Bourbaki

Exercise 8, § 3

Lie Groups and Lie Algebras, V

Let (aij)(a_{ij}) be a real symmetric matrix with nn rows and nn columns.

a) Put sk=∑i=1naiks_k = \sum_{i=1}^n a_{ik}. Then, for all ξ1,…,ξn∈R\xi_1, \ldots, \xi_n \in \mathbf{R},

∑i,kaikξiξk=∑kskξk2−12∑i,kaik(ξi−ξk)2. \sum_{i,k} a_{ik} \xi_i \xi_k = \sum_k s_k \xi_k^2 - \frac{1}{2} \sum_{i,k} a_{ik} (\xi_i - \xi_k)^2.

b) Let ζ1,…,ζn∈R∗\zeta_1, \ldots, \zeta_n \in \mathbf{R}^*. Put ∑iζiaik=tk\sum_i \zeta_i a_{ik} = t_k. Then

∑i,kaikξiξk=∑ktkξk2ζk−12∑i,kζiζkaik(ξiζi−ξkζk)2. \sum_{i,k} a_{ik} \xi_i \xi_k = \sum_k \frac{t_k \xi_k^2}{\zeta_k} - \frac{1}{2} \sum_{i,k} \zeta_i \zeta_k a_{ik} \left( \frac{\xi_i}{\zeta_i} - \frac{\xi_k}{\zeta_k} \right)^2.

(In a), replace ξi\xi_i by ξiζi\frac{\xi_i}{\zeta_i} and aika_{ik} by ζiζkaik\zeta_i \zeta_k a_{ik}.)

c) If there exist numbers ζ1,…,ζn>0\zeta_1, \ldots, \zeta_n > 0 such that ∑iζiaik=0\sum_i \zeta_i a_{ik} = 0 (k=1,2,…,nk = 1, 2, \ldots, n), and if aij⩽0a_{ij} \leqslant 0 for i≠ji \neq j, then the quadratic form ∑i,kaikξiξk\sum_{i,k} a_{ik} \xi_i \xi_k is degenerate and positive. (Use b).)

d) Let ∑i,jqijξiξj\sum_{i,j} q_{ij} \xi_i \xi_j be a quadratic form on Rn\mathbf{R}^n such that qij⩽0q_{ij} \leqslant 0 for i≠ji \neq j. Assume that there is no partition I∪JI \cup J of {1,2,…,n}\{1, 2, \ldots, n\} such that qij=0q_{ij} = 0 for i∈I,j∈Ji \in I, j \in J. Then the form is non-degenerate and positive if and only if there exist ζ1>0,…,ζn>0\zeta_1 > 0, \ldots, \zeta_n > 0 such that ∑iζiqik=0\sum_i \zeta_i q_{ik} = 0 (k=1,…,nk = 1, \ldots, n).

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.