Jordan Form & Generalized Eigenvectors Practice Quiz with a Step-by-Step Interactive Lesson
Use the question set below to practice Jordan form and generalized eigenvectors: Jordan blocks \(J_k(\lambda)\), nilpotent parts, ordinary and generalized eigenvectors, Jordan chains, algebraic versus geometric multiplicity, generalized eigenspaces \(\ker((A-\lambda I)^k)\), minimal polynomial exponents, diagonalizability criteria, nilpotency index, trace, determinant, and field issues. Open the lesson for compact worked examples and quick checks.
How this Jordan form practice works
- 1. Take the practice set: answer questions about blocks, chains, kernels, minimal polynomials, nilpotent powers, and diagonalizability.
- 2. Open the lesson: review the definitions, recognition tests, worked examples, and single-answer checks.
- 3. Retry: return to the question set and first decide whether the problem asks for a block size, a chain relation, a multiplicity, or a polynomial exponent.
What you will learn in the Jordan form and generalized eigenvectors lesson
Jordan blocks
- Block shape: \(J_k(\lambda)\) has \(\lambda\) on the diagonal and \(1\) on the superdiagonal
- Nilpotent part: \(J_k(\lambda)-\lambda I\) dies at power \(k\)
- Diagonal form: all blocks have size \(1\)
Generalized eigenvectors
- Generalized: \((A-\lambda I)^k v=0\) for some \(k\ge1\)
- Chain: \((A-\lambda I)v_1=0\) and \((A-\lambda I)v_i=v_{i-1}\)
- Rank: if \(Nv≠0\) but \(N^2v=0\), the vector sits above an eigenvector
Multiplicities and kernels
- Algebraic multiplicity: total size of all \(\lambda\)-blocks
- Geometric multiplicity: number of \(\lambda\)-blocks
- Kernels: \(\dim\ker(A-\lambda I)\) counts eigenvector directions
Minimal polynomial and traps
- Largest block: exponent of \(X-\lambda\) in \(m_A(X)\)
- Diagonalizable: geometric multiplicity equals algebraic multiplicity for every eigenvalue
- Field: full Jordan form is guaranteed over an algebraically closed field such as \(\mathbb{C}\)
Practice set
Jordan Form & Generalized Eigenvectors practice questions with instant score
Answer all 10 questions below, then get your final score and a mistake review at the end so you know exactly what to improve.
A generalized eigenvector for eigenvalue \(\lambda\) satisfies:
Correct answer: D. \((A-\lambda I)^k v=0\) for some \(k\ge1\)
Explanation: Generalized eigenvectors are killed by some power of \(A-\lambda I\).
A Jordan block for eigenvalue \(\lambda\) has \(\lambda\) on the diagonal and usually what above it?
Correct answer: B. Ones
Explanation: A Jordan block has ones on the superdiagonal.
A matrix is diagonalizable exactly when all Jordan blocks have size:
Correct answer: A. \(1\)
Explanation: Diagonal matrices are Jordan forms with only \(1\times1\) blocks.
For \(J=\begin{pmatrix}\lambda&1\\0&\lambda\end{pmatrix}\), what is its only eigenvalue?
Correct answer: C. \(\lambda\)
Explanation: The diagonal entries of this Jordan block are both \(\lambda\).
For \(J=\begin{pmatrix}2&1\\0&2\end{pmatrix}\), is \(J\) diagonalizable?
Correct answer: B. No
Explanation: This is a nontrivial Jordan block of size \(2\), so it is not diagonalizable.
The size of the largest Jordan block for \(\lambda\) is the exponent of \((X-\lambda)\) in:
Correct answer: B. The minimal polynomial
Explanation: The minimal polynomial records the largest Jordan block size for each eigenvalue.
If \((A-\lambda I)v=0\), then \(v\) is:
Correct answer: D. An eigenvector if \(v\ne0\)
Explanation: This is the usual eigenvector equation.
If \((A-\lambda I)^2v=0\) but \((A-\lambda I)v\ne0\), then \(v\) is:
Correct answer: C. A generalized eigenvector
Explanation: It is a generalized eigenvector of rank \(2\), not an ordinary eigenvector.
The trace of a Jordan matrix is:
Correct answer: C. The sum of its diagonal eigenvalues
Explanation: The trace is the sum of diagonal entries, i.e. eigenvalues counted with algebraic multiplicity.
A nilpotent Jordan block has which eigenvalue?
Correct answer: A. \(0\)
Explanation: Nilpotent matrices have only eigenvalue \(0\).
Result
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