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Polynomials Associated with a Linear
Operator
Definition 1: Annihilating Polynomial Let T : V → V be a linear operator on a vector space V over a field F. A polynomial p(x) ∈ F[x] is called an annihilating polynomial of T if
p(T ) = 0.
It “annihilates” the operator when substituted into it. There can be infinitely many annihilating polynomials for a given operator.
Definition 2: Characteristic Polynomial For a square matrix A representing T , the characteristic polynomial of T (or A) is defined as χT (x) = det(xI − A).
Its roots are the eigenvalues of T. It is always monic and has degree dim V.
Definition 3: Minimal Polynomial The minimal polynomial mT (x) of T is the monic polynomial of least degree such that mT (T ) = 0.
It divides every other annihilating polynomial and contains all eigenvalues with the largest Jordan block multiplicity.
5 Key Differences Between Them
- Existence: Annihilating polynomials: infinite, characteristic: unique, minimal: unique.
- Degree: characteristic = dim V , minimal ≤ dim V , annihilating ≥ minimal degree.
- Monicity: characteristic and minimal are always monic; annihilating need not be.
- Roots: characteristic: eigenvalues with algebraic multiplicity, min- imal: eigenvalues with largest Jordan block size, annihilating: may include eigenvalues with arbitrary multiplicity.
- Divisibility: minimal divides every annihilating polynomial; charac- teristic is divisible by minimal.
Example for Understanding
Consider the matrix A =
- Characteristic polynomial:
χA(x) = det(xI − A) = det
x − 2 − 1 0 x − 2
= (x − 2)^2
- Minimal polynomial: The minimal polynomial must annihilate A and be monic. Here, mA(x) = (x − 2)^2
- Annihilating polynomial: Any polynomial divisible by (x − 2)^2 annihi- lates A, e.g.
p(x) = (x − 2)^3 , q(x) = (x − 2)^2 (x + 1)
Observation: - The characteristic polynomial gives eigenvalues 2 with algebraic multiplicity 2. - Minimal polynomial also reflects the largest Jordan block size (2 here). - Annihilating polynomials can be many, as long as they contain the minimal polynomial as a factor.
Algebraic Multiplicity (AM) and
Geometric Multiplicity (GM) of
Eigenvalues
- An eigenvalue may appear multiple times (AM¿1) but have only 1 independent eigenvector (GM=1).
- For distinct eigenvalues, GM = AM = 1 automatically.
- Example
A =
- Characteristic polynomial:
χA(x) = det(xI − A) = (x − 2)^2 =⇒ AM(2) = 2
- Eigenspace: Solve (A − 2 I)v = 0: 0 1 0 0
x y
=⇒ y = 0
So eigenspace: span{(1, 0)T^ }, hence GM(2) = 1
- Observation: AM > GM =⇒ A is not diagonalizable.
- Summary Table Eigenvalue AM (root multiplicity) GM (dimension of eigenspace) 2 2 1
Key takeaway: GM ≤ AM, and equality for all eigenvalues ⇐⇒ matrix diagonalizable.
Eigenspace of a Matrix or Linear
Operator
- Definition of Eigenspace Let A be a square matrix of order n (or T : V → V a linear operator) and let λ be an eigenvalue of A. The eigenspace corresponding to λ is the set of all eigenvectors associated with λ, together with the zero vector:
Eλ = {v ∈ Fn^ : (A − λI)v = 0}
It is a subspace of Fn, and its dimension is called the geometric multiplic- ity of λ.
- Properties of Eigenspace
- Eigenspace Eλ is always a subspace of Fn.
- Dimension of Eλ = geometric multiplicity of λ: 1 ≤ dim(Eλ) ≤ alge- braic multiplicity.
- If λ 1 , λ 2 ,... , λk are distinct eigenvalues, their eigenspaces intersect triv- ially: Eλi ∩ Eλj = { 0 } for i ̸= j.
- A matrix is diagonalizable ⇐⇒ the sum of dimensions of all eigenspaces = n.
- All eigenvectors corresponding to λ form a basis of Eλ.
- Hidden Concepts / Important Points
- Eigenspace contains the zero vector even though zero is not considered an eigenvector.
- Geometric multiplicity = number of linearly independent eigenvectors for a given eigenvalue.
