Proof: $g\mbox{ is self-adjoint endomorphism}\leftrightarrow \forall i,j \in \{1,2,...,n\}(f(g(b_i),b_j)=f(b_i,g(b_j)))$

Question

let $g \in End_K(E)$, and $f: (E \times E)\to \Bbb{R}^1$ a symmetric bilinear form positive definite, and $(b_1,b_2,...,b_n )$ a basis, then $$f \mbox{ is self-adjoint endomorphism } \leftrightarrow \forall i,j \in \{1,2,...,n\}(f(g(b_i),b_j)=f(b_i,g(b_j)))$$ I must proof: $$1)g \mbox{ is self-adjoint endomorphism } \to \forall i,j \in \{1,2,...,n\}(f(g(b_i),b_j)=f(b_i,g(b_j)))$$ $$\wedge $$$$2)\forall i,j \in \{1,2,...,n\}(f(g(b_i),b_j)=f(b_i,g(b_j))) \to g \mbox{ is self-adjoint endomorphism }$$ $1)$ $b_1,b_2,...,b_n \in E$ therefore $\forall i \in \{1,2,...,n\}(b_i \in E)$ therefore $\forall i,j \in \{1,2,...,n\}(b_i,b_j \in E )$ and by hypothesis I have "$g \mbox{ is self-adjoint endomorphism} $" then $f(g(b_i),b_j)=f(b_i,g(b_j))$ is true!!

$2)$ if "$g \mbox{ is self-adjoint endomorphism} $" then CLIC, let $v,w \in E$ and by hypothesis I have $$v=\alpha_1b_1+\alpha_2b_2+...+\alpha_nb_n$$ $$\mbox{ and }$$ $$w=\beta_1b_1+\beta_2b_2+...+\beta_nb_n$$ $g$ is endomorphism then $$g(v)=\alpha_1g(b_1)+\alpha_2g(b_2)+...+\alpha_ng(b_n)$$ $$\mbox{ and }$$ $$g(w)=\beta_1g(b_1)+\beta_2g(b_2)+...+\beta_ng(b_n)$$ therefore I must proof: $$f(g(v),w)=f(v,g(w))$$ and $f(g(v),w)=f(\alpha_1g(b_1)+\alpha_2g(b_2)+...+\alpha_ng(b_n),w)$ but $f$ is a bilinear form therefore $$f(g(v),w)=...=\alpha_1f(g(b_1),w)+\alpha_2f(g(b_2),w)+...+\alpha_nf(g(b_n),w)$$

and $f(v,g(w)=f(v,\beta_1g(b_1)+\beta_2g(b_2)+...+\beta_ng(b_n)$ but $f$ is a bilinear form therefore $$f(v,g(w))=...=\beta_1f(v,g(b_1))+\beta_2f(v,g(b_2))+...+\beta_nf(v,g(b_n))$$ but by hypothesis I have $$v=\alpha_1b_1+\alpha_2b_2+...+\alpha_nb_n$$ $$\mbox{ and }$$ $$w=\beta_1b_1+\beta_2b_2+...+\beta_nb_n$$ therefore $$f(g(v),w)=...=\alpha_1f(g(b_1),\beta_1b_1+\beta_2b_2+...+\beta_nb_n)+\alpha_2f(g(b_2),\beta_1b_1+\beta_2b_2+...+\beta_nb_n)+...+\alpha_nf(g(b_n),\beta_1b_1+\beta_2b_2+...+\beta_nb_n)$$ $$\mbox{and}$$ $$f(v,g(w))=...=\beta_1f(\alpha_1b_1+\alpha_2b_2+...+\alpha_nb_n,g(b_1))+\beta_2f(\alpha_1b_1+\alpha_2b_2+...+\alpha_nb_n,g(b_2))+...+\beta_nf(\alpha_1b_1+\alpha_2b_2+...+\alpha_nb_n,g(b_n))$$ but $f$ is a bilinear form therefore $$f(g(v),w)=...=...=\sum_{i=1}^{n}\alpha_1\beta_if(g(b_1),b_i)+\sum_{i=1}^{n}\alpha_2\beta_if(g(b_2),b_i)+...+\sum_{i=1}^{n}\alpha_n\beta_if(g(b_n),b_i)$$ $$\mbox{and}$$$$f(v,g(w))=...=...= \sum_{i=1}^{n}\beta_1\alpha_if(b_1,g(b_i))+\sum_{i=1}^{n}\beta_2\alpha_if(b_2,g(b_i))+...+\sum_{i=1}^{n}\beta_n\alpha_if(b_n, g(b_i))$$ but I don't know to continue.. How can I do? thanks in advance!

Answer 1

• The if condition is straightforward by the definition of the self-adjoint endomorphism.
• For the only if condition: Let $\displaystyle x=\sum_{i=1}^n\alpha_i b_i$ and $\displaystyle y=\sum_{j=1}^n\beta_j b_j$ then we have by bi-linearity of $f$ and linearity of $g$

$$f(g(x),y)=\sum_{i=1}^n\sum_{j=1}^n\alpha_i\beta_jf(g(b_i),b_j)=\sum_{i=1}^n\sum_{j=1}^n\alpha_i\beta_jf(b_i,g(b_j))=f(x,g(y))$$ hence $g$ is self-adjoint.