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$f : \mathbb R^3 \to \mathbb R, f(x_1, x_2, x_3) = 3x_1^2 - x_2^2 - 2x_3^2 - 4x_1x_2 - 2x_1x_3 + 6x_2x_3$

I am trying to find $f$'s canonical form using Jacobi's method and I don't know how to interpret the results.

Let $A$ be the matrix of $f$ in the canonical base of $\mathbb R^3$. $A$ will be $$ A = \begin{pmatrix} 3 & -2 & -1 \\ -2 & -1 & 3 \\ -1 & 3 & -2 \end{pmatrix} $$

The Jacobi method says that the canonical form of $f$ is : $f(x) = \Delta_1x_1 + \frac{\Delta_1}{\Delta_2}x_2 + \frac{\Delta_2}{\Delta_3}x_3$

where

$$ \Delta_1 = \begin{vmatrix} 3 \end{vmatrix} = 3 $$

$$ \Delta_2 = \begin{vmatrix} 3 && -2 \\ -2 && -1 \end{vmatrix} = 1 $$

$$ \Delta_3 = \begin{vmatrix} 3 & -2 & -1 \\ -2 & -1 & 3 \\ -1 & 3 & -2 \end{vmatrix} = 0 $$

What does $\Delta_3 = 0$ mean ? Clearely I cannot get the canonical form this way, but I can get it any other way ?

I also tried another method, which is finding $\lambda_1, \lambda_2, \lambda_3$ by solving $\det(A-\lambda I_3) = 0$ and the canonical form is supposed to be $f(x) = \lambda_1x_1^2 + \lambda_2x_2^2 + \lambda_3x_3^2$. However by solving the equation I get two complex solution.

Bottom line, can I get the canonical form of $f$ in some way?

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The Jordan canonical form is $\begin{pmatrix} 0 & 0 & 0 \\ 0& \sqrt{21}&0\\ 0 &0 &-\sqrt{21} \\ \end{pmatrix}$. For this particular matrix, since the characteristic equation has n=3 distinct roots, we know that this must be the Jordan canonical form –  Calvin Lin Dec 30 '12 at 17:34
    
More detail is necessary. I am not finding a consistent meaning for Jacobi form. Where did you find this? Note encyclopediaofmath.org/index.php/Jacobi_method –  Will Jagy Dec 30 '12 at 19:46
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