Vector space/Inner product/Endomorphism/Sesquilinear form/Fact/Proof

Proof

1. We have

   $\begin{array}{cc}{\psi }_{\phi }(a{v}_1+b{v}_2,w)& =⟨\phi (a{v}_1+b{v}_2),w⟩\\ & =⟨a\phi (v_1)+b\phi (v_2),w⟩\\ & =a⟨\phi (v_1),w⟩+b⟨\phi (v_2),w⟩\\ & =a{\psi }_{\phi }(v_1,w)+b{\psi }_{\phi }(v_2,w)\end{array}$

   and

   $\begin{array}{cc}{\psi }_{\phi }(v,a{w}_1+b{w}_2)& =⟨\phi (v),a{w}_1+b{w}_2⟩\\ & =\bar{a}⟨\phi (v),w_1⟩+\bar{b}⟨\phi (v),w_2⟩\\ & =\bar{a}{\psi }_{\phi }(v,w_1)+\bar{b}{\psi }_{\phi }(v,w_2),\end{array}$

   that is, the assignment is linear in the first component, and antilinear in the second component. Therefore, ${\Psi }_{\phi }$ 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 ${\Psi }_{\phi }=0$ is the zero form, then $⟨\phi (v),w⟩=0$ for all $v,w$. In particular, $⟨\phi (v),\phi (v)⟩=0$, which implies $\phi (v)=0$.
3. If $\phi$ is not bijective, then let $v\in \mathrm{kern}\phi$, $v\ne 0$. Then, ${\Psi }_{\phi }(v,-)$ is the zero mapping in the second component, and the form is degenerate. To prove the converse, suppose that ${\Psi }_{\phi }(-,-)$ is degenerate. Then there exists a vector $v\in V$, $v\ne 0$, such that $⟨\phi (v),-⟩$ is the zero-mapping. Since an inner product is nondegenerate, this implies $\phi (v)=0$, and $\phi$ is not bijective.
4. In the self-adjoint case, we have

   ${\Psi }_{\phi }(v,w)=⟨\phi (v),w⟩=⟨v,\phi (w)⟩=\overline{⟨\phi (w),v⟩}=\overline{{\Psi }_{\phi }(w,v)}.$

   The converse follows from

   $⟨\phi (v),w⟩={\Psi }_{\phi }(v,w)=\overline{{\Psi }_{\phi }(w,v)}=\overline{⟨\phi (w),v⟩}=⟨v,\phi (w)⟩.$

To fact