Lemma 4

Let $Q = (q_{ij})$ be a real, symmetric, square matrix of order $n$ such that: a) $q_{ij} \leq 0$ for $i \neq j$; b) there does not exist a partition of $\{1, 2, \ldots, n\}$ into two non-empty subsets I and J such that $(i, j) \in I \times J$ implies $q_{ij} = 0$; c) the quadratic form $q(x_1, \ldots, x_n) = \sum_{i,j} q_{ij} x_i x_j$ on $\mathbf{R}^n$ is positive.

Then: (i) The kernel N of q is of dimension 0 or 1. If $\dim N = 1$, N is generated by a vector all of whose coordinates are > 0. (ii) The smallest eigenvalue of Q is of multiplicity 1 and a corresponding eigenvector has all its coordinates > 0 or all its coordinates < 0.

Since q is a positive quadratic form, the kernel N of q is the set of isotropic vectors for q (Algebra, Chap. IX, § 7, no. 1, Cor. of Prop. 2). Let $a_1, \ldots, a_n$ be the canonical basis of $\mathbf{R}^n$. If $\sum_i c_i a_i \in N$, Lemma 3 shows that we also have $\sum_i |c_i| a_i \in N$, and hence $\sum_i q_{ji} |c_i| = 0$ for all j. Let I be the set of i such that $c_i \neq 0$. If $j \notin I$, then $q_{ji} |c_i| \leq 0$ for $i \in I$ and $q_{ji} |c_i| = 0$ for $i \notin I$, so $q_{ji} = 0$ for $j \notin I$ and $i \in I$. Assumption b) thus implies that either $I = \varnothing$ or $I = \{1, \ldots, n\}$. Consequently, every non-zero vector in N has all its coordinates $\neq 0$. If $\dim N \geq 2$, the intersection of N with the hyperplane with equation $x_i = 0$ would be of dimension $\geq 1$, contrary to what we have just shown. The preceding argument also shows that, if $\dim N = 1$, then N contains a vector all of whose coordinates are > 0. This proves (i).

On the other hand, we know that the eigenvalues of Q are real (Algebra, Chap. IX, § 7, no. 3, Prop. 5) and positive since q is positive. Let $\lambda$ be the smallest of them. The matrix $Q' = Q - \lambda I_n$ is then the matrix of a degenerate positive form $q'$ and the off-diagonal elements of $Q'$ are the same as those of Q. Consequently, $Q'$ satisfies conditions a), b) and c) of the statement of the lemma. Since the kernel $N'$ of $q'$ is the eigenspace of Q corresponding to the eigenvalue $\lambda$, assertion (ii) follows from (i).