Minimal Polynomials & Cayley-Hamilton Practice Quiz with a Step-by-Step Interactive Lesson
Use the question set below to practice minimal polynomials and the Cayley-Hamilton theorem: recognizing when a polynomial cancels a matrix, finding the monic polynomial of least degree, using \(m_A\mid p_A\), substituting \(A\) into its characteristic polynomial, reducing powers like \(A^2\) or \(A^3\), deriving inverse formulas, reading eigenvalue information from roots, and applying the repeated-root test for diagonalizability. The lesson keeps examples small enough to follow mentally while still covering the high-level structure.
Answer the question set and review your mistakes at the end.
How this minimal polynomial practice works
1. Take the practice set: answer questions about \(p_A(A)=0\), \(m_A(A)=0\), divisibility, diagonalizability, and polynomial relations.
2. Open the lesson: review the definitions, Cayley-Hamilton reductions, standard examples, and common traps.
3. Retry: return to the question set and first ask which polynomial relation the matrix satisfies.
What you will learn in the minimal polynomials lesson
Definitions and divisibility
Annihilating polynomial: a polynomial \(q\) with \(q(A)=0\)
Minimal polynomial: the monic annihilating polynomial of least degree
Divisibility: \(m_A\) divides every annihilating polynomial, especially \(p_A\)
Cayley-Hamilton use
Every square matrix satisfies \(p_A(A)=0\)
Constants become multiples of \(I\), such as \(A^2-5A+6I=0\)
Use the resulting equation to reduce high powers or express \(A^{-1}\)
Diagonalization and roots
A split minimal polynomial with no repeated factor characterizes diagonalizability
For diagonalizable matrices, each distinct eigenvalue appears once in \(m_A\)
Repeated factors signal Jordan-block behavior and prevent diagonalizability
Standard matrix examples
Scalar matrix: \(A=\lambda I\) has \(m_A(X)=X-\lambda\)
Purpose: Learn how polynomial equations control a square matrix. The minimal polynomial \(m_A\) is the smallest monic polynomial with \(m_A(A)=0\), while Cayley-Hamilton gives one guaranteed annihilating polynomial: the characteristic polynomial \(p_A\). The skill is to move between equations like \(A^2-3A+2I=0\), divisibility statements, eigenvalue roots, and structural conclusions such as diagonalizability.
Success criteria
State Cayley-Hamilton as \(p_A(A)=0\) for every square matrix \(A\).
Define an annihilating polynomial and the minimal polynomial \(m_A\).
Use the divisibility facts \(m_A\mid q\) whenever \(q(A)=0\), and \(m_A\mid p_A\).
Substitute \(A\) into a characteristic polynomial with constants multiplied by \(I\).
Find \(m_A\) for scalar matrices, zero matrices, nilpotent matrices, projections, and involutions.
Use minimal polynomial roots to infer possible eigenvalues.
Apply the test: diagonalizable exactly when \(m_A\) splits with no repeated factor.
Reduce powers or derive inverse relations from polynomial equations.
Key vocabulary
Annihilating polynomial: a polynomial \(q(X)\) such that \(q(A)=0\).
Minimal polynomial: the monic annihilating polynomial of least degree.
Characteristic polynomial: \(p_A(X)=\det(XI-A)\), depending on the chosen convention but satisfying \(p_A(A)=0\).
Splits: factors into linear factors over the field being used.
Repeated factor: a power such as \((X-\lambda)^2\), which detects nontrivial Jordan-block behavior.
Quick pre-check
Pre-check 1: What does Cayley-Hamilton say about a square matrix \(A\)?
Hint: The theorem substitutes the matrix into its own characteristic polynomial.
Pre-check 2: The minimal polynomial of \(A\) is:
Hint: Look for the shortest monic polynomial that cancels the matrix.
A polynomial relation is a matrix equation
Learning goal: Read equations such as \(A^2=0\), \(A^2=A\), or \(A^2-3A+2I=0\) as polynomial cancellations.
Key idea
To evaluate a polynomial at a matrix, replace \(X^k\) by \(A^k\), replace \(X\) by \(A\), and replace a scalar constant \(c\) by \(cI\). Thus \(q(X)=X^2-3X+2\) gives \(q(A)=A^2-3A+2I\). If this equals \(0\), then \(q\) annihilates \(A\). Similarly, \(A^2-4A+3I=0\) comes from \(X^2-4X+3\) and rearranges to \(A^2=4A-3I\).
