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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.

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1 Answer 1

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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.

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