Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 0 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.