Diagonalisability and a Krylov basis in a parametrised vibration model
A three-degree-of-freedom vibration model is reduced, after nondimensionalisation, to the real matrix A(α) = [[1,1,0],[0,2,α],[0,0,2]], with α real. The engineer needs to know whether the modes can be decoupled by a change of basis, and how the system behaves at the critical parameter value α=1. The characteristic polynomial is requested first, then the geometric multiplicity of each eigenvalue. A colleague claims that because the eigenvalues are 1 and 2, with 2 repeated, the matrix is automatically diagonalisable for every α; the engineer is not convinced. At α=1 the team also needs a Jordan form, the minimal polynomial, and a single vector whose iterates under A generate the whole space, so that a Krylov-type computation can be set up. All answers must be justified over C, and the dependence on α must be stated explicitly rather than checked at one value only.
1. Determine all real α for which A(α) is diagonalisable over C. Show the geometric-multiplicity calculation.
Read the worked answer
Since A(α) is upper triangular, det(A-λI)=(1-λ)(2-λ)², so the eigenvalues are 1 and 2, with algebraic multiplicities 1 and 2. For λ=1, solving (A-I)v=0 gives v2=0, v2+αv3=0 and v3=0, hence v2=v3=0 with v1 free; the geometric multiplicity is 1, matching the algebraic multiplicity. For λ=2, A-2I = [[-1,1,0],[0,0,α],[0,0,0]]. If α≠0, the first two rows are independent, so rank=2 and nullity=3-2=1, which is less than 2; the matrix is not diagonalisable. If α=0, the rank is 1 and the nullity is 2, with independent eigenvectors (1,1,0) and (0,0,1). Therefore A(α) is diagonalisable exactly when α=0.
2. At α=1, give a Jordan form J and the minimal polynomial of A(1).
Read the worked answer
At α=1 the eigenvalue 2 has geometric multiplicity 1, so its Jordan block has size 2. The eigenvalue 1 is simple. Hence J is diag(1,J2(2)) up to ordering, where J2(2)=[[2,1],[0,2]]. The minimal polynomial must contain (x-1) once and (x-2) to the power equal to the largest 2-block, namely 2, so m_A(x)=(x-1)(x-2)². The exponent 2 is necessary: (A-I)(A-2I) is nonzero, while (A-I)(A-2I)²=0. Thus J and m_A together show one 2×2 block and one 1×1 block.
3. Does A(1) admit a cyclic vector? If yes, exhibit one and prove that the generated set is a basis.
Read the worked answer
Yes. Since deg m_A=3 equals the size of the matrix, A(1) is cyclic. Let N=A(1)-2I = [[-1,1,0],[0,0,1],[0,0,0]] and take v=e3=(0,0,1). Then Nv=(0,1,0)=e2 and N²v=(-1,0,0)=-e1. Hence {v,Nv,N²v} is a basis. Because A=2I+N, the span of {v,Av,A²v} equals the span of {v,Nv,N²v}, so e3 is a cyclic vector and {e3,e2,e1} is a Krylov basis. In this basis A has the companion form associated with m_A.