Given two quadratic forms, $$q_1, q_2: \mathbb{R}^n \to \mathbb{R}$$, and the fact that the following set is a linear subspace of $$\mathbb{R}^n$$,

$$L = \{v\in\mathbb{R}^n\mid q_1(v)\geq q_2(v)\}$$

We have to prove that either of these statements hold: $$\text{1.}\quad \forall v(v\in \mathbb{R}^n \to q_1(v)\geq q_2(v)) \\ \text{or}\\ \text{2.}\quad \forall v(v\in \mathbb{R}^n \to q_1(v)\leq q_2(v))$$

It's easy to prove statement $$1$$ holds iff $$L=\mathbb{R}^n$$. So, assuming $$L\subsetneq \mathbb{R}^n$$, we have to prove that statement $$2$$ is true.

I've tried defining a new quadratic form $$p = q_1 - q_2$$ and then proving by contradiction that $$p(v) \leq 0 \space \text{for each}\space v\in\mathbb{R}^n$$, but I couldn't develop this further.

So we basically want to show that $$L$$ is a trivial subspace.

So if $$p$$ is your quadratic form, first put it into diagonal form.

Then the diagonal matrix representing the form has some positive and some negative eigenvalues (if it has only positive/negative eigenvalues, we are done since $$p$$ is definite).

Let $$e$$ be an eigenvector corresponding to a positive eigenvalue and $$f$$ corresponding to a negative one (so $$e \in L$$ but $$f \notin L$$).

Also, let $$B$$ be the bilinear form corresponding to $$p$$.

Then for any real $$t$$, \begin{align*} p(e+tf) &= B(e+tf,e+tf) \\ &= B(e,e) + B(e,tf) + B(tf,e) + B(tf,tf) \\ &= B(e,e) + 2tB(e,f) + t^2B(f,f) \\ &= B(e,e)+t^2B(f,f) \\ &= p(e)+t^2p(f) \end{align*}

$$B(e,f)=0$$ since the matrix is symmetric (or more generally, normal), so the eigenvectors are orthogonal.

Now $$p(e)$$ is positive and $$p(f)$$ is negative by definition. But if we let $$t \downarrow 0$$ then $$p(e+tf) > 0$$, meaning $$e+tf$$ belongs to $$L$$, which shouldn't happen. $$\blacksquare$$

• Just a quick note - no need to actually diagonalize the matrix (it is not possible anyway, since $q_1,q_2$ are unknown), it is only worth noting that it is symmetric and therefore diagonalizable by a unitary matrix – matan129 Sep 26 '18 at 16:54