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K − 1, so that k−1 Aj1 B1 uj . 12) j=0 We deﬁne Sx = k−1 j=0 Aj2 B2 uj . This deﬁnition does not depend on the particular k −1 j choice of the vectors u0 , . . , uk−1 . Indeed, let k ≥ n and x = j=0 A1 B1 uj . By adding zero vectors if necessary we may assume without loss of generality that k = k . Let u and u be the vectors in (Rm )k with components uj and uj , respectively. Then Hk (u − u ) = Γ1 (k)Λ1 (k)(u − u ) = Γ1 (k)(x − x) = 0. As Hk = Γ2 (k)Λ2 (k) it follows that Γ2 (k)Λ2 (k)(u − u ) = 0, and as Θ2 is observable k−1 k−1 this implies that Λ2 (k)(u − u ) = 0, that is j=0 Aj2 B2 uj = j=0 Aj2 B2 uj .
The pair (A, B) is called stabilizable if there exists a matrix F such that A + BF is stable. In this case any matrix F such that A + BF is stable is called a stabilizing feedback matrix . 9). The next result shows under which conditions stabilization is possible. 3. The pair (A, B) is stabilizable if and only if every (A, B)-uncontrollable eigenvalue of A lies in the open unit disc. 48 Chapter 4. Stability Proof. Let (A, B) be stabilizable and let λ be an uncontrollable eigenvalue of A, so that xT A − λI B = 0 for some x = 0.
Then by using the Cayley–Hamilton theorem (0 = a0 + a1 A + · · · + an An ) one sees that ⎛ ⎜ ⎜ AS = S ⎜ ⎝ 0 ⎞ 1 .. −a0 . 1 . . . −an−1 ⎟ ⎟ ⎟, ⎠ ⎛ ⎞ 0 ⎜ .. ⎟ ⎜ ⎟ SB = ⎜ . ⎟ . ⎝0⎠ 1 So, we may as well assume that A and B have the form given above. Deﬁne F = a0 − p0 a1 − p1 . . an−1 − pn−1 , then ⎛ 0 ⎜ ⎜ A + BF = ⎜ ⎝ −p0 ⎞ 1 .. 1 . . . −pn−1 ⎟ ⎟ ⎟. ⎠ It is easily seen that A + BF has characteristic polynomial p(λ). Conversely, if (A, B) is not controllable, then by an appropriate choice of basis we may write A= A11 0 A12 , A22 B= B1 .