- AM = GM for all eigenvalues =⇒ matrix diagonalizable. If AM > GM for any eigenvalue, matrix is defective.
- The dimension of an eigenspace determines how “much freedom” we have in choosing eigenvectors for that eigenvalue.
- Eigenspaces of distinct eigenvalues are linearly independent subspaces.
- Example
A =
Each Jki (λ) is a Jordan block of size ki:
Jk(λ) =
λ 1 0 · · · 0 0 λ 1 · · · 0 0 0 λ
0 0 · · · 0 λ
k×k —
- Rules and Conditions
- Every eigenvalue λ appears on the diagonal of JCF.
- The number of Jordan blocks corresponding to λ = geometric multi- plicity (GM) of λ.
- The sum of sizes of Jordan blocks for λ = algebraic multiplicity (AM) of λ.
- If AM = GM for all eigenvalues, then JCF = diagonal matrix.
- Size of largest Jordan block for λ = degree of λ in the minimal poly- nomial.
—
- Stepwise Method to Find JCF
- Find the characteristic polynomial χA(λ) = det(A − λI).
- Find all eigenvalues and their algebraic multiplicities (AM).
- For each eigenvalue λ, find the nullity of (A − λI) =⇒ geometric multiplicity (GM).
- Construct Jordan blocks: - Number of blocks = GM. - Total size of blocks = AM.
- Determine sizes of blocks by studying dim ker((A − λI)k) for k = 1 , 2 ,... until it stabilizes.
- Assemble blocks in block-diagonal form = JCF.
- Important Points and Hidden Concepts
- JCF is unique up to reordering of Jordan blocks.
- The minimal polynomial of A tells the largest Jordan block size for each eigenvalue.
- Diagonalizability ⇐⇒ every Jordan block is 1 × 1.
- JCF gives complete structural information about a linear operator: spectrum, AM, GM, minimal polynomial.
- Even if A is not diagonalizable, JCF provides the “nearest diagonal form”.
—
- Example
A =
Step 1: Characteristic polynomial:
det(A − λI) = (5 − λ)^3
Eigenvalue: λ = 5, Algebraic multiplicity = 3. Step 2: Geometric multiplicity: Solve (A − 5 I)v = 0:
(A − 5 I) =
Rank = 2, so nullity = 1. Thus GM = 1. Step 3: Jordan block structure: AM = 3, GM = 1 =⇒ only 1 block of size 3. So JCF is:
J =
- Summary
- Algebraic multiplicities (AM) must satisfy
AM(1) + AM(2) = 4,
with AM(1) ≥ GM(1) = 1 and AM(2) ≥ GM(2) = 2. The only possibilities are
(AM(1), AM(2)) = (1, 3) or (2, 2).
(Values like (3, 1) are impossible because AM(2) ≥ 2.)
- Now determine Jordan block structures consistent with the AM and GM data. Recall: - The number of Jordan blocks for eigenvalue λ equals GM(λ). - The sum of sizes of those blocks equals AM(λ).
- Case 1: (AM(1), AM(2)) = (1, 3). Since GM(1) = 1 and AM(1) = 1, the blocks for eigenvalue 1 consist of one 1 × 1 block J 1 (1) = [1]. For eigenvalue 2: GM(2) = 2 so there are two Jordan blocks whose sizes sum to 3. The only possibility (up to ordering) is block sizes 2 and 1. Thus the Jordan form is
J ∼ diag
J 1 (1), J 2 (2), J 1 (2)
i.e. (upper-triangular) matrix
J =
(rows/columns permuted so blocks appear in this order).
- Case 2: (AM(1), AM(2)) = (2, 2). Here GM(1) = 1 but AM(1) = 2, so there must be a single Jordan block of size 2 for eigenvalue 1 (one block because GM=1, total size=2). For eigenvalue 2, AM(2) = 2 and GM(2) = 2, so there are two 1 × 1 blocks for 2. Thus the Jordan form is
J ∼ diag
J 2 (1), J 1 (2), J 1 (2)
i.e.
J =
(after suitable permutation of basis).
Final Answer (all possible upper-triangular Jordan forms).
Exactly two similarity-types (up to reordering of Jordan blocks) are possible:
diag
J 1 (1), J 2 (2), J 1 (2)
or diag
J 2 (1), J 1 (2), J 1 (2)
(Here Jk(λ) denotes the k × k Jordan block with λ on the diagonal and 1’s on the superdiagonal.)
