= Solution
<Linearization> gives $\theta''+\pi^2\lambda\theta=0$. The two homogeneous Dirichlet conditions select $\boxed{\lambda=n^2,\quad\theta=A\sin(n\pi\xi),\quad n=1,2,\ldots}$. For any small trial <amplitude> in this mode, $\int_0^1\sin^2(n\pi\xi)=\int_0^1\cos^2(n\pi\xi)=1/2$, so
$$
\boxed{U[A\sin(n\pi\xi)]=\frac{\pi^2EIA^2}{4L}(n^2-\lambda)+O(A^4).}
$$
For $\lambda<1$ every nonzero quadratic <variation> is positive by the sharp <Poincaré inequality>. For $\lambda>1$, the first mode gives negative energy at sufficiently small <amplitude>, so zero cannot be a <local minimum>.
\b[The printed strict “only if” conclusion omits the critical case.] At $\lambda=1$ the quadratic test is degenerate, not negative. In fact zero is still the unique minimum of the full displayed <functional>: use $1-\cos\theta\leq\theta^2/2$ and $\int\theta'^2\geq\pi^2\int\theta^2$ to get $U\geq0$, with strict inequality for a nonzero continuous function because $1-\cos t<t^2/2$ when $t\ne0$. In particular on the critical first mode,
$$
\boxed{U[A\sin\pi\xi]=\frac{\pi^2EI}{64L}A^4+O(A^6)>0\quad\text{for small }A\ne0.}
$$
Thus strict positivity of the <second variation> holds for $\lambda<1$, while local minimality of the full energy holds also at $\lambda=1$. This is the distinction captured by <critical-load minimality of the angular Euler strut>.
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