Practice set
Jordan Form & Generalized Eigenvectors practice quiz 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.
Over \(\mathbb{C}\), every square matrix has:
Correct answer: C. A Jordan normal form
Explanation: The Jordan normal form exists over an algebraically closed field such as \(\mathbb{C}\).
A \(3\times3\) Jordan block for \(\lambda\) has minimal polynomial:
Correct answer: C. \((X-\lambda)^3\)
Explanation: The exponent equals the block size.
The number of Jordan blocks for eigenvalue \(\lambda\) equals:
Correct answer: B. \(\dim\ker(A-\lambda I)\)
Explanation: Each block contributes one dimension to the eigenspace.
For one eigenvalue \(\lambda\), its algebraic multiplicity equals:
Correct answer: B. The sum of the sizes of its Jordan blocks
Explanation: In Jordan form, it is the total size of all Jordan blocks with eigenvalue \(\lambda\).
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 one eigenvalue \(\lambda\), its geometric multiplicity equals:
Correct answer: A. The number of Jordan blocks for \(\lambda\)
Explanation: Each Jordan block contributes one independent eigenvector.
For \(J=\begin{pmatrix}\lambda&1\\0&\lambda\end{pmatrix}\), what is \((J-\lambda I)^2\)?
Correct answer: D. \(0\)
Explanation: The nilpotent part of a \(2\times2\) Jordan block squares to zero.
In a Jordan chain \(v_1,v_2,v_3\), \((A-\lambda I)v_1\) equals:
Correct answer: C. \(0\)
Explanation: The first vector is an ordinary eigenvector.
If a nilpotent matrix has a block of size \(4\), its nilpotency index is at least:
Correct answer: A. \(4\)
Explanation: The largest block determines the nilpotency index.
If algebraic and geometric multiplicities are equal for every eigenvalue, the matrix is:
Correct answer: C. Diagonalizable
Explanation: There are enough eigenvectors to form a basis.
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Jordan Form & Generalized Eigenvectors Practice Questions with Answers and a Guided 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 \(\mathbb{C}\)

