A Parameterized Transition Matrix That Refuses to Diagonalize
A forecasting team uses a three-state transition matrix A_t = [[2,1,0],[0,t,1],[0,0,2]] to project annual changes among three categories. The parameter t is a policy lever that can be any real number. The team wants a basis of eigenvectors so it can write A_t = P D P^{-1} and forecast by powers of D. A junior analyst claims that because the matrix is upper triangular, it is always diagonalizable and the diagonal entries 2, t, 2 are automatically the eigenvalues. A senior analyst warns that repeated diagonal entries can create a defective matrix. The team needs to know for which t a real eigenbasis exists, what happens at the borderline value t=2, and how the minimal polynomial reflects the failure of diagonalizability. No numerical software is to be used; the judgment must come from rank and nullity computations.
1. For which real t is A_t diagonalizable over R? Explain using eigenvalue multiplicities and eigenspace dimensions.
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Characteristic polynomial: (2-λ)(t-λ)(2-λ). If t≠2, λ=2 has algebraic multiplicity 2. Solve (A_t-2I)v=0. A_t-2I = [[0,1,0],[0,t-2,1],[0,0,0]]. Since t-2≠0, the first two rows are independent, so rank=2 and nullity=3-2=1. Thus the geometric multiplicity of λ=2 is 1, not 2, so A_t is defective. If t=2, A_2-2I = [[0,1,0],[0,0,1],[0,0,0]] has rank 2 and nullity 1, while λ=2 has algebraic multiplicity 3. Again the eigenspace is too small. Therefore no real t makes A_t diagonalizable over R.
2. At t=2, compute the minimal polynomial and explain why the matrix is not diagonalizable.
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At t=2, let N=A_2-2I = [[0,1,0],[0,0,1],[0,0,0]]. Compute N^2 = [[0,0,1],[0,0,0],[0,0,0]] ≠0 and N^3=0. Hence (A_2-2I)^2≠0 but (A_2-2I)^3=0. The minimal polynomial is (x-2)^3. A matrix is diagonalizable over R exactly when its minimal polynomial splits into distinct linear factors; here the factor (x-2) is repeated, so A_2 is not diagonalizable. Equivalently, the eigenspace for λ=2 has dimension 1, less than the algebraic multiplicity 3.
3. If t=3, determine the Jordan canonical form over R. Explain the size of the Jordan block for λ=2.
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For t=3, eigenvalues are 2,3,2. The eigenvalue 3 has algebraic multiplicity 1, so its Jordan block is [3]. For λ=2, A_3-2I = [[0,1,0],[0,1,1],[0,0,0]]. Its rank is 2, so nullity is 1: only one eigenvector for λ=2. The generalized eigenspace for λ=2 has dimension 2 because the remaining eigenvalue is 3. Therefore λ=2 contributes one Jordan block of size 2, J_2(2)=[[2,1],[0,2]]. The Jordan form over R is J_2(2) ⊕ [3] (up to ordering of blocks).