1. ## Canonical form of a quadratic form

Write the quadratic form $x_1x_2 +x_2x_3 +x_3x_1$ in canonical form over $\mathbb{C}$.

I started off by finding a symmetric matrix A for the quadratic form $Q(\vec{x},\vec{x})=\vec{x}^TA\vec{x}$, which is
$\begin{pmatrix} 0 & \frac{1}{2} & \frac{1}{2}\\ \frac{1}{2} & 0 & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & 0\end{pmatrix}$ with eigenvalues of $\lambda=-.5, -.5, 1$ and eigenvectors $(-1,1,0), (-1,0,1), (1, 1, 1)$ respectively.

I don't see how it's now possible to rewrite the expression as $Q(\vec{x},\vec{x})=\lambda_1x_1^2+\lambda_2x_2^2+\ lambda_3x_3^2$. And where do complex numbers have anything to do with the problem?

2. Originally Posted by davesface
... with eigenvalues of $\lambda=-.5, -.5, 1$ and eigenvectors $(-1,1,0), (-1,0,1), (1, 1, 1)$ respectively.
Right.

I don't see how it's now possible to rewrite the expression as $Q(\vec{x},\vec{x})=\lambda_1x_1^2+\lambda_2x_2^2+\ lambda_3x_3^2$.
You need an orthormal basis of eigenvectors $B=\{e_1,e_2,e_3\}$ (with the usual inner product). If $P=[e_1\;e_2\;e_3]$ then,

$P^tAP=\textrm{diag}(-1/2,-1/2,1)$ .

And where do complex numbers have anything to do with the problem?
Absolutely nothing. Any cuadratic form on $\mathbb{R}$ is diagonalizable on $\mathbb{R}$ .

Regards.

Fernando Revilla

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