Determining the adjoint of $S(\sum_{i=1}^n \alpha_i e_i) = \alpha_2 e_1 - \alpha_1 e_2$.

Question

If $\{e_1,e_2,\dotsc,e_n\}$ is an orthonormal basis of a vector space $V$ and $S \in L(V)$ is defined by $$ S\Big( \alpha_1 e_1 + \alpha_2 e_2 + \dots + \alpha_n e_n \Big) = \alpha_2 e_1 - \alpha_1 e_2, $$ then find the adjoint of $S$, i.e. $S^*$.

Answer 1

Result: Suppose $\{e_1,e_2,\dots,e_n\}$ is an orthonormal basis of $V$ and $v \in V$ .Then $$v=\langle v,e_1 \rangle e_1+\langle v,e_2 \rangle e_2+\dots+\langle v,e_n \rangle e_n$$

Now to find the matrix of $S$ with respect to the basis $\{e_1,e_2,\dots,e_n\}$.

Start with

$S(e_1)=S\Big(\langle e_1,e_1 \rangle e_1+\langle e_1,e_2 \rangle e_2+...+\langle e_1,e_n \rangle e_n\Big)$

$\hspace{1cm}$ $=S(e_1)$

$\hspace{1cm}$ $=S(1.e_1+0.e_2+...+0.e_n)=0.e_1-1.e_2$

Therefore, the first column of $[S]_\Bbb{B}$ is $$\begin{pmatrix} 0 \\-1\\0\\0\\\vdots \\0\end{pmatrix}$$ and so the first row of $[S^*]=[(\overline {S})^T]$ is $$\begin{pmatrix} 0 &-1&0&0&\dots &0\end{pmatrix}$$

Similarly to find $S(e_i)$ where $i=2,3,...,n$ and express it as a linear combination of $e_i's$ and make the coefficients as the columns to find $[S]$ and then find $[S^*]$