In linear algebra, a quadratic form is defined as $Q(x)=x^TAx$ for some (non-singular) matrix $A$ and any $x\in V$, where $V$ is a vector space. Actually, quadratic form can be any one satisfying $Q(\lambda x)=|\lambda|^2Q(x)$ for any scalar $\lambda$. Is there a prove that the representation form of quadratic form in linear algebra is unique? That is for any quadratic form, we can associate it with a matrix $A$?
I have a very crude non-rigorous proof on this. Expand quadratic form as polynomial series and prove that only $A_{ij}x^ix^j$ term meets the condition.
Best Answer
Your version of quadratic form is not enough. One normally includes the parallelogram law: $$ Q(\lambda x)=|\lambda|^{2}Q(x),\;\;\; Q(x+y)+Q(x-y)=2Q(x)+2Q(y), $$ where $x$, $y$ are vectors in a real or complex vector space $X$. There are subtle issues concerning quadratic forms on general linear spaces that are related to the axiom of choice; I won't bother you with those here. However, the subtle issues disappear if $Q(x) \ge 0$ for all $x$, for example, or if $|Q(x)| \le C|x|^{2}$ for some constant $C$, where $|\cdot|$ has the properties of a norm except for strict positivity.
Matrix Representation: Because of your assertion about matrices and a lack of complex conjugate notation, I think you intend to work in $V=\mathbb{R}^{n}$. The first thing to know about real quadratic forms is that, even on $\mathbb{R}^{2}$, a representation $Q(x)=x^{T}Ax$ is not unique because of matrices $B$ such that $B^{T}=-B$. For example, $$ B= \left[\begin{matrix}0 & -1 \\ 1 & 0\end{matrix}\right] \implies \left[\begin{matrix}x & y\end{matrix}\right]B\left[\begin{matrix}x \\ y\end{matrix}\right] =0\mbox{ for all }x, y. $$ You can directly verify using transpose that $B^{T}=-B$ implies $x^{T}Bx=0$ for all $x$. If there is a representation $Q(x)=x^{T}Ax$, then $Q(x)=x^{T}A^{T}x$ holds, as does $Q(x)=x^{T}(\frac{1}{2}(A^{T}+A))x$. In other words, you can find a symmetric $A'=\frac{1}{2}(A^{T}+A)$ such that $Q(x)=x^{T}A'x$; a symmetric representing matrix $A$ is unique because the $0$ matrix is the only antisymmetric and symmetric matrix. Quadratic forms are familiar from Calculus. For $\mathbb{R}^{3}$, a general quadratic form has the symmetric representation $$ a_{11}x^{2}+2a_{12}xy+2a_{13}xz+a_{22}y^{2}+2a_{23}yz+a_{33}z^{2} = \left[\begin{matrix} x & y & z\end{matrix}\right] \left[\begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{12} & a_{22} & a_{23} \\ a_{13} & a_{23} & a_{33}\end{matrix}\right]\left[\begin{matrix}x \\y\\z \end{matrix}\right] $$ This generalizes to higher dimensions.