The characteristic polynomial (Definition 7.4.7) encodes the eigenvalues and their algebraic multiplicities. In this section we show that every operator satisfies its own characteristic polynomial, and we study the “smallest” polynomial relation satisfied by an operator.
7.9.1 The Cayley–Hamilton Theorem
Theorem 7.9.1 (Cayley–Hamilton). Let be a finite-dimensional vector space over any field , and let with characteristic polynomial . Then That is, substituting for in gives the zero operator. Equivalently, for every square matrix .
Proof. We give two proofs. The first works over an algebraically closed field and explains the theorem; the second works over every field.
Proof 1 ( algebraically closed): Let be the distinct eigenvalues of and . By Theorem 7.6.14, , so and the factors commute (Proposition 7.2.2). Let . The restriction of to is a nilpotent operator on a space of dimension , so its -th power is zero (Proposition 7.7.2). Applying the factor first, we get . Since (Theorem 7.6.11), .
Proof 2 (any field, using the adjugate): Let for a basis . Sending an operator to its matrix in is a ring isomorphism (Corollary 5.2.10), and it sends to ; so it suffices to show .
Let , and let be its adjugate. The adjugate identity (Theorem 5.4.16), which holds for matrices over the commutative ring (Example 5.4.25), gives Each entry of is the determinant of an submatrix of , which by the Leibniz formula is a polynomial in of degree at most . So we can write Write also (with ), and set . Expanding, Both sides of the adjugate identity are matrices of polynomials, so their coefficients agree: Multiply the -th equation on the left by and add: The sum telescopes: the term subtracted at index is added back at index , so all that remains is . Hence . ∎
Note. The Cayley–Hamilton Theorem does not follow from “plugging in for in .” That substitution produces the scalar , while is a matrix; the two have nothing to do with each other a priori. The theorem requires proof.
Proof 2 is not the only route to arbitrary fields. Every field is contained in an algebraically closed field (a theorem we do not prove here), and the polynomial and the matrix are computed the same way whether the entries of are regarded in or in the larger field. So Proof 1, applied over the larger field, also gives .
Corollary 7.9.2 (Expression for the inverse). If is invertible, then can be expressed as a polynomial in .
Proof. Write . Then (Proposition 7.4.8(a) and Theorem 5.4.8).
By Cayley–Hamilton: .
Rearranging: .
Thus . ∎
7.9.2 The Minimal Polynomial
Definition 7.9.3 (Minimal polynomial). The minimal polynomial of , finite-dimensional, is the monic polynomial of smallest degree such that . (It exists and is unique by the next proposition.) For a square matrix , is defined in the same way.
Proposition 7.9.4 (Existence, uniqueness, and divisibility). Let with finite-dimensional.
- There exists a nonzero polynomial with .
- Among all such polynomials, there is a unique monic one of minimal degree.
- divides every polynomial satisfying .
Proof. (a) Cayley–Hamilton gives , and is monic, hence nonzero.
(b) Let be the minimal degree of a nonzero polynomial annihilating ; dividing by the leading coefficient, there is a monic one. If and are both monic of degree with , then and . By minimality, .
(c) Given with , divide by with remainder: where or . (Since is monic of degree , if has degree and leading coefficient , then has smaller degree; repeating this produces and .) Then . By minimality of , we must have , so . ∎
Theorem 7.9.5 (Characteristic and minimal polynomials). Let , finite-dimensional, over any field .
- divides .
- For : is an eigenvalue of . So and have the same roots in .
Proof. (a) By Cayley–Hamilton, . By Proposition 7.9.4(c), .
(b) The second equivalence is Proposition 7.4.8(d). If is an eigenvalue, there exists with . For any polynomial , . Taking gives , so . Conversely, if , then because divides . ∎
Over an algebraically closed field, the minimal polynomial can be read off from the Jordan form.
Proposition 7.9.6 (Minimal polynomial and Jordan block sizes). Let be algebraically closed, , and let be the distinct eigenvalues of . For each , let be the index of nilpotency of , i.e., the smallest with . Then and is the size of the largest Jordan block with eigenvalue .
