Endomorphism/Bilinear form/Introduction/Section

Let ${}V$ be a ${}{\mathbb {K} }$-vector space, endowed with an inner product. An endomorphism

$\varphi \colon V\longrightarrow V$

induces, with the help of the inner product, a form ${}\Psi _{\varphi }$ defined by

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

The following properties hold for this form.

Lemma

Let ${}V$ be a ${}{\mathbb {K} }$-vector space, endowed with an

inner product. Then the following statements hold.

1. The assignment

   $\varphi \longmapsto \Psi _{\varphi }$

   assigns to an endomorphism a sesquilinear form. Hence,

   $\operatorname {End} _{}{\left(V\right)}\longrightarrow \operatorname {Sesq} _{}{\left(V\right)},\varphi \longmapsto \Psi _{\varphi }.$

2. This assignment is linear; it is bijective if ${}V$ has finite dimension.
3. Let ${}V$ be finite-dimensional. The endomorphism ${}\varphi$ is bijective if and only if ${}\Psi _{\varphi }$ is not degenerate.
4. Let ${}V$ be finite-dimensional. The endomorphism ${}\varphi$ is self-adjoint if and only if ${}\Psi _{\varphi }$ is Hermitian.

Proof

1. We have

   ${}{\begin{aligned}\psi _{\varphi }(av_{1}+bv_{2},w)&=\left\langle \varphi (av_{1}+bv_{2}),w\right\rangle \\&=\left\langle a\varphi (v_{1})+b\varphi (v_{2}),w\right\rangle \\&=a\left\langle \varphi (v_{1}),w\right\rangle +b\left\langle \varphi (v_{2}),w\right\rangle \\&=a\psi _{\varphi }(v_{1},w)+b\psi _{\varphi }(v_{2},w)\end{aligned}}$

   and

   ${}{\begin{aligned}\psi _{\varphi }(v,aw_{1}+bw_{2})&=\left\langle \varphi (v),aw_{1}+bw_{2}\right\rangle \\&={\overline {a}}\left\langle \varphi (v),w_{1}\right\rangle +{\overline {b}}\left\langle \varphi (v),w_{2}\right\rangle \\&={\overline {a}}\psi _{\varphi }(v,w_{1})+{\overline {b}}\psi _{\varphi }(v,w_{2}),\end{aligned}}$

   that is, the assignment is linear in the first component, and antilinear in the second component. Therefore, ${}\Psi _{\varphi }$ is a sesquilinear form.
2. The linearity follows from the linearity of the inner product in the first component. In the finite-dimensional case, we have on the left-hand side and on the right-hand side vector spaces of the dimension ${}{\left(\dim _{K}{\left(V\right)}\right)}^{2}$; therefore, it is enough to show injectivity. If ${}\Psi _{\varphi }=0$ is the zero form, then ${}\left\langle \varphi (v),w\right\rangle =0$ for all ${}v,w$. In particular, ${}\left\langle \varphi (v),\varphi (v)\right\rangle =0$, which implies ${}\varphi (v)=0$.
3. If ${}\varphi$ is not bijective, then let ${}v\in \operatorname {kern} \varphi$, ${}v\neq 0$. Then, ${}\Psi _{\varphi }(v,-)$ is the zero mapping in the second component, and the form is degenerate. To prove the converse, suppose that ${}\Psi _{\varphi }(-,-)$ is degenerate. Then there exists a vector ${}v\in V$, ${}v\neq 0$, such that ${}\left\langle \varphi (v),-\right\rangle$ is the zero-mapping. Since an inner product is nondegenerate, this implies ${}\varphi (v)=0$, and ${}\varphi$ is not bijective.
4. In the self-adjoint case, we have

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

   The converse follows from

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

$\Box$