Try
A matrix can describe both an action and a list of relations. Over F=Q take A with rows (0,−2),(1,3), acting on V=Q². Its columns say Ae₁=e₂ and Ae₂=−2e₁+3e₂. Introduce a formal symbol t for “apply A.” The same statements become the relations te₁−e₂=0 and 2e₁+(t−3)e₂=0.
Put these relations in columns. Their matrix is M(t)=tI−A, with rows (t,2),(−1,t−3). The ambient space is now Q[t]², whose entries are polynomials, and the relation subgroup is M(t)Q[t]². Predict how reducing polynomial vectors modulo these columns could recover the original two-dimensional vector space with its operator.
Predict
Throughout the general proof F is any field, n is a nonnegative integer and A is an n-by-n matrix over F. Write V_A for Fⁿ regarded as an F[t]-module by p(t)·v=p(A)v. This is the operator-module construction. The symbols t and A are not interchangeable kinds of objects: t is a polynomial scalar in the presentation, while A is the operator defining scalar action on the target.
Every p∈F[t]ⁿ has a unique finite expression p(t)=Σ from k=0 to d of tᵏvₖ, with coefficient vectors vₖ∈Fⁿ. Define E(p)=Σ Aᵏvₖ. This evaluates a polynomial vector by applying powers of A to its vector coefficients. Predict E(te₁−e₂) and E(2e₁+(t−3)e₂) in the worked example. Both are zero, exactly because they encode the two matrix columns.
The difficult direction is the converse: do these n relations generate every polynomial vector that evaluates to zero? Checking only the listed columns does not answer it. We will produce an explicit relation coefficient vector for every element of the kernel.
Experiment
For the worked A, compute A², then evaluate p(t)=t²e₁+te₂+(1,−1). Keep vector coefficients and polynomial coordinates separate.
| Piece of p | Its value under E |
|---|
| t²e₁ | A²e₁=(−2,3) |
| te₂ | Ae₂=(−2,3) |
| Constant (1,−1) | (1,−1) |
| Total | (−3,5) |
Thus p has polynomial coordinates (t²+1,t−1), but its constant remainder should be (−3,5). Verify the exact certificate
p(t) − (−3,5) = (tI−A)q(t), q(t)=(t,2).
Indeed the right side is (t²+4,t−6), the difference on the left. This proves equality of two classes in the polynomial quotient, not equality of their polynomial coordinates. Multiplying the class by t corresponds under E to applying A; for instance A(−3,5)=(−10,12). Predict that value by evaluating tp instead, then compare.
Understand
Evaluation is a module map onto V_A
E is additive and F-linear by its finite-sum formula. Also E(tp)=Σ Aᵏ⁺¹vₖ=A E(p). Induction yields E(tʲp)=AʲE(p); taking an F-linear combination proves E(qp)=q(A)E(p) for every scalar polynomial q. This is precisely F[t]-linearity with the declared target action. E is onto because the constant polynomial vector v maps to v. Neither an eigenbasis nor a splitting field is required.
For any polynomial vector q, E((tI−A)q)=0: evaluation of tq gives A E(q), and evaluation of the vector Aq gives the same result, since A commutes with each of its powers. Thus the relation image is contained in ker E.
A telescoping identity proves the entire kernel
For v∈Fⁿ and k≥1, multiply out
tᵏv − Aᵏv = (tI−A) Σ from j=0 to k−1 of t^(k−1−j) Aʲv.
The product consists of a first sum with terms t^(k−j)Aʲv and a subtracted sum with terms t^(k−1−j)Aʲ⁺¹v. All intermediate terms cancel; the endpoints are tᵏv and −Aᵏv. These are polynomial-vector identities. For k=0 the difference is zero and the corresponding sum is empty.
Apply this to each coefficient of p. It gives
p(t) − E(p) = (tI−A)Q(t),
where Q is the sum of the displayed finite expressions for all k≥1. If E(p)=0, this is an explicit expression of p in the relation image. Therefore ker E=(tI−A)F[t]ⁿ. The module first isomorphism theorem now gives
F[t]ⁿ / (tI−A)F[t]ⁿ ≅ V_A, via [p]↦E(p).
