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Thread: Canonical form of a quadratic form

  1. #1
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    Canonical form of a quadratic form

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


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

    I don't see how it's now possible to rewrite the expression as $\displaystyle Q(\vec{x},\vec{x})=\lambda_1x_1^2+\lambda_2x_2^2+\ lambda_3x_3^2$. And where do complex numbers have anything to do with the problem?
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by davesface View Post
    ... with eigenvalues of $\displaystyle \lambda=-.5, -.5, 1$ and eigenvectors $\displaystyle (-1,1,0), (-1,0,1), (1, 1, 1)$ respectively.
    Right.

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

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

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

    Regards.

    Fernando Revilla
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