Vector space/Finite-dimensional/Inner product/Adjoint endomorphism/Existence/Fact/Proof

Proof

Let

$\varphi \colon V\longrightarrow V$

be given, and let ${}w\in V$ be fixed. Then, the mapping

$V\longrightarrow {\mathbb {K} },v\longmapsto \left\langle \varphi (v),w\right\rangle ,$

is a linear form on ${}V$. Therefore, there exists (due to fact in the real case; for the complex case see exercise) a right gradient ${}r=:{\hat {\varphi }}(w)$ in ${}V$ (uniquely determined by ${}\varphi$ and ${}w$) fulfilling

${}\left\langle v,{\hat {\varphi }}(w)\right\rangle =\left\langle \varphi (v),w\right\rangle \,.$

We have to show that the assignment

$w\longmapsto {\hat {\varphi }}(w)$

is linear. We have

${}{\begin{aligned}\left\langle v,{\hat {\varphi }}(w_{1}+w_{2})\right\rangle &=\left\langle \varphi (v),w_{1}+w_{2}\right\rangle \\&=\left\langle \varphi (v),w_{1}\right\rangle +\left\langle \varphi (v),w_{2}\right\rangle \\&=\left\langle v,{\hat {\varphi }}(w_{1})\right\rangle +\left\langle v,{\hat {\varphi }}(w_{2})\right\rangle \\&=\left\langle v,{\hat {\varphi }}(w_{1})+{\hat {\varphi }}(w_{2})\right\rangle .\end{aligned}}$

As this holds for all ${}v\in V$, we have

${}{\hat {\varphi }}(w_{1}+w_{2})={\hat {\varphi }}(w_{1})+{\hat {\varphi }}(w_{2})\,.$

Moreover,

${}{\begin{aligned}\left\langle v,{\hat {\varphi }}(sw)\right\rangle &=\left\langle \varphi (v),sw\right\rangle \\&={\overline {s}}\left\langle \varphi (v),w\right\rangle \\&={\overline {s}}\left\langle v,{\hat {\varphi }}(w)\right\rangle \\&=\left\langle v,s{\hat {\varphi }}(w)\right\rangle .\end{aligned}}$

As this holds for all ${}v\in V$, we get

${}{\hat {\varphi }}(sw)=s{\hat {\varphi }}(w)\,.$

To fact