Projectivity to send a conic in its canonical form

Question

I have the projective conic $Q=\{x_0^2+x_1x_2+x_3x_4+x_5x_6=0\} \subset \mathbb{P}^6$, which is of maximal rank. Hence $Q$ is projectively equivalent to its canonical form that is $Q'=\{ x_0^2+\ldots+x_6^2=0\}$. I want to find a projectivity $T: \mathbb{P}^6 \to \mathbb{P}^6$ such that $T(Q)=Q'$.

I know that $Q$ is the zero-locus of a polynomial $q(x)=x^tMx$, while $Q'$ is the zero-locus of $q'(x)=x^tM'x$; moreover $M=B^tMB$ for a certain matrix $B$ (that is the one given by $T=B^{-1}$). How can I find $B$ in my case?

UPDATE: My working field is $\mathbb{C}$.

Answer 1

To start, you'll have to express $q(x)=x_0^2 + x_1 x_2 + x_3 x_4 + x_5 x_6$ as $q(x) = x^t M x$ for a real symmetric matrix $M$.

Since $M$ is real symmetric, you can diagonalize $M$ in an orthonormal basis of eigenvectors in $\mathbb R^7$. Those will form the columns of an orthogonal matrix $B$ such that $M'=B^t M B$ is a diagonal matrix.

Finally $q'(x) = q(Bx) = x^t M' x$ will be in canonical form.

In short: Make sure your $M$ is symmetric, find a basis of eigenvectors of $M$, orthonormalize for each eigenspace.

Answer 2

There is an algorithm over the rational that takes us between a symmetric matrix and a diagonal matrix. You seem to be requiring a sum of squares, which you can accomplish by more transformations, this time diagonal and involving complex numbers

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