“…Proof: First of all, if the linear subsystem (2) satisfies H1)-H3), then it is invertible, weakly minimum phase, and with CB symmetric positive definite, and it is possible to choose the matrix K 2 M k; p (IR) and the symmetric positive definite matrix P 2 M p; p (IR) such that y T Qy = 0 ) Cy = 0: (8) Indeed, as done in [6], one can assume, without loss of generality, that (2) is in the special coordinate basis (see [7]) _ y 01 = A 01 y 01 + A 11 y 1 _ y 02 = A 02 y 02 + A 12 y 1 _ y1 = D01y01 + D02y02 + D1y1 + CBũ y = y 1 with A 01 Hurwitz, A 02 + A T 02 = 0, and take K = (K 01 ; K 02 ; K 1 ) Hence, y T Q y = ky 01 k 2 + ky 1 k 2 and so y T Q y = 0 ) Cy = y 1 = 0: Assume now that (5) is of L-T, and set X(x) = f(x; 0); x 2 IR n .…”