \alpha(v_i), w_j \rangle= \langle \sum_{h=1}^k a_{hj} w_h, w_j \rangle=a_{ji}$ and also $\langle \alpha^*(w_j),v_i \rangle= \overline{\langle v_i, \alpha^*(w_j) \rangle}= \overline{\langle \alpha(v_i), w_j \rangle}=\overline{a_{ji}}$ as needed.

The proof for this proposition is similar to the proof of Part 1.

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