Recognition checklist
Move all terms to one side so the right side is \(0\).
Replace \(A^k\) by \(X^k\), \(A\) by \(X\), and \(cI\) by \(c\).
Factor the polynomial when useful.
Remember: a polynomial can cancel \(A\) without being minimal.
Worked example
Example: If \(A^2=A\), which polynomial cancels \(A\)?
Move terms to one side: \(A^2-A=0\). Therefore \(q(X)=X^2-X=X(X-1)\) satisfies \(q(A)=0\). The minimal polynomial divides \(X(X-1)\), and if \(A\) is neither \(0\) nor \(I\), both factors are needed.
Try it
Try it 1: If \(A^2-3A+2I=0\), which polynomial cancels \(A\)?
Hint: The constant term \(2\) becomes \(2I\) after substitution.
Try it 2: What is the minimal polynomial of \(A=3I\)?
Hint: \(A-3I=0\), and no lower-degree monic polynomial cancels \(A\).
The minimal polynomial divides every cancellation
Learning goal: Use divisibility to narrow down possible minimal polynomials before doing any heavy computation.
Key idea
The set of polynomials that annihilate \(A\) is an ideal in the polynomial ring. Its monic generator is \(m_A\). Therefore, whenever \(q(A)=0\), the minimal polynomial divides \(q\). Cayley-Hamilton says \(p_A(A)=0\), so \(m_A\mid p_A\). This gives a small list of candidates from the factors of \(p_A\).
Worked example
Example: If \(p_A(X)=(X-2)^3\), what can \(m_A\) look like?
The minimal polynomial must divide \((X-2)^3\), and it must have \(2\) as a root if \(2\) is an eigenvalue. The candidates are \(X-2\), \((X-2)^2\), or \((X-2)^3\). Which one is correct depends on how large the nilpotent part is.
Common traps
\(m_A\mid p_A\), not usually \(p_A\mid m_A\).
The minimal polynomial can be equal to the characteristic polynomial, but it can also be much smaller.
A candidate minimal polynomial must still annihilate \(A\).
Roots of \(m_A\) are exactly the eigenvalues of \(A\) over a splitting field.
Try it
Try it 1: If \(q(A)=0\), what must the minimal polynomial \(m_A\) do to \(q\)?
Hint: The minimal polynomial is the monic generator of all polynomial relations satisfied by \(A\).
Try it 2: If \(p_A(X)=(X-2)^3\), what must \(m_A\) divide?
Hint: Cayley-Hamilton makes the characteristic polynomial one annihilating polynomial.
Use \(p_A(A)=0\) to reduce powers
Learning goal: Convert a characteristic polynomial into a matrix equation and use it for reductions.
Key idea
If \(p_A(X)=X^2-5X+6\), then Cayley-Hamilton gives \(A^2-5A+6I=0\). This lets you replace \(A^2\) by \(5A-6I\). For higher powers, multiply the relation by \(A\) and reduce again. If the constant term is nonzero and \(A\) is invertible, the same equation can often give a formula for \(A^{-1}\).
Worked example
Example: Suppose \(A^2-3A+2I=0\) and \(A\) is invertible. Express \(A^{-1}\) using \(A\) and \(I\).
Multiply the equation by \(A^{-1}\): \(A-3I+2A^{-1}=0\). Hence \(2A^{-1}=3I-A\), so \(A^{-1}=\dfrac{3I-A}{2}\).
Try it
Try it 1: If \(p_A(X)=X^2-5X+6\), what equation does Cayley-Hamilton give?
Hint: The scalar constant \(6\) becomes \(6I\).
Try it 2: If \(A^2-3A+2I=0\) and \(A\) is invertible, which relation follows?
Hint: Multiply the equation by \(A^{-1}\) and isolate the inverse term.
Repeated factors are the obstruction
Learning goal: Use the minimal polynomial, not just the characteristic polynomial, to decide diagonalizability.
Key idea
A matrix is diagonalizable over a field exactly when its minimal polynomial splits into distinct linear factors over that field. If \(A\) is diagonalizable with distinct eigenvalues \(\lambda_1,\dots,\lambda_r\), then \(m_A(X)=(X-\lambda_1)\cdots(X-\lambda_r)\). A repeated factor such as \((X-\lambda)^2\) means the matrix has nontrivial nilpotent behavior on the \(\lambda\)-part, so it is not diagonalizable.
Worked example
Example: If \(A\) is diagonalizable with eigenvalues \(1\) and \(2\), what is \(m_A(X)\)?
