+ U (v) = Constant and this implies stability: any trajectory stays in a bounded neighborhood of the origin since U (v) remains bounded. Recall that the differential equation d 2 v/dt 2 +ω 2 v = 0 has solutions v = a exp(±iωt). These solutions can only be bounded (as the time runs over R) if ω is a real number. Indeed exp(i(α + iβ)t) = exp(iαt) exp(−bt) is bounded if an only if b = 0. "Therefore", S has only real (and positive) eigenvalues. If S is not positive definite, just consider S + λId with a large positive real number λ.Au reste, quoiqu'il soit difficile, peut-être impossible, de déterminer en général les racines de l'équation P = 0, on peut cependant s'assurer, par la nature même du problème, que ces racines sont nécessairement toutes réelles [...]; car sans cela les valeurs de y , y , y , . . . pourraient croître à l'infini, ce qui serait absurde [99].Cauchy's proof (1829) [38]. Let λ be a complex eigenvalue of S and let v be a non zero eigenvector in C n . Of course,v is an eigenvector with eigenvalueλ. We still denote by the same symbol , the extension to C n of the scalar product as a bilinear form (not as a hermitian form). We haveSince v = 0, we have v,v = 0. This implies that λ =λ so that λ is indeed real. Amazingly, this proof, which looks crystal clear today, was not so convincing at the beginning of the nineteenth century, probably because complex numbers still sounded mysterious (they were called imaginary or even impossible).A more algebraic proof. If n 3, one uses the fact that any matrix with real entries has an invariant 2 dimensional subspace E. Of course, since S is symmetric, the orthogonal of E is also invariant and the proof goes by induction. One could argue that the standard proof of the existence of an invariant 2 dimensional subspace uses complex numbers so that this proof is very close to Cauchy's proof.Just for fun, let us describe a magic proof given by Sylvester (1852) [157]. Consider the characteristic polynomial P (X) = det(XI n − S). Note that P (X)P (−X) = det(X 2 I n − S 2 ). Since S is symmetric, all diagonal entries of S 2 are sums of squares and are therefore non negative. Let us write P (X)P (−X) = (−X 2 ) n + a 1 (−X 2 ) n−1 + a 2 (−X 2 ) n−2 + . . . + a n .