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$q(x_1, x_2, x_3) = x_2x_3+{x_3}^2-{x_1}^2$ is a quadratic form on $\mathbb{R}^3$.

I need to find basis of $\mathbb{R}^3$ when $q$ is: $q = \delta_1{y_1}^2+\delta_2{y_2}^2 + \delta_3{y_3}^2$ when $|\delta_i| = 1$ for $i = 1, 2, 3$.

how I start to solve it?

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1 Answer 1

Let $\mathbf{x}=(x_1,x_2,x_3)^T$. You may first find a symmetric matrix $A$ such that $q(x_1,x_2,x_3)=\mathbf{x}^TA\mathbf{x}$. Since $A$ is symmetric, it is orthogonally diagonalizable. So, find three unit eigenvectors $u,v,w$ such that correspond to the three eigenvalues of $A$. Then $Q=[u,v,w]$ will automatically be an orthogonal matrix and $A=QDQ^T$, where $D$ is the diagonal matrix whose diagonal entries are the eigenvalues of $A$. Now, let $\mathbf{z}=(z_1,z_2,z_3)^T = Q^T\mathbf{x}$. Express $q(x_1,x_2,x_3)$ in terms of $z_1,z_2,z_3$. What do you see? If you have carried out the above steps correctly, by now you shall be very close to the solution.

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