2(b) Letβ={1, x, x2}, thenT(1) = 2T(x) = 2x-1T(x2) = 2x2-2x,soA= [T]β=2-1002-2002,soλ= 2 and dim(Kλ) = 3.To find a basis forKλ, first check the dimension of the eigenspace:Ax= 2x⇒x=c100⇒dim(Eλ) = 1.Note a basis ofKλcannot consist of an eigenvector and a two cycle of generalized eigenvectors, becausethe initial vector of the cycle would be an eigenvector, so these three vectors would not be linearlyindependent. Instead, the basis ofKλmust be a three cycle of generalized eigenvectors{(A-2I)2v,(A-2I)v, v}such that the initial vector is an eigenvector:(A-2I)3v= 0⇒v=a001+b010+c100.Takev=001⇒(A-2I)v=0-20,(A-2I)2v=200.Therefore{2,-2x, x2}is a Jordan canonical basis forTandJ=2000-20001-12-1002-20022000-20001=210021002is a Jordan canonical form ofT.