A 4×4 matrix A has been diagonalized as A = PDP⁻¹. A colleague says computing A¹⁰⁰ requires multiplying A by itself 100 times. What is the correct approach?
AMultiply A by itself 100 times — diagonalization only helps for A²
BCompute PD¹⁰⁰P⁻¹, where D¹⁰⁰ is found by raising each diagonal entry to the 100th power
CRaise each entry of A to the 100th power
DRaise each eigenvalue to the 100th power to get the eigenvalues of A¹⁰⁰, but the full matrix cannot be recovered
The key payoff of diagonalization is Aⁿ = PDⁿP⁻¹. Since D is diagonal, Dⁿ is trivial — just raise each diagonal entry to the nth power. This reduces 100 full matrix multiplications to raising scalars to a power, then doing two matrix multiplications (with P and P⁻¹). Option D is close to true but incomplete — it correctly identifies the eigenvalues of A¹⁰⁰ but doesn't reconstruct the full matrix.
Question 2 Multiple Choice
Which condition is sufficient (but not necessary) to guarantee that an n × n matrix A is diagonalizable?
AAll n eigenvalues of A are distinct (no repeated eigenvalues)
BA is invertible (nonzero determinant)
CA is upper triangular
DA has all positive eigenvalues
Distinct eigenvalues guarantee diagonalizability because eigenvectors from distinct eigenvalues are always linearly independent, so you automatically get n linearly independent eigenvectors to form the columns of P. This condition is sufficient but not necessary — a matrix can still be diagonalizable with repeated eigenvalues, provided each eigenvalue's geometric multiplicity equals its algebraic multiplicity. Invertibility and triangular structure do not imply diagonalizability.
Question 3 True / False
A matrix with a repeated eigenvalue can seldom be diagonalized.
TTrue
FFalse
Answer: False
This is a common misconception. Repeated eigenvalues do not automatically prevent diagonalization. What matters is whether the geometric multiplicity of each eigenvalue (the dimension of its eigenspace) equals its algebraic multiplicity (how many times it appears as a root of the characteristic polynomial). If these match for every eigenvalue, the matrix is diagonalizable even with repeats. A non-diagonalizable example with a repeated eigenvalue is [[1,1],[0,1]] — its eigenspace for λ = 1 has dimension 1, but λ = 1 has algebraic multiplicity 2.
Question 4 True / False
If A = PDP⁻¹ where D is diagonal with entries λ₁, λ₂, …, λₙ, then A¹⁰ = PD¹⁰P⁻¹ where D¹⁰ has entries λ₁¹⁰, λ₂¹⁰, …, λₙ¹⁰.
TTrue
FFalse
Answer: True
This follows directly from the algebra of the factorization. A² = (PDP⁻¹)(PDP⁻¹) = PD(P⁻¹P)DP⁻¹ = PD²P⁻¹, because P⁻¹P = I. By induction, Aⁿ = PDⁿP⁻¹. For a diagonal matrix, raising it to a power just raises each diagonal entry to that power — there are no off-diagonal interactions. This is precisely what makes diagonalization computationally powerful.
Question 5 Short Answer
Explain in your own words why diagonalizing a matrix makes computing its powers far more efficient.
Think about your answer, then reveal below.
Model answer: Diagonalization rewrites A = PDP⁻¹ where D is diagonal. Powers simplify as Aⁿ = PDⁿP⁻¹. Raising a diagonal matrix to a power is trivial — just raise each diagonal entry (an eigenvalue) to the nth power. This replaces n−1 full matrix multiplications with scalar exponentiation plus two fixed matrix multiplications.
The key is that P and P⁻¹ cancel in the middle during repeated multiplication: (PDP⁻¹)(PDP⁻¹) = PD(P⁻¹P)DP⁻¹ = PD²P⁻¹. Once you see this cancellation, the formula Aⁿ = PDⁿP⁻¹ follows immediately. The deeper insight: in the eigenvector basis, A acts by simple scaling — each basis vector gets multiplied by its eigenvalue. Repeated application just multiplies the scale factors, which is why powers are trivial in this basis.