# Real and Complex Canonical Form

I have to convert some quadratic forms into real and complex canonical forms.

One of these forms is as such:

$$q_1\begin{pmatrix} x \\ y \\ z \end{pmatrix} = x^2 +3y^2 +z^2 +2xy−2xz−2yz$$

From the matrix $\begin{Bmatrix} 1 & 1 & -1 \\1 & 3 & -1 \\-1 & -1 & 1 \end{Bmatrix}$

I got $\begin{Bmatrix} 1 & 0 & 0 \\0 & 2 & 0 \\0 & 0 & -2 \end{Bmatrix}$ using column and row operations but I am not sure that this is correct? Is it the real form? If so, how do I get the complex form?

I also have to compare this equation to others and say which are equivalent over $\mathbb{R}$ and $\mathbb{C}$. How would I go about this?

Edit: I worked out the both the real and complex form for this one to have $1,1,0$ on the diagonal. I have two other quadratic forms $q_2$ and $q_3$ for which I worked out the diagonals for the real form as $1,1,-1$ and $-1,-1,-1$ respectively. Can I just change the $-1$s to $+1$s to make them complex?

I am also asked which of $q_1,q_2,q_3$ are equivalent over $\mathbb{R}$ and $\mathbb{C}$. If the real form is unique then how can they be equivalent? Am I right in thinking that none of the real forms are equivalent but the complex forms of $q_2$ and $q_3$ are?

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The real canonical form should have diagonal entries purely $\{\pm 1\}$, and the complex canonical form should have diagonal entries all $+1$. The goal should not be just to find the canonical form (since the complex canonical form of any quadratic form over $\mathbb{C}$ is the same), but to find the transformation / change of variables that carries out the reduction. – Willie Wong Nov 28 '12 at 13:25
I did it again and I have 1,1,0 on the diagonal. So the complex canonical form will just be the same? So do I compare transformation matrices to see if they are the same as other forms? – user51014 Nov 28 '12 at 13:47
Yes, if the real canonical form has no negative entries, the real and complex ones are the same and can be arrived at via the same transformation. You should think that working over $\mathbb{C}$ is more flexible, so any operation that is "legal in $\mathbb{R}$" will be itself a "legal operation in $\mathbb{C}$". (A real valued matrix is automatically a complex valued matrix, but not vice versa.) – Willie Wong Nov 28 '12 at 13:50
thank you. I still am not really sure how to check if they are equivalent though. I have worked out another form that has 1,0,-1 on diagonal for real and 1,0,1 for complex. Is it just enough to look at them and say that neither are the same as q1? thanks for your help. – user51014 Nov 28 '12 at 13:59
Hm. there should be only one real canonical form. Please edit the question statement to show your work. If you are finding three different answers you are doing something wrong. (Usually one can find the canonical forms by "completing the square".) – Willie Wong Nov 28 '12 at 14:11