For a diagonalizable matrix, each distinct eigenvalue appears exactly once in the minimal polynomial. Therefore \(m_A(X)=(X-1)(X-2)\), regardless of repeated entries in the characteristic polynomial.
Try it
Try it 1: If \(A\) is diagonalizable with eigenvalues \(1,2,3\), what is \(m_A(X)\)?
Hint: Use each distinct eigenvalue once.
Try it 2: If \(m_A(X)=(X-1)^2\), what does this suggest over a split field?
Hint: Diagonalizable matrices have minimal polynomials with no repeated factor.
Nilpotent, projection, and involution examples
Learning goal: Recognize minimal polynomials from common matrix relations.
Key idea
Small polynomial identities often determine \(m_A\) almost completely. If \(A^k=0\), then \(A\) is nilpotent and \(m_A\) is a power of \(X\). If \(A^2=A\), then \(m_A\mid X(X-1)\). If \(A^2=I\), then \(m_A\mid (X-1)(X+1)\). If \(A^2=4I\), then \(m_A\mid (X-2)(X+2)\), so possible eigenvalues are \(2\) and \(-2\). The minimal polynomial is the smallest monic divisor that still fits the matrix.
Worked example
Example: If \(A^2=I\) and \(A≠ I,-I\), find \(m_A(X)\).
The relation \(A^2-I=0\) gives \((A-I)(A+I)=0\), so \(m_A\mid (X-1)(X+1)\). Since \(A≠ I\), \(X-1\) alone does not cancel \(A\). Since \(A≠ -I\), \(X+1\) alone does not cancel \(A\). Thus \(m_A(X)=(X-1)(X+1)\).
Try it
Try it 1: If \(A≠0\) and \(A^2=0\), what is \(m_A(X)\)?
Hint: \(X\) alone would force \(A=0\), so the square is needed.
Try it 2: If \(A^2=A\) and \(A\) is neither \(0\) nor \(I\), what is \(m_A(X)\)?
Hint: Both roots \(0\) and \(1\) are needed in the nontrivial projection case.
Summary
Scalar \(A=\lambda I\): \(m_A=X-\lambda\), so \(I\) gives \(X-1\) and \(-I\) gives \(X+1\).
Zero matrix: \(m_A=X\).
Nonzero square-zero matrix: \(m_A=X^2\).
Nontrivial projection: \(m_A=X(X-1)\).
Nontrivial involution: \(m_A=(X-1)(X+1)\).
The field changes what splitting means
Learning goal: Track which field is being used and avoid confusing real roots with complex roots.
Key idea
The minimal polynomial can fail to split over the field you are working in. For example, \(X^2+1\) has no real root, so a real matrix with minimal polynomial \(X^2+1\) has no real eigenvalues and cannot be diagonalized over \(\mathbb{R}\). Over \(\mathbb{C}\), it splits as \((X-i)(X+i)\), and the diagonalization question changes.
Common traps
Always ask whether the field is \(\mathbb{R}\), \(\mathbb{C}\), or another field.
Roots of \(m_A\) give eigenvalues only after the polynomial is viewed over a field where roots may exist.
A degree \(1\) minimal polynomial \(X-\lambda\) means \(A=\lambda I\).
A repeated factor blocks diagonalization only after the polynomial has split into linear factors.
Worked example
Example: Over \(\mathbb{R}\), what does \(m_A(X)=X^2+1\) say about real eigenvalues?
Real eigenvalues must be real roots of the minimal polynomial. Since \(X^2+1\) has no real roots, \(A\) has no real eigenvalues. Over \(\mathbb{C}\), the roots would be \(i\) and \(-i\).
Try it
Try it 1: If \(m_A(X)=X^2+1\) over \(\mathbb{R}\), what follows?
Hint: Real eigenvalues would have to be real roots of \(X^2+1\).
Try it 2: If the minimal polynomial has degree \(1\), then \(A\) is:
Hint: \(m_A=X-\lambda\) means \(A-\lambda I=0\).
A reliable checklist for quiz problems
Learning goal: Finish with a compact method for deciding what the problem is asking and which polynomial relation to use.
Method
Identify the given relation: characteristic polynomial, minimal polynomial, or direct equation in \(A\).
If given \(p_A\), write \(p_A(A)=0\) and remember constants multiply \(I\).
If given \(q(A)=0\), conclude \(m_A\mid q\).
For a relation such as \(A^3=A\) or \(A^2=4I\), move all terms to \(0\), then factor the matching polynomial.
