2
$\begingroup$

For a quadratic form $q(\mathbf{v})$, when you change the basis do you always change the quadratic form? Can you have the same quadratic form with respect to different basis? Or is the quadratic form unique to the basis.

Also, if you're given a quadratic form say $q(\mathbf{v}) = 3x^2 + y^2 - 2z^2 + 4xy - 2xz$, $\:$ and you can clearly deduce the matrix from this $$\begin{pmatrix}3&2&-1\\2&1&0\\-1&0&-2\end{pmatrix}$$ what is the basis for this quadratic form? Is it unique? Can you deduce it from this? Perhaps with a different basis is there a different 'method' of deducing the matrix?

Thanks!

$\endgroup$
  • $\begingroup$ Could you define "basis for a quadratic form"? $\endgroup$ – Calle Mar 31 '12 at 17:11
1
$\begingroup$

The basis for your quadratic form is the natural basis, i.e., $(1, 0,\cdots, 0), \cdots, (0, \cdots, 0, 1)$. The matrix of quadratic form depends on the basis. Canonical form of a symmetric matrix is a diagonal matrix. You may compare these two forms...

$\endgroup$
0
$\begingroup$

Your question is not too clear, but in any case a change of basis does not necessarily change the matrix describing the quadratic form, and so conversely changing the basis while keeping the same matrix does not always change the form. An extreme case is the null matrix which always defines the null form, no matter what the basis is. But more interestingly, if you take the standard quadratic form, $x^2+y^2+z^2$, it will have the same (identity) matrix on another than the standard basis if and only if that basis is orthonormal (for the standard scalar product). And there are many orthonormal bases other than the standard basis.

$\endgroup$
0
$\begingroup$

This link may help you. See proposition 2.3. I think that implies that matrix of quad form changes on changing basis. That is there are many other basis too (other than that of std basis).

$\endgroup$

Your Answer

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