A parameterised matrix that is never diagonalisable
A student is given A_a = [[2,1,0],[0,2,1],[0,0,a]] with real parameter a. They compute the characteristic polynomial det(A_a − λI) = (2−λ)²(a−λ). The student argues: if a≠2, the eigenvalues are 2 with algebraic multiplicity 2 and a with algebraic multiplicity 1; since there are two distinct eigenvalue values, the matrix must be diagonalisable. If a=2, they note that the only eigenvalue is 2 and check whether A_2−2I is the zero matrix; it is not, so they conclude that A_2 is not diagonalisable. They are unsure about the case a≠2 and ask whether any real a makes A_a diagonalisable. The instructor asks them to compute eigenspaces rather than compare only algebraic multiplicities, and to consider the rank of A_a−2I for both cases.
1. Compute the rank and nullity of A_a−2I for a≠2 and for a=2. What is the geometric multiplicity of eigenvalue 2 in each case?
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For a≠2, A_a−2I = [[0,1,0],[0,0,1],[0,0,a−2]]. The first two rows are independent, and the third is (a−2) times the second, so rank=2. By rank-nullity in R³, nullity=3−2=1. Hence the eigenspace for λ=2 is one-dimensional: geometric multiplicity 1. For a=2, A_2−2I = [[0,1,0],[0,0,1],[0,0,0]], also rank 2 and nullity 1.
2. Decide for which real a the matrix A_a is diagonalisable over R, and justify using algebraic versus geometric multiplicity.
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For a≠2, λ=2 has algebraic multiplicity 2 but geometric multiplicity 1, so the matrix is defective. The simple eigenvalue a has geometric multiplicity 1. Total geometric multiplicity is 2<3, so A_a is not diagonalisable. For a=2, the only eigenvalue 2 has algebraic multiplicity 3 and geometric multiplicity 1, so it is also not diagonalisable. Therefore no real value of a makes A_a diagonalisable over R. Distinct eigenvalues are sufficient only when all eigenvalues are simple; here 2 is repeated.
3. For a≠2, write a Jordan form of A_a; for a=2, write a Jordan form. What does this say about the minimal polynomial?
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For a≠2, the Jordan form can be diag(J₂(2), [a]), with one 2×2 Jordan block for λ=2 and a 1×1 block for λ=a. The minimal polynomial is (λ−2)²(λ−a). For a=2, A_2 is already the single Jordan block J₃(2), so the minimal polynomial is (λ−2)³. The exponent in the minimal polynomial equals the largest Jordan block size, confirming defectiveness.