Every quotient class has a constant representative E(p), and that representative is unique: if a constant v lies in the relation image, applying E gives v=0. The quotient therefore has F-dimension n, although the polynomial module presenting it has infinitely many F-linearly independent monomials when n>0.
Why the polynomial Smith form has no free summand
The determinant det(tI−A) is monic of degree n. In the permutation expansion its degree-n term comes from choosing t on every diagonal; any other choice has lower degree. For n=0 the empty determinant is 1. Thus the determinant is never the zero polynomial. Over the rational-function field F(t), the adjugate formula makes tI−A invertible, so its rank is n.
The polynomial Smith theorem therefore has n nonzero monic diagonal entries dᵢ, some possibly equal to 1. Its cokernel is a direct sum of F[t]/(dᵢ); there are no zero rows to contribute free F[t] summands. Unit factors contribute the zero module. A nonzero free F[t] summand would also contradict finite F-dimension, because its independent vectors 1,t,t²,… already give infinitely many F-directions.
This does not say the module has dimension zero over F. For a monic polynomial d of positive degree s, division gives unique representatives of degree below s, so F[t]/(d) has F-basis 1,t,…,t^(s−1). The sum of the nonunit degrees is n. Independently, determinants in a Smith certificate have only nonzero constant unit multipliers, so the monic product of all dᵢ equals the monic polynomial det(tI−A). These degree statements concern F-vector spaces; freeness here concerns the larger scalar ring F[t].
Polynomial equivalence and constant similarity
Suppose polynomial unimodular matrices U(t),V(t) satisfy U(t)(tI−A)V(t)=tI−B. Left multiplication by U induces the quotient isomorphism [p]↦[Up]; right multiplication by V changes only the generating columns of the relation module. Compose this quotient map with the evaluation identifications. The resulting map S:V_A→V_B is an invertible F[t]-linear map, hence an invertible F-linear map satisfying SA=BS. In constant matrices, B=SAS⁻¹. An explicit formula is Sx=E_B(U(t)x) for constant vectors x; evaluate its basis columns using the proved rule.
Conversely, if B=SAS⁻¹ for an invertible constant matrix S, then S(tI−A)S⁻¹=tI−B, a polynomial equivalence. The existence of polynomial U and V therefore establishes similarity through the module argument. It does not mean U(t) itself is a constant change-of-basis matrix, or that U and V are inverse to one another. Nor does arbitrary substitution t=0 extract the required S; evaluation here uses B acting on vector coefficients.
Apply
Change to B=[[1,1],[0,1]] over Q and p(t)=t³e₂−2te₁. Find its constant representative and an explicit polynomial q with p−E_B(p)=(tI−B)q. Do not just compare dimensions or determinants.
Check the changed telescoping certificate
Bᵏe₁=e₁ and Bᵏe₂=ke₁+e₂ for k≥0, by induction. Hence E_B(p)=(3,1)−(2,0)=(1,1). The telescoping construction gives q=t²e₂+tBe₂+B²e₂−2e₁=(t,t²+t+1). Multiplication by tI−B, whose rows are (t−1,−1),(0,t−1), gives (−2t−1,t³−1). That is exactly p−(1,1). The quotient class is not zero, despite its representative involving higher powers of t. Its polynomial presentation has full rank over Q(t), but the quotient remains a two-dimensional Q-vector space.
Explain why the zero-dimensional case causes no exception to the presentation theorem: source, target and cokernel are zero, the relation matrix is empty, and the determinant is 1. There are no nonunit Smith factors. Also explain why the proof never replaced A by a number or assumed A was diagonalizable.
Continue
Continue to rational canonical form. The polynomial quotient factors will supply actual F-bases and companion matrices. This lesson establishes their presentation and the bridge from module equivalence to similarity; the next lesson proves the canonical block classification.
For related operator tools, revisit minimal polynomials and primary components. Those are available without assuming a split characteristic polynomial. Keep exact algebraic certificates separate from claims about numerical stability or noisy matrix data.
