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**Example text**

12 to a . 11) if the columns of span a deflating subspace of a , A - BB. For the reduced pencil a . 11). g. ) For the discrete case again we get analogous results. 18) E * X E = C*QC + A * X A - ( A * X B + C * S ) ( R + B * X B ) -1 ( A * X B + C'S)*. 19 THEOREM. 18). 20) [ V' = i XE - ( 1 t + B * X B ) - 1 ( A * X B + C'S)* ] span an n-dimensional subspace of C 2"+m , which is a deflating subspace for aA' - fiB'. PROOF: With V' as above we obtain '1 "l A'V' = -A*X -B*XJ [A- B(R + B*XB)-'(A*XB 13'V e = -A*X -B*XJ E, and thus the result follows.

V) If ( E , A, B ) is strongly stabilizable and Tt positive definite ~hen a A - fil3 has no eigenvalues on the imaginary axis. 34. Z3 0. Hence B ' z 2 = 0 and "ffz~A = --flz2E. 17. 39 COROLLARY. ,4',B' be as above and let z = (o~,~) z2 E C2'~+m\{0}, a,/3 C C, Z3 # (o,o), ~ueh that ( ~ A ' - ~B')~ Z3 ii) I f a = 0 -- 0. 1 then E z l = O . o. 1~ ~ 1 iv) i~ ~ ~ o, I~1~ = Izt ~ then ~ [ z , ] = O, z~B = O, ~z~ A t Z3 J -~z~ E . v) If (E, A, B ) is strongly stabillzable and 7~ positive det~nite, then a A ' - fiB' has no eigenvaJues on the unit circle.

To every Jordan block to an eigenva~ue 0 there is a paired block to an eigenvalue c o , but not conversely. Also there is not necessarily a pairing for blocks to eigenvalues on the unit circle. 17. A,B be as above, and let z = z2 ~ C"+'~\{0}, Z3 (~, ~) # (0, 0) ~u~h that ~A~ =/~B~. 7 we have [aA - Zs] [zl] z2 ~- 0 Z3 and [z~ z~ z~ ] [~A + 3//1 = 0. This implies [a[2(z~A* zl + z ~ C * Q C z l -k z ] S * C z l ) = -c~'fl z ~ E z , z; B = - ( ~ ; c * s + z;R). 35). 36 COROLLARY. 4'~ = ~B'~. 18 we have [Zz2] * ~z~ [-~A'- ~B'] = O.