# Matrix of quadratic form has to be symmetric?

On Wikipedia it is stated that any $n\times n$ real symmetric matrix A determines a quadratic form. But isn't $ax^2 + bxy + cxy + dy^2$, the quadratic form given by $v^T A v$ with $A=\begin{bmatrix}a & b \\ c & d\end{bmatrix}$ and $v=\begin{bmatrix} x \\ y\end{bmatrix}$, a quadratic form even when $c \neq b$?

Why does the Wikipedia article state that the matrix has to be symmetric?

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You have to combine the $bxy$ and $cxy$ terms... – Ted Jun 15 '13 at 8:24
The section you linked to is about forms over $\mathbb{R}$, which has characteristic not equal to 2. Over such fields, we do not get any new quadratic form by considering non-symmetric matrices because we can divide by 2. So $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ gives the same quadratic form as $\begin{pmatrix} a & \frac{b+c}{2} \\ \frac{b+c}{2} & d \end{pmatrix}$. – Michael Wijaya Jun 15 '13 at 8:27
@MichaelWijaya Thank you, this answers my question. Now I understand that symmetry is only required when division by 2 is not possible. – newb Jun 15 '13 at 8:30

## 1 Answer

It simply means that the matrix $A$ can be made symmetric without loss of generality. Simply define the off-diagonal elements of $A$ in your example to be $(b+c)/2$.

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