Bourbaki

Remark 17

Tag 0DMQ, permanent. Cite it as [Bourbaki, Tag 0DMQ].

Algebra, chapter IX, § 7. Formes hermitiennes et corps ordonnés

Suppose the conditions of No. 3 are satisfied; for every ξ∈A\xi \in A, denote by ∣ξ∣|\xi| the element ρ⩾0\rho \geqslant 0 of K such that ρ2=ξξ‾\rho^2 = \xi \overline{\xi} (absolute value of ξ\xi). For every square matrix M=(αij)M = (\alpha_{ij}) over A, put f(M)=max⁡i∣αii∣f(M) = \max_i |\alpha_{ii}|, g(M)=max⁡i,j∣αij∣g(M) = \max_{i,j} |\alpha_{ij}|, and denote by φ(M)\varphi(M) the largest absolute value of the eigenvalues of MM (one will show that this definition has a meaning when A is the field of quaternions over K).

a) Let A,B,DA, B, D be three square matrices of order nn over A. Show that if DD is diagonal, one has

g2(ADB∗)≤f2(D)f(AA∗)f(BB∗) g^2(ADB^*) \leq f^2(D)f(AA^*)f(BB^*)
(use inequality (1) of No. 1). Deduce that
g(ABB∗A∗)≤f(AA∗)φ(BB∗) g(ABB^*A^*) \leq f(AA^*)\varphi(BB^*)
(apply Prop. 5 to the hermitian matrix BB∗BB^*). Deduce that, for mm arbitrary square matrices AiA_i (1≤i≤m1 \leq i \leq m) over A, one has
g2(A1A2…Am)≤f(A1A1∗)φ(A2A2∗)…φ(Am−1Am−1∗)f(Am∗Am) g^2(A_1A_2\ldots A_m) \leq f(A_1A_1^*)\varphi(A_2A_2^*)\ldots\varphi(A_{m-1}A_{m-1}^*)f(A_m^*A_m)
φ2(A1A2…Am)≤φ(A1A1∗)φ(A2A2∗)…φ(AmAm∗) \varphi^2(A_1A_2\ldots A_m) \leq \varphi(A_1A_1^*)\varphi(A_2A_2^*)\ldots\varphi(A_mA_m^*)
g2(A1A2…Am)≤φ(A1A1∗)…φ(AmAm∗). g^2(A_1A_2\ldots A_m) \leq \varphi(A_1A_1^*)\ldots\varphi(A_mA_m^*).
(For (3), reason by induction from (2). For (4), use Ex. 12; finally deduce (5) from (3).)

Show that inequality (3) no longer necessarily holds when Am∗AmA_m^*A_m is replaced there by AmAm∗A_mA_m^* (observe that one may have f(A∗A)≠f(AA∗)f(A^*A) \neq f(AA^*)), or when φ(AiAi∗)\varphi(A_iA_i^*) is replaced by f(AiAi∗)f(A_iA_i^*) for 2≤i≤m−12 \leq i \leq m-1 (take all the AiA_i equal to the square matrix whose all elements are 1).

b) Let uu be a normal endomorphism of E; if λ\lambda is an eigenvalue of uu (element of K(i)K(i) when A=KA = K), λ‾\overline{\lambda} is an eigenvalue of u∗uu^*u. For every normal matrix MM (matrix of a normal endomorphism of E with respect to an orthonormal basis), one therefore has φ2(M)=φ(MM∗)\varphi^2(M) = \varphi(MM^*). Deduce that one also has g(M)≤φ(M)g(M) \leq \varphi(M), and more generally, if M1,…,MmM_1, \ldots, M_m are normal,

g(M1…Mm)≤φ(M1)…φ(Mm) g(M_1\ldots M_m) \leq \varphi(M_1)\ldots\varphi(M_m)
φ(M1…Mm)≤φ(M1)…φ(Mm) \varphi(M_1\ldots M_m) \leq \varphi(M_1)\ldots\varphi(M_m)
(use (4) and (5)).

c) Show that if HH is a positive hermitian matrix over A, one has f(H)=g(H)≤φ(H)f(H) = g(H) \leq \varphi(H) (use (3) and b), by writing H=AA∗H = AA^*.

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

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