Proof. Let . On , the factor is zero; since the factors of commute, vanishes on each , hence on . Thus .
Now fix and write with (divide by as long as possible, Lemma 4.3.14). Since is algebraically closed, is a constant times a product of factors with (Proposition 4.3.17), and each is invertible on the invariant subspace (Proposition 7.6.8(d)); so is invertible on . From we get on , so . Factoring into linear factors and using Lemma 7.4.11, occurs at least times for each , so . As divides and both are monic, .
We have because , and by Proposition 7.7.2. Finally, in a Jordan basis the blocks with eigenvalue form a chain basis of for (proof of Theorem 7.8.5), and kills a chain of length if and only if . So the index of nilpotency of is the length of the longest chain, i.e., the size of the largest block. ∎
Example 7.9.7 (Minimal polynomial examples).
- For : , but since .
- For : . Since , we need .
- For : , but since and, by Theorem 7.9.5(b), both and must be roots of .
7.9.3 Diagonalizability and the Minimal Polynomial
The minimal polynomial detects diagonalizability over every field. The proof uses a dimension inequality.
Lemma 7.9.8 (Kernel of a product). Let , finite-dimensional. Then . Consequently, .
Proof. Let . The map sends into , and the kernel of is . By Rank–Nullity for , The second statement follows by induction on . ∎
Theorem 7.9.9 (Diagonalizability via the minimal polynomial). Let , finite-dimensional, over any field . Then is diagonalizable if and only if is a product of distinct linear factors: with distinct.
Proof. () Let be diagonalizable with distinct eigenvalues , and let . Then : on the factor vanishes, the factors commute, and these eigenspaces span . Hence . On the other hand, every is a root of by Theorem 7.9.5(b). Write (Lemma 4.3.14); for , gives , and repeating, divides . Two monic polynomials that divide each other are equal, so .
() Suppose with distinct . By Theorem 7.9.5(b), the are exactly the eigenvalues of . Since , Lemma 7.9.8 gives The reverse inequality holds by Corollary 7.4.6, so , and is diagonalizable by Theorem 7.4.13. ∎
Note. Over an algebraically closed field this can also be read off from the Jordan form: by Proposition 7.9.6, has distinct linear factors if and only if every , i.e., if and only if all Jordan blocks have size .
7.9.4 Exercises
Question 7.9.10 (Exercises on Jordan form and minimal polynomials).
- Let have characteristic polynomial . List all possible Jordan normal forms for .
- Prove that if and are similar matrices, then .
- Find the Jordan normal form and minimal polynomial of
- Let satisfy , with finite-dimensional over any field. Prove that is diagonalizable and find its possible eigenvalues.
- Suppose has minimal polynomial . What can you conclude about ?
Show solution
(a) The eigenvalues are and , each with algebraic multiplicity 2, so for each eigenvalue the Jordan blocks have sizes summing to 2. Possibilities for each:
- One block of size 2:
- Two blocks of size 1:
Combining independently for each eigenvalue, the possible Jordan forms (up to the order of the blocks) are:
- (diagonal matrix)
(b) If , then for all , so for any polynomial : So iff . The two matrices are annihilated by the same polynomials, so the monic annihilating polynomial of least degree is the same for both.
(c) The matrix is already in Jordan form: .
Characteristic polynomial: .
Minimal polynomial: We need the smallest with . Compute and . So , in agreement with Proposition 7.9.6 (one block, of size 3).
(d) The condition means . Thus annihilates , and divides .
A monic divisor of of degree 1 has the form with a root of , so the monic divisors are , , , and . Each is a product of distinct linear factors. By Theorem 7.9.9, is diagonalizable.
The eigenvalues are roots of (Theorem 7.9.5), which divides . So the eigenvalues are among .
(Such operators are called projections or idempotents. They decompose and project onto along ; compare the exercises of Section 6.1.)
(e) Since is a product of three distinct linear factors:
- is diagonalizable (Theorem 7.9.9).
- The eigenvalues are exactly (the roots of , by Theorem 7.9.5).
- (at least one eigenvector for each eigenvalue, and these are linearly independent).
- .
The eigenspace dimensions can be any positive integers summing to ; each is at least 1.