I've looked into this statement for a while (in fact, maybe for too long), and I thought to myself that I had to prove it
If $Q \colon \mathbb{R}^3 \to \mathbb{R}$ is a quadratic form with only cross product terms, then $Q$ is indefinite.
I've been really scratching my head over this one. I constructed a general symmetric matrix that corresponds to this quadratic form. $$A = \begin{bmatrix}0 & \frac{a_1}{2} & \frac{a_3}{2} \\\ \frac{a_1}{2} & 0 & \frac{a_2}{2} \\\ \frac{a_3}{2} & \frac{a_2}{2} & 0 \end{bmatrix}$$
Then I computed the characteristic polynomial to get $$\operatorname{char}_A(\lambda) = -\lambda^3\Biggl(\lambda + \frac{\sqrt{a_1^2 + a^2_2 + a^2_3}}{2}\Biggr)\Biggl(\lambda – \frac{\sqrt{a_1^2 + a^2_2 + a^2_3}}{2}\Biggr) + \frac{a_1a_2a_3}{4}$$
The last constant term is causing a lot of issues for me. If one of the entries is 0, then we're left with a good polynomial and we clearly see that the eigenvalues of $A$ alternate. We've proved the statement!..almost 🙁
If all of the coefficients are nonzero, then this seems impossible to factor into a product of linear factors. Is there anyway around this?
Best Answer
An easy way to prove that a matrix $Q$ is indefinite is to exhibit vectors $\mathbf x$ and $\mathbf y$ such that $\mathbf x^{\mathsf T} Q \mathbf x > 0$ and $\mathbf y^{\mathsf T}Q \mathbf y < 0$. This is a much less constrained search than searching for eigenvectors or eigenvalues of $Q$.
(In general, another good approach is Sylvester's criterion, but here it's inconclusive because the first principal minor is $0$.)
In the case of a matrix with only off-diagonal terms, we do not have to look far. In fact, the full power of that condition is not necessary: it's enough to have indices $i$ and $j$ such that $Q_{ii} = Q_{jj} = 0$ while $Q_{ij} = Q_{ji} \ne 0$. When $i=1$ and $j=2$, we have $$Q = \begin{bmatrix}0 & a & * \\ a & 0 & * \\ * & * & *\end{bmatrix}$$ where $*$ means "I don't care". Let $\mathbf x = (1,1,0)$ and $\mathbf y = (1,-1,0)$. Then $$ \begin{bmatrix}1 & 1 & 0\end{bmatrix}\begin{bmatrix}0 & a & * \\ a & 0 & * \\ * & * & *\end{bmatrix} \begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix} = 2a $$ while $$ \begin{bmatrix}1 & -1 & 0\end{bmatrix}\begin{bmatrix}0 & a & * \\ a & 0 & * \\ * & * & *\end{bmatrix} \begin{bmatrix}1 \\ -1 \\ 0\end{bmatrix} = -2a. $$ We do not know if $a$ is positive or negative, but either way, $\mathbf x^{\mathsf T} Q \mathbf x$ and $\mathbf y^{\mathsf T}Q \mathbf y$ have opposite signs: one is positive and one is negative. This proves that $Q$ is indefinite.