= Positive-definiteness threshold in a two-by-two family
{title2=$A_t=\begin{pmatrix}3&t\\t&3\end{pmatrix},\quad |t|<3$}
Vary the coupling in the opening example. The <eigenvectors> $e_+=(1,1)^T/\sqrt2$ and $e_-=(1,-1)^T/\sqrt2$ form a fixed <orthonormal eigenbasis>, with <eigenvalues> $3+t$ and $3-t$. Writing a <vector> as $ue_++ve_-$ gives
$$
q_t=3x^2+2txy+3y^2=(3+t)u^2+(3-t)v^2.
$$
For $|t|<3$, both directions cost energy and $A_t$ is a <positive-definite matrix>. At $t=3$, the <quadratic form> is $6u^2$: the entire line spanned by $e_-$ has zero energy, and $A_t$ is a singular <positive semidefinite matrix>. At $t=-3$, the flat line is instead spanned by $e_+$. For $|t|>3$, the <quadratic form> is an <indefinite quadratic form>; one direction lowers the energy without bound as its displacement grows. At $t=4$, for instance, $q_t(e_+)=7$ while $q_t(e_-)=-1$.
\Image[/quadratic-energy-eigenvalue-threshold.png]
{title=Eigenvalues and directional energies across the positivity threshold}
The geometric transition is precise. For $|t|<3$, $q_t=1$ is an <ellipse>. At $t=3$, it consists of the two parallel lines $u=\pm1/\sqrt6$, extending freely in the zero-energy direction. For $t>3$, it is a <hyperbola>. The boundary case retains nonnegative energy but loses both the bounded positive <level sets> and the unique zero-energy <vector>.
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