Bourbaki

Corollary 2

Tag 080T, permanent. Cite it as [Bourbaki, Tag 080T].

Topological Vector Spaces, chapter V, § 3. Tensor product of hilbertian spaces

— Let (aij)1≤i,j≤n(a_{ij})_{1 \leq i,j \leq n} be a hermitian matrix, with complex elements, and with determinant D. Suppose that the inequality

∑i,j=1naijz‾izj≥0 \sum_{i,j=1}^n a_{ij} \overline{z}_i z_j \geq 0

holds for all complex numbers z1,...,znz_1, ..., z_n. Then

0≤D≤a11...ann. 0 \leq D \leq a_{11} ... a_{nn}.

Suppose D is non-null; the equality D=a11...annD = a_{11} ... a_{nn} holds if and only if aij=0a_{ij} = 0 for all i≠ji \neq j.

Let Φ\Phi be the hermitian form on the vector space Cn\mathbf{C}^n given by

Φ(z,z′)=∑i,j=1naijz‾izj′ \Phi(\mathbf{z}, \mathbf{z}') = \sum_{i,j=1}^n a_{ij} \overline{z}_i z'_j

for z=(z1,...,zn)\mathbf{z} = (z_1, ..., z_n) and z′=(z1′,...,zn′)\mathbf{z}' = (z'_1, ..., z'_n) in Cn\mathbf{C}^n. By hypothesis, Φ\Phi is positive.

First assume that Φ\Phi is separating, that is, that D is non-null. If (e1,...,en)(e_1, ..., e_n) is the canonical basis of Cn\mathbf{C}^n, we have Φ(ei,ej)=aij\Phi(e_i, e_j) = a_{ij}, and cor. 2 follows immediately from cor. 1, a) by putting xi=eix_i = e_i.

Since aii=Φ(ei,ei)≥0a_{ii} = \Phi(e_i, e_i) \geq 0, we also have inequality (40) if D=0D = 0.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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