# example of quadratic form with nonunique +ve and -ve definite subspaces

What is an example of a quadratic form $Q$ on real vector space $V$ such that the maximal positive definite subspace and the maximal negative definite subspace are not uniquely determined?

By Sylvester's law of inertia, the dimensions of these subspaces are uniquely determined. I think (?) this means they both have to have dimension at least 3.

Should we do this by completing the square in two different ways (e.g. $x_1^2 + x_2^2 + x_3^2 - x_4^2 - x_5^2 - x_6^2 = (x_1 + x_2 + x_3)^2 - (x_1 + x_2)^2 + ... - ...$ ), or by considering bases and subspaces explicitly?

Many thanks for any help with this!

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"+ve, -ve" means "postive, negative"... though I wish this practice were extinct. –  The Chaz 2.0 Apr 10 '12 at 17:25

## 1 Answer

As long as $Q$ is not positive or negative definite, the answer is: any $Q$ is such an example. There are many ways to diagonalize a symmetric matrix by congruence...

For instance, suppose that in the basis $B=\{v_i\}$ the expression of your $Q$ is $x_1^2+\cdots+x_j^2-x_{j+1}^2-\cdots-x_m^2.$ Replace $v_j$ by $2v_j+v_{j+1}$ and $v_{j+1}$ by $v_j+2v_{j+1}.$ If I am not wrong, the expression of $Q$ in this new basis is $y_1^2+\cdots+y_{j-1}^2+3y_j^2-3y_{j+1}^2-y_{j+2}^2-\cdots-y_m^2.$ There you have two different maximal subspaces. You can work out easily another change of basis for the semidefinite case, to the same effect.

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Many thanks for this. –  Harry Macpherson Apr 10 '12 at 18:46