Page 169 - Textos de Matemática Vol. 39

Thus,
N = T ′ − 1 A T ′ = ⎢⎣ . . . . . . . . . 0 ··· 1
STABILITY OF INVARIANT SUBSPACES 159
In fact, item (a) is a consequence of In = e1 Ae1 · · · An−1e1 (i.e. e1 is a cyclic vector for A). And item (b) expresses the continuity of the eigenvalues with respect to the elements of a matrix.
Now, if p(λ) = λn − a1λn−1 − . . . − an is the characteristic polyno- mial of A′, as Λ(A′) ⊂ Bρ(0), we have that ∥ a ∥< ν1. Furthermore, if (λ1,...,λn)aretheeigenvaluesofA′ (withrepetitions,possibly)andwewrite q(λ)=(λ−λ1)·...·(λ−λd)=λd−c1λd−1−...−cd anddefine
⎡⎢ 0 · · · 0 c d ⎤⎥ (3.3) R=⎢1 ··· 0 cd−1⎥
Now, recalling the definition of zi we have that zn−1+a1zn = zn(R − a1Id) + a1zn = znR
⎣. ... . ⎦ 0 ··· 1 c1
then ∥ L−R ∥< ν2. This implies that if Z is the matrix of (3.2) with R the matrix of (3.3) then ∥ X − Z ∥< η2.
Ontheotherhand,as∥In−T∥<η1 weconcludethat∥X−TZ∥<ε. And putting X′ = TZ we have that ∥ X−X′ ∥< ε. In other words, we have seen thatforanyε>0thereisδ>0suchthatif∥A−A′ ∥<δthen∥X−X′ ∥<ε. It only remains to prove that X′ ∈ S(A′). That is to say, A′X′ = X′L′ for some matrix L′ ∈ Cd×d. But A′X′ = X′L′ if and only if T −1A′T T −1X′ = T −1X′L′.
On the one hand T−1X′ = Z, and on the other hand
⎡⎢ 0 · · · 0 ⎢1 ··· 0
a n ⎤⎥ an−1⎥
⎥⎦ . a1
⎡⎢ anzn ⎤⎥ N Z = ⎢z1 + an−1 zn ⎥ .
⎣ . ⎦ zn−1 + a1zn
zn−2+a2zn = zn(R2 − a1R − a2Id) + a2zn = zn(R − a1Id)R = zn−1R .
z1+an−1zn = zn(Rn−1 − a1Rn−2 − . . . − an−1Id) + an−1zn = = zn(Rn−2 − a1Rn−3 − . . . − an−2Id)R = z2R.
Hence, we have that NZ = ZR except for the first row. But R is a nonderoga- tory matrix and its characteristic polynomial is q(λ) which satisfies q(λ)|p(λ).