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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.

Can you provide some insight about this problem?

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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. $\blacksquare$

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  • $\begingroup$ 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 $\endgroup$ – matan129 Sep 26 '18 at 16:54

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