3
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ You have to combine the $bxy$ and $cxy$ terms... $\endgroup$ – Ted Jun 15 '13 at 8:24
  • 6
    $\begingroup$ 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 $\begin{pmatrix} a & \frac{b+c}{2} \\ \frac{b+c}{2} & d \end{pmatrix}$. $\endgroup$ – Michael Wijaya Jun 15 '13 at 8:27
  • $\begingroup$ @MichaelWijaya Thank you, this answers my question. Now I understand that symmetry is only required when division by 2 is not possible. $\endgroup$ – newb Jun 15 '13 at 8:30
1
$\begingroup$

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 $(b+c)/2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.