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The question goes as follows:

Let $V = \mathbb R^{2n}$ be a vector space, and $q:V\to \mathbb R$ a real quadratic form. Let $\xi \in T_2^{sym}(V)$ be a bilinear functional for which $q(v) = \xi (v,v)$. Let us assume that $\xi$ is nondegenerate.

Prove that $q$ has a signature $0$ iff there exists a basis $[e]$ such that $$q(v) =\sum_{i=0}^n x_iy_i$$ where $[v]^{[e]} = (x_1,y_1,...x_n,y_n)$.

I couldn't figure how to prove any of the directions of the "iff". This were my thoughts:

Since $q(v) = \xi (v,v)$, we get that $\xi$ is the polarization of $q$, and since $\xi$ is nondegenerate we get that the rank of $q$ must be $2n$, or equaly $n_{-} + n_{+} = 2n$.

Now, for the first direction we assume that the signature is $0$, therefore $n_- = n_+ = n$, but I don't know how to continue from there.

For the second direction, we get that in the basis $[e]$ we have the matrix representing $q$ as having $\frac{1}{2}$ on every even entry above and below the main diagonal (and all the rest are $0$s), but I don't know how to continue from here as well.

Would appreciate your help on this question, preferably hints to the direction and comments on my approach.

Thanks!

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There is less to this than may appear. All that is going on is $$ \left( \begin{array}{rr} 1 & 1 \\ - \frac{1}{2} & \frac{1}{2} \end{array} \right) \left( \begin{array}{rr} 0 & 1 \\ 1 & 0 \end{array} \right) \left( \begin{array}{rr} 1 & - \frac{1}{2} \\ 1 & \frac{1}{2} \end{array} \right) = \left( \begin{array}{rr} 2 & 0 \\ 0 & - \frac{1}{2} \end{array} \right) $$

If you really want diagonal entries $1,-1$ you can then multiply on left and right by $$ \left( \begin{array}{rr} \frac{1}{\sqrt 2} & 0 \\ & \sqrt 2 \end{array} \right), $$ this matrix being its own transpose.

Other direction, you can construct a matrix $P$ of nonzero determinant such that $$ P^T C P = F, $$ where we are given $$ C = \left( \begin{array}{rr} a & 0 \\ 0 & - b \end{array} \right) $$ with real $a,b > 0,$ while $$ F = \left( \begin{array}{rr} 0 & 1 \\ 1 & 0 \end{array} \right) $$

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