Page 168 - Textos de Matemática Vol. 39

158 ION ZABALLA SinceAX=XLwehavethatx1L=0andxi =xi+1L.Hence,xiLi =0,
i = 1,...,n, and
⎡xLn−1 ⎤
⎢ . ⎥ ( 3 . 1 ) X = ⎢⎣ x L ⎥⎦
x
wherex=xn andXLn =0.AsListherestrictionofAto<X >,theminimal polynomial of L is a divisor of λn; namely, it is λd. Thus Ld = 0 and in order for the matrix X to have full column rank, x must be a vector such that xLd−1 ̸= 0. In other words, the matrices X which are solution of equation AX = XL are those with the form (3.1) with x any row vector satisfying xLd−1 ̸= 0. We will assume from now on that such a vector has been chosen and let its components be x = x1 x2 · · · xdT .
Let ε > 0 be any positive real number. For this ε > 0 there are real numbers η1 > 0 and η2 > 0 such that if ∥ In − T ∥< η1 and ∥ X − Z ∥< η2 thenZ∈M∗n,d and∥X−TZ∥<ε.
Let a = (a1,...,an) ∈ Cn and R ∈ Cd×d an arbitrary matrix. Put zn = x, zj =x(Rn−j −a1Rn−j−1−···−an−jId),j=1,...,n−1,and
⎡⎢ z 1 ⎤⎥ (3.2) Z = ⎢z2⎥.
⎣ . ⎦ zn
Notice that
xj −zj =xLn−j −y(Rn−j −a1Rn−j−1 −···−an−jId),
andsotherearerealnumbersν1 >0andν2 >0suchthatif∥a∥<ν1 and ∥L−R∥<ν2 then∥X−Z∥<η2.
Let ρ>0 be a real number such that if λ1,..., λn are complex numbers in the open ball centered at 0 and with radius ρ, then the vectors a=(a1,...,an) a n d c = ( c 1 , . . . , c d ) o f t h e c o e ffi c i e n t s o f t h e p o l y n o m i a l s p ( λ ) = (λ−λ 1) · . . . · (λ−λ n) and q(λ) = (λ−λ1)·...·(λ−λd) satisfy ∥ a ∥< ν1 and ∥ c ∥< ν2. Such a real ρ there always exists because of the continuity of the coefficients of a polynomial with respect to its roots.
Finally, if e1 = 1 0 ···0T, there is a real number δ > 0 such that if ∥A − A′∥ < δ, ∥ e1 − b′ ∥< δ and T = b′ A′b′ · · · A′n−1b′ then
(a) T is invertible (b′ is a cyclic vector with respect to A′) and ∥In−T ∥< η1; (b) Λ(A′) ⊂ Bρ(0).