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I'm given a quadratic form $\Phi:\mathbb{R}^3\longrightarrow\mathbb{R}$, for which we know that:

  1. $(0,1,0)$ and $(0,1,-1)$ are conjugated by the function

  2. $(1,0,-1)$ belongs to the kernel

  3. $\Phi(0,0,1)=1$

  4. The trace is $0$

From here, I know the matrix must be symmetric, so it will have up to six unique numbers, I have four conditions that I can apply, and I will get four equations that relate the numbers of the matrix, but I still need two to full determine it. Applying the above I get that the matrix must be of the form:

$$A=\pmatrix{2c-1 & b & c\\b&-2c&-2c\\c & -2c & 1}$$

How do I determine $b$ and $c$?

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1 Answer 1

up vote 1 down vote accepted

Ok, solved it, the last two equations came from knowing the vector that was in the kernel, so it should be that $f_p[(1,-0,-1),(x,y,z)]=0$, being $f_p$ the polar form of $\Phi$: $f_p=\frac{1}{2}[\Phi(x+y)-\Phi(x)-\Phi(y)]$

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