Diagonal matrix/Orthogonal basis/Adjoint endomorphism/Example

Suppose that for the linear mapping

${\displaystyle \varphi \colon {\mathbb {K} }^{n}\longrightarrow {\mathbb {K} }^{n},}$

there exists an orthonormal basis (with respect to the standard inner product) ${\displaystyle {}u_{1},\ldots ,u_{n}}$ consisting of eigenvectors; that is, the describing matrix with respect to this basis is in diagonal form

${\displaystyle {\begin{pmatrix}\lambda _{1}&0&\cdots &\cdots &0\\0&\lambda _{2}&0&\cdots &0\\\vdots &\ddots &\ddots &\ddots &\vdots \\0&\cdots &0&\lambda _{n-1}&0\\0&\cdots &\cdots &0&\lambda _{n}\end{pmatrix}}.}$

Then the adjoint endomorphism is described by the complex-conjugated matrix

${\displaystyle {}\psi ={\begin{pmatrix}{\overline {\lambda }}_{1}&0&\cdots &\cdots &0\\0&{\overline {\lambda }}_{2}&0&\cdots &0\\\vdots &\ddots &\ddots &\ddots &\vdots \\0&\cdots &0&{\overline {\lambda }}_{n-1}&0\\0&\cdots &\cdots &0&{\overline {\lambda }}_{n}\end{pmatrix}}\,.}$

Indeed, on one hand we have

${\displaystyle {}\left\langle \varphi (u_{i}),u_{j}\right\rangle =\left\langle \lambda _{i}u_{i},u_{j}\right\rangle =\lambda _{i}\left\langle u_{i},u_{j}\right\rangle \,,}$

and on the other hand we have

${\displaystyle {}\left\langle u_{i},\psi (u_{j})\right\rangle =\left\langle u_{i},{\overline {\lambda _{j}}}u_{j}\right\rangle =\lambda _{j}\left\langle u_{i},u_{j}\right\rangle \,.}$

For ${\displaystyle {}i\neq j}$, we have ${\displaystyle {}0}$ on both sides, and for ${\displaystyle {}i=j}$, we have ${\displaystyle {}\lambda _{i}}$ on both sides.