Remark. Which of these two actually occurs depends on the algebraic multiplicities; the data given (only GM for λ = 1 and range-dimension for T − 2 I) permit both possibilities.
Question & Solution
Problem. Let M be a 7 × 7 real matrix with characteristic polynomial
cM (x) = (x − 1)α(x − 2)β^ (x − 3)^2 , α > β.
Suppose
rank(M − I 7 ) = rank(M − 2 I 7 ) = rank(M − 3 I 7 ) = 5.
If mM (x) is the minimal polynomial of M , find the value of mM (5).
Solution.
We proceed stepwise.
- From the ranks given, for each λ ∈ { 1 , 2 , 3 } we have
dim ker(M − λI) = 7 − rank(M − λI) = 7 − 5 = 2.
Hence the geometric multiplicities are
GM(1) = GM(2) = GM(3) = 2.
Question & Answer (with
explanation)
Problem. Let T : C^7 → C^7 be a C-linear operator whose eigenvalues are 2, 3 , 5. Define
W := { v ∈ C^7 : (T − 5 I)kv = 0 for some k > 0 }
(so W is the generalized eigenspace for the eigenvalue 5). Suppose
(T − 2 I)^2 (T − 3 I)^2 (T − 5 I)^2 = 0.
Which of the following statements are necessarily true?
- T has at least four linearly independent eigenvectors.
- dim W ≥ 2.
- ker((T − 2 I)^2025 ) = ker((T − 2 I)^2026 ).
- (T − 2 I)(T − 3 I) is a nilpotent operator.
Solution and reasoning. Write the algebraic multiplicities of the eigenvalues 2, 3 , 5 as a 2 , a 3 , a 5. They are positive integers with
a 2 + a 3 + a 5 = 7.
The relation (T − 2 I)^2 (T − 3 I)^2 (T − 5 I)^2 = 0 implies that the size of every Jordan block for each eigenvalue is at most 2. In other words, for each eigenvalue λ ∈ { 2 , 3 , 5 } every Jordan block has size 1 or 2.
Key observation (blocks vs geometric multiplicity). If an eigenvalue λ has algebraic multiplicity aλ and each Jordan block has size ≤ 2, then the minimum possible number of Jordan blocks for λ is la λ 2
m ,
because we can pack at most two eigenvalues into a single 2×2 Jordan block. The geometric multiplicity (number of linearly independent eigenvectors) for λ equals the number of Jordan blocks for λ.
(1) T has at least four linearly independent eigenvectors. — True. The total number of linearly independent eigenvectors (sum of geometric multiplicities) is at least (^) X
λ∈{ 2 , 3 , 5 }
la λ 2
m .
We must minimize this sum subject to a 2 + a 3 + a 5 = 7 and ai ≥ 1. Evaluate possibilities: (a 2 , a 3 , a 5 )
P
⌈ai/ 2 ⌉ example (3, 2 , 2) 2 + 1 + 1 = 4 gives 4 (4, 2 , 1) 2 + 1 + 1 = 4 gives 4 (5, 1 , 1) 3 + 1 + 1 = 5 gives 5
One checks there is no distribution summing to 7 with all ai ≥ 1 that yields a sum < 4. Hence the total geometric multiplicity is ≥ 4. Thus statement (1) is necessarily true.
(2) dim W ≥ 2. — Not necessarily true. The dimension dim W equals the algebraic multiplicity a 5 of the eigenvalue
- From a 2 + a 3 + a 5 = 7 with each ai ≥ 1, it is possible to have a 5 = 1 (for instance try (a 2 , a 3 , a 5 ) = (4, 2 , 1)). That choice is compatible with the block-size constraint (one can take two 2×2 blocks for eigenvalue 2, one 2× 2 block for 3, and one 1 × 1 block for 5, etc.). Hence dim W might be 1, so (2) is not necessarily true. (Counterexample: choose algebraic multiplicities (4, 2 , 1).)
(3) ker((T − 2 I)^2025 ) = ker((T − 2 I)^2026 ). — True. For any eigenvalue λ, the chain of subspaces ker((T − λI)k) stabilizes once k reaches the size of the largest Jordan block for λ. Here every Jordan block for eigenvalue 2 has size ≤ 2, so the kernel stabilizes at exponent 2. Therefore for all integers m ≥ 2 we have
ker((T − 2 I)m) = ker((T − 2 I)^2 ).
