Bourbaki

Exercise 12, § 6

Tag 02VP, permanent. Cite it as [Bourbaki, Tag 02VP].

Spectral Theories, III, p. 133

Let AA be a real matrix of type (I,I)(I,I), irreducible and positive. For every nonzero x∈RIx\in \mathbf{R}^I, put

r=r= infxxii∈≠0I(Ax)ixi_{xxi_i\in\not=0I}\frac{(Ax)_i}{x_i},

where (AxAx)i=∑k∈Iaikxk_i=\sum_{k\in I}a_{ik}x_k is the coordinate of AxAx of index ii.

a) If x≽0x\succcurlyeq 0 and x≠0x\not = 0, then rxr_x is the greatest real number ϱ\varrho such that ϱx≼Ax\varrho x\preccurlyeq Ax. b) Suppose that x≽0x\succcurlyeq 0 and put y=(1I+y= (1_I+ A)x)x. Then rx⩽ryr_x\leqslant r_y.

c) The restriction to (R+∗)I(\mathbf{R}_+^*)^I of the function x↦rxx\mapsto r_x is continuous.

d) Deduce that there exists z∈RIz\in \mathbf{R}^I such that z≽0z\succcurlyeq 0 and such that rz=r_z= supx≽0rx_{x\succcurlyeq 0}r_x. (Use the compact sets M={x∈R+I∣∑i∈Ixi=1}M =\{x\in \mathbf{R}_+^I|\sum_{i\in I}x_i= 1\} as well as (1I+(1_I+ A)n−1^{n-1}(M), then Exercise 10.)

We note r=rzr=r_z. An element zz of RI\mathbf{R}^I such that z≽0z\succcurlyeq 0 and rz=rr_z=r will be called extremal.

e) One has r>0r >0.

f) If zz is an extremal vector, then Az=rzAz=rz. (If u=Az−rz≽0u=Az-rz\succcurlyeq 0 were nonzero, then (1I+(1_I+ A)n−1u^{n-1}u would be ≻0\succ 0 according to Exercise 10 and, for x=(1I+x= (1_I+ A)n−1z)^{n-1}z, one would have Ax≻rx.)Ax\succ rx.)

g) Every extremal element is ≻0\succ 0.

For every vector yy in CI\mathbf{C}^I, we denote by y+y^+ the vector (∣yi∣)i∈I(|y_i|)_{i\in I}; thus one has y+∈RIy^+\in \mathbf{R}^I and y+≽0y^+\succcurlyeq 0.

h) Let α∈C\alpha \in \mathbf{C} be an eigenvalue of A, viewed as a complex matrix, and let yy in CI\mathbf{C}^I be a corresponding eigenvector. One has ∣α∣y+≼Ay+|\alpha |y^+\preccurlyeq Ay^+ and ∣α∣⩽r|\alpha |\leqslant r.

i) Deduce that if α=r\alpha =r, all the coordinates of yy are nonzero, then that the eigenspace (in CI\mathbf{C}^I) of A associated with the eigenvalue rr has dimension 1.

We denote by Δ(X) the determinant of X⋅1I−AX\cdot 1_I-A and by B(X) the transpose of the matrix of the cofactors of X⋅1I−AX\cdot 1_I-A. We denote B(X) = (Bij(X)B_{ij}(X))i,j∈I_{i,j\in I}. We therefore have B(X)(X⋅1I−A)=(X⋅1I−A(X)(X\cdot 1_I-A) = (X\cdot 1_I-A)B(X)=Δ(X)⋅1E(X) = \Delta (X)\cdot 1_E (A, III, p. 99, prop. 12). j) For every i∈Ii\in I, the column of index ii of B(r)(r) is nonzero. (If this is not the case, construct an eigenvector of A associated with the eigenvalue rr and whose coefficient of index ii is zero, in contradiction with question f).) k) All the coefficients of B(r)(r) have the same sign. (First prove, using the relation (r1I−A(r1_I-A)B(r)=0(r) = 0, that for every i∈Ii\in I, all the coefficients of the column of index ii of B(r)(r) are nonzero and have the same sign, then reason with the transposes.)

l) We have Δ′(r)>0\Delta '(r)>0 and Bij(r)>0B_{ij}(r)>0 for every (i,j)∈I2(i, j)\in I^2. (Use the formula Δ′(X)=∑iBii(X)\Delta '(X) =\sum_iB_{ii}(X) and observe that Δ′\Delta ' does not change sign on the interval [r,+∞].[r,+\infty ].)

m) Conclude that rr is a simple root of the characteristic polynomial of A.

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Machine translation of the English, by gpt-5-6. Not checked by a person.