If asked for eigenvalues, look at roots of \(m_A\) or \(p_A\).
If asked about diagonalizability, inspect the split minimal polynomial for repeated factors.
If asked to reduce powers or find \(A^{-1}\), rearrange the polynomial equation.
Worked example
Example: Suppose \(m_A(X)=X(X-2)\). Is \(A\) diagonalizable?
The minimal polynomial splits and has no repeated factor. Therefore \(A\) is diagonalizable over the field containing \(0\) and \(2\). The possible eigenvalues are \(0\) and \(2\), but each appears only once in \(m_A\).
Try it
Try it 1: If \(m_A(X)=X(X-2)\), is \(A\) diagonalizable?
Hint: A split minimal polynomial with no repeated factor is the diagonalizability test.
Try it 2: If \(A^3=A\), which polynomial cancels \(A\)?
Hint: Move all terms to one side.
Final recap
Cayley-Hamilton: \(p_A(A)=0\).
The minimal polynomial \(m_A\) is the monic polynomial of least degree with \(m_A(A)=0\).
If \(q(A)=0\), then \(m_A\mid q\); in particular \(m_A\mid p_A\).
For diagonalizable matrices, \(m_A\) is the product of distinct linear factors for the distinct eigenvalues.
A repeated factor in a split minimal polynomial prevents diagonalizability.
Polynomial equations reduce powers and can produce inverse formulas.
Field choice matters for splitting and eigenvalues.
Next step: Close this lesson and try the quiz again. For each item, decide first whether you are using Cayley-Hamilton, the definition of \(m_A\), a divisibility fact, or the square-free diagonalizability test.
Practice set
Minimale veeltermen en Cayley-Hamilton 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.
0/10answered
Question 1Not answered
Wat zegt de stelling van Cayley-Hamilton over een vierkante matrix \(A\)?
Correct answer: A. \(p_A(A)=0\)
Explanation: Een vierkante matrix voldoet aan haar eigen karakteristieke veelterm.
Question 2Not answered
De minimale veelterm van \(A\) is:
Correct answer: A. De monische veelterm van kleinste graad waarvoor \(m(A)=0\)
Explanation: Dat is de monische veelterm van kleinste graad waarvoor \(m(A)=0\).
Question 3Not answered
Hoe verhoudt de minimale veelterm zich tot de karakteristieke veelterm?
Correct answer: D. Het deelt die altijd
Explanation: De minimale veelterm deelt altijd de karakteristieke veelterm.
Question 4Not answered
Wat is de minimale veelterm van \(A=3I\)?
Correct answer: B. \(X-3\)
Explanation: Omdat \(A-3I=0\), is de kleinste monische veelterm \(X-3\).
Question 5Not answered
Als \(A\) diagonaliseerbaar is met eigenwaarden \(1\) en \(2\), wat is dan de minimale veelterm?
Correct answer: D. \((X-1)(X-2)\)
Explanation: Bij een diagonaliseerbare matrix komt elke verschillende eigenwaarde één keer voor in de minimale veelterm.
Question 6Not answered
Als de minimale veelterm van \(A\) gelijk is aan \(X^2\), wat moet dan gelden?
Correct answer: B. \(A^2=0\) en \(A\ne0\)
Explanation: Invullen van \(A\) geeft \(A^2=0\), en de graad laat zien dat \(A\ne0\).
Question 7Not answered
Als een matrix een gesplitste minimale veelterm heeft zonder herhaalde factor, wat volgt daaruit?
Correct answer: B. De matrix is diagonaliseerbaar
Explanation: Een gesplitste minimale veelterm met enkelvoudige nulpunten kenmerkt diagonaliseerbaarheid.
Question 8Not answered
Wat is de minimale veelterm van de nulmatrix?
Correct answer: B. \(X\)
Explanation: De nulmatrix wordt opgeheven door \(X\), en geen monische veelterm van lagere graad werkt.
Question 9Not answered
Als \(p_A(X)=X^2-5X+6\), welke vergelijking geeft Cayley-Hamilton dan?
Correct answer: C. \(A^2-5A+6I=0\)
Explanation: Vervang \(X\) door \(A\) en de constante term door een veelvoud van \(I\).
Question 10Not answered
Als \(A\) diagonaliseerbaar is en alleen de eigenwaarde \(4\) heeft, wat is \(A\)?
Correct answer: C. \(4I\)
Explanation: In een eigenbasis is de diagonale matrix \(4I\), dus \(A=4I\).