In particular, ker((T − 2 I)^2025 ) = ker((T − 2 I)^2026 ). Thus (3) is necessarily true.
(4) (T − 2 I)(T − 3 I) is nilpotent. — Not necessarily true. The operator (T − 2 I)(T − 3 I) is a polynomial in T. An operator is nilpotent iff 0 is its only eigenvalue. If v is an eigenvector of T with eigenvalue λ ∈ { 2 , 3 , 5 }, then (T − 2 I)(T − 3 I)v = (λ − 2)(λ − 3) v.
- Step 1: Compute eigenvalues - Solve det(λI −A) = 0 - Eigenvalues and their algebraic multiplicities must match.
- Step 2: Compute geometric multiplicities - For each eigenvalue λ, compute dim ker(A − λI) - Number of Jordan blocks for λ = geometric multiplicity.
- Step 3: Determine nilpotency / Jordan block sizes - For each eigenvalue λ, compute the sequence:
dim ker((A − λI)k), k = 1, 2 ,...
- The ”jumps” in dimensions determine the sizes of Jordan blocks. - For nilpotent matrices (all eigenvalues = 0), the largest Jordan block size = nilpotency index (smallest p such that Ap^ = 0).
- Step 4: Check minimal polynomial - Minimal polynomial gives the size of the largest Jordan block for each eigenvalue.
- Step 5: Compare rank (optional check) - Rank can help quickly identify Jordan structure for nilpotent matrices.
— Solution: Step-by-step check of options Target matrix T : - Eigenvalues = 0 (AM = 3) - Nilpotency index = 2 (T 2 = 0) - Rank = 1
(^) M 12 = 0, rank = 1 → Same JCF
(^) M 22 = 0, rank = 1 → Same JCF
(^) M 32 = 0, rank = 1 → Same JCF
(^) M 42 ̸= 0, nilpotency index = 3 → Does
NOT match JCF
Conclusion:
Matrices with the same Jordan canonical form as T are: M 1 , M 2 , M 3.
Exam Tip:
- For nilpotent matrices, always check Ap^ = 0 to determine the largest Jordan block.
- Eigenvalues alone are not enough to determine JCF.
- Rank and minimal polynomial help quickly verify JCF for nilpotent matrices.
Question & Detailed Solution
Question. Let T : R^4 → R^4 be a linear transformation with
χT (x) = (x − 2)^4 , mT (x) = (x − 2)^2.
Which one of the following matrices can represent the Jordan canonical form of T?
(A)
(two 2 × 2 Jordan blocks)
(B)
(one 2 × 2 block and two 1 × 1 blocks)
This is exactly two 2 × 2 Jordan blocks J 2 (2) ⊕ J 2 (2). - Largest block size = 2 ⇒ minimal polynomial has factor (x − 2)^2. - Algebraic multiplicity = 4 (two blocks of size 2). Hence (A) is valid.
(B) (^)
This is one 2 × 2 block plus two 1 × 1 blocks: J 2 (2) ⊕ 2 ⊕ 2. - Largest block size = 2 ⇒ minimal polynomial (x − 2)^2. - Algebraic multiplicity = 4. Hence (B) is valid.
(C) (^)
This is a single 4 × 4 Jordan block J 4 (2). - Largest block size = 4 ⇒ minimal polynomial would be (x − 2)^4 , which contradicts mT = (x − 2)^2. Hence (C) is not valid.
(D)
2 I 4 =
This is four 1 × 1 Jordan blocks. - Largest block size = 1 ⇒ minimal polyno- mial would be (x − 2), contradicting mT = (x − 2)^2. Hence (D) is not valid.
Final answer: (A) and (B) only.
Exam tips / hidden checks:
- χT fixes algebraic multiplicities; mT fixes the maximal Jordan-block size.
- Compare degrees: deg mT = size of largest Jordan block for each eigen- value.
- If the minimal polynomial degree equals the degree of the characteristic polynomial for that eigenvalue, a single largest block of that size is forced.
- Always rule out (i) a single large block that would make deg mT too big, and (ii) a purely diagonal form that would make deg mT too small.
