# Symmetric Bilinear Form/ Hermitian Form unique Matrix representation

Given $$\mathbb{K} = \mathbb{R}$$ or $$\mathbb{C}$$ I have $$\varphi:\mathbb{K^n}\times\mathbb{K^n}\rightarrow \mathbb{K}$$ a symmetric Bilinear Form or complex Hermitian Form/Symmetric sesquilinear form (depending on $$\mathbb{K}$$) and I have to prove that there exists a unique matrix $$A = (a_{i,j})\in \mathbb{K^{n\times n}}$$ with $$\varphi(x,y) = x^TA\overline{y}$$

My attempt:

• I'll take $$\{{e_1,....,e_n}\}$$ so the standard Basis which is also an orthonormal Basis for $$\mathbb{K^n}$$ and I write $$\varphi(e_i,e_j) = B_{i,j}$$.
• I define $$\varphi_A(x,y) = \langle{x,Ay}\rangle$$ where $$\langle{.,.}\rangle$$ denotes the standard dot product and I prove it is indeed a Bilinear or complex Hermitian Form in respect to $$\mathbb{K}$$.

Then it follows that $$Aej = \sum_{k=1}^{n} B_{k,j}e_k$$ and so $$\varphi_A(e_i,e_j)=\langle{x,Ae_j}\rangle = \sum_{k=1}^{n}B_{k,j}\langle{e_i,e_k}\rangle = B_{i,j}$$ because $$\langle{e_i,e_k}\rangle \neq 0$$ only when $$k = i$$ and so $$\varphi_A$$ and $$\varphi$$ are generally equal.

• As for uniqueness I suppose there exist $$A_1$$,$$A_2\in \mathbb{K}^{n\times n}$$ with $$\langle{x,A_1y}\rangle =\varphi(x,y) = \langle{x,A_2y}\rangle \forall x,y\in \mathbb{K^n}$$ in which case $$\langle{x,A_1y-A_2y}\rangle = \langle{x,A_1y}\rangle - \langle{x,A_2y}\rangle = \varphi(x,y) - \varphi(x,y) = 0 \forall x,y\in \mathbb{K^n}$$ and for $$x = A_1y-A_2y$$ $$0 =\langle{x,x}\rangle = \langle{A_1y-A_2y,A_1y-A_2y}\rangle$$ = $$||x||^2 = ||A_1y-A_2y||^2 = ||(A_1-A_2)y||^2 \Rightarrow A_1 - A_2 =0 \iff A_1 = A_2$$

Please let me know if there is something I am missing, if this proof isn't formal enough or if there is any easier/more obvious proof that I clearly missed.

By choosing $$x=e_i$$ and $$y=e_j$$, if you plug in $$\phi(x,y)=x^TAy$$, you can see that $$a_{ij}=\phi(e_i,e_j)$$ so the elements of the matrix are uniquely defined by $$\phi$$. I think this is a shorter proof of uniqueness. You have different matrices for different choices of the basis, of course.