A 3×3 matrix A has eigenvalues 2, 2, and 5. Is A necessarily diagonalizable?
AYes — A has real eigenvalues, which is sufficient for diagonalizability
BNo — A has a repeated eigenvalue, so it cannot be diagonalized
CIt depends — A is diagonalizable if and only if the eigenvalue 2 has two linearly independent eigenvectors
DIt depends — A is diagonalizable if and only if the eigenvalue 5 has a non-zero eigenvector
The condition for diagonalizability is having n linearly independent eigenvectors, not having distinct eigenvalues. A repeated eigenvalue (here, λ=2 with multiplicity 2) may or may not provide two independent eigenvectors. If it does (geometric multiplicity equals algebraic multiplicity), A is diagonalizable. If it does not (only one independent eigenvector for λ=2), A is not diagonalizable and requires Jordan normal form instead. Option B is the common misconception — repeated eigenvalues do not automatically prevent diagonalization.
Question 2 Multiple Choice
What is the primary computational advantage of diagonalizing a matrix A = PDP⁻¹ before computing A¹⁰⁰?
AIt reduces the matrix to a smaller size, making storage more efficient
BIt allows A¹⁰⁰ = PD¹⁰⁰P⁻¹, where D¹⁰⁰ requires only raising scalar diagonal entries to the 100th power
CIt ensures the result has only integer entries, simplifying exact computation
DIt converts the problem to solving a system of linear equations, which is faster
The key identity is Aⁿ = PDⁿP⁻¹. A diagonal matrix Dⁿ is trivial to compute: just raise each diagonal entry to the n-th power (scalar exponentiation). Without diagonalization, computing A¹⁰⁰ would require 99 matrix multiplications, each O(n³). With diagonalization, it requires computing P and P⁻¹ once, then one multiplication PDⁿP⁻¹. This is the payoff for all the eigenvalue machinery — it reduces matrix exponentiation (hard) to scalar exponentiation (trivial) plus two matrix multiplications.
Question 3 True / False
If a matrix has a repeated eigenvalue, it can seldom be diagonalized.
TTrue
FFalse
Answer: False
This is a common misconception. A matrix is diagonalizable if and only if it has n linearly independent eigenvectors. Distinct eigenvalues guarantee this (since eigenvectors for distinct eigenvalues are always independent), so a matrix with n distinct eigenvalues is always diagonalizable. But a repeated eigenvalue *may* still yield enough independent eigenvectors — the identity matrix I has only one eigenvalue (λ=1, with multiplicity n) yet is perfectly diagonalizable (it's already diagonal). The question is whether the geometric multiplicity (dimension of the eigenspace) equals the algebraic multiplicity (multiplicity as a root of the characteristic polynomial).
Question 4 True / False
Two similar matrices A and B, related by B = P⁻¹AP for some invertible P, always have the same eigenvalues because they represent the same linear transformation in different coordinate systems.
TTrue
FFalse
Answer: True
Similarity is a change of basis: A and B describe the same linear transformation, just expressed in different bases (related by P). Since eigenvalues capture the intrinsic scaling behavior of a transformation — not the coordinate system used to describe it — they are invariants of the transformation itself, not of any particular matrix representation. Similar matrices therefore share all eigenvalues, the same characteristic polynomial, the same determinant, and the same trace. This is why diagonalization can be thought of as finding the 'simplest' basis to represent a transformation.
Question 5 Short Answer
Explain what it means geometrically that the columns of P are eigenvectors. Why does this choice of P make the factorization A = PDP⁻¹ hold?
Think about your answer, then reveal below.
Model answer: Each column of P is an eigenvector vᵢ of A, meaning Avᵢ = λᵢvᵢ — A acts on vᵢ purely by scaling (no rotation or mixing). The equation AP = PD expresses this simultaneously for all eigenvectors: multiplying each column by A is the same as multiplying it by its eigenvalue (which is what D does — it scales the i-th column by λᵢ). So PDP⁻¹ applies three steps: convert to eigenvector coordinates (P⁻¹), scale each coordinate by its eigenvalue (D), convert back (P). The composition of these three steps is exactly what A does.
The geometric insight is that eigenvectors are the 'natural axes' of a linear transformation — the directions where the transformation acts simplest (pure scaling). In the eigenvector basis, A looks diagonal because all cross-terms vanish: each basis vector is only scaled, never mixed with others. Diagonalization is just the algebraic expression of this geometric fact. This is also why not every matrix is diagonalizable: if A rotates or shears space in a way that mixes directions, there may not be enough invariant directions (eigenvectors) to form a complete basis.