Try
Work over the rational field ℚ. Let A have rows (0,−1,0), (1,0,0), (0,0,2). It rotates the first two coordinates and multiplies the last by 2. Can one vector and its successive images nevertheless give a basis of all three coordinates?
Try v=(1,0,1). Compute Av and A²v before continuing. A vector is cyclic when its successive images span the space. Cyclic does not mean that the operator has an eigenbasis, or even that its characteristic polynomial splits over the field.
Recall the polynomial Smith form and the presentation by tI−A. Throughout, F is a field, vectors are columns, and multiplication by t on the operator module means applying A. Polynomial matrix operations change generators and relations of that module; they are not themselves constant changes of vector-space basis.
Predict
Predict the last column of the matrix in the basis v, Av, A²v if
A³v = 2v − Av + 2A²v.
The first two columns should shift the basis vectors forward. The last column should record the three coefficients in this equation, in that order. Explain why placing these coefficients in the last row would use a different convention.
Also predict whether equal characteristic polynomials must imply similarity. Keep two candidate operators in mind: one cyclic block for (t²+1)² and two cyclic blocks for t²+1. Both act on four-dimensional rational spaces. We will find an invariant that separates them without searching for rational eigenvalues.
Experiment
Compute the following orbit independently, then compare the coordinates. This is an exact paper experiment, with no numerical eigenvalue approximation.
| Vector | Original coordinates | Coordinates in the orbit basis |
|---|
| v | (1,0,1) | (1,0,0) |
| Av | (0,1,2) | (0,1,0) |
| A²v | (−1,0,4) | (0,0,1) |
| A³v | (0,−1,8) | (2,−1,2) |
The basis matrix P has rows (1,0,−1), (0,1,0), (1,2,4), and det P=5. Hence P is invertible over ℚ. With C having rows (0,0,2), (1,0,−1), (0,1,2), direct multiplication gives AP=PC. Verify its last column: 2v−Av+2A²v=(0,−1,8). Thus P⁻¹AP=C.
Here q(t)=(t²+1)(t−2)=t³−2t²+t−2. The factors are coprime over ℚ: division of t²+1 by t−2 leaves 5. The polynomial Chinese remainder theorem identifies F[t]/(q) with F[t]/(t²+1) ⊕ F[t]/(t−2). Under the original coordinates, the element 1 corresponds to v, so the explicit orbit basis realizes that abstract cyclic merger. The nonzero remainder 5 matters: this same basis computation is not a certificate in characteristic 5.
Understand
Build a companion block from division. Let f=tˢ+cₛ₋₁tˢ⁻¹+⋯+c₁t+c₀ be monic of degree s≥1. Polynomial division gives every class in F[t]/(f) a unique representative of degree below s. Existence gives spanning by 1,t,…,tˢ⁻¹; uniqueness gives independence, since a nonzero multiple of f cannot have smaller degree. In this ordered F-basis, multiplication by t has ones immediately below the diagonal, zeros elsewhere except its last column (−c₀,−c₁,…,−cₛ₋₁). This is our companion matrix C(f).
The polynomial f annihilates multiplication by t because f is zero in the quotient. Conversely, if g annihilates the operator, apply g to the class of 1: the result is the class of g, so f divides g. Thus the monic minimal polynomial is exactly f. The existing Cayley–Hamilton theorem says the characteristic polynomial annihilates the matrix; division by the minimal polynomial shows that f divides it. Both are monic of degree s, so the characteristic polynomial is f. This uses Cayley–Hamilton, not a substitution of a matrix into a polynomial determinant identity whose coefficients are themselves matrices.
Pass from a presentation to a vector-space basis. For an n by n matrix A, det(tI−A) is monic of degree n, hence nonzero. Its polynomial Smith diagonal therefore has no zero entries. Unit entries contribute zero quotient modules; the remaining monic entries of positive degree are f₁,…,fᵣ with f₁ dividing f₂ dividing ⋯ dividing fᵣ. The proved presentation theorem gives
V ≅ F[t]/(f₁) ⊕ ⋯ ⊕ F[t]/(fᵣ).
There is no free F[t] summand. Such a summand would also contradict finite F-dimension, since its powers of t are F-linearly independent. Transport the division bases of the displayed summands to V. Their union is an F-basis, and A in that basis is the block diagonal matrix C(f₁) ⊕ ⋯ ⊕ C(fᵣ), called its rational canonical form. “Rational” here means defined over F; F need not be ℚ.
The characteristic polynomial is the product of the fᵢ, by the block determinant formula. A polynomial kills the direct sum exactly when it kills every summand, hence exactly when it is divisible by every fᵢ. The minimal polynomial is their monic least common multiple, which the divisibility chain makes fᵣ.
Prove the similarity classification in both directions. If A and B have the same ordered nonunit Smith factors, both are similar over F to the same companion-block matrix, hence to each other. Conversely, if B=P⁻¹AP for a constant invertible F-matrix P, then tI−B=P⁻¹(tI−A)P. These constant matrices are also unimodular over F[t], so Smith uniqueness gives the same factors. Equivalently, a module isomorphism is an invertible F-linear map that intertwines the two operators. Polynomial matrices U(t),V(t) in a Smith calculation are not generally the constant similarity matrix P: obtain P by transporting an actual F-basis as above.
For the zero-dimensional vector space, the characteristic polynomial is the empty determinant 1. Every polynomial annihilates its unique operator, so the monic generator of its annihilator ideal is also 1. There are no nonunit factors; empty product and empty least common multiple are both 1. These conventions make the formulas valid at n=0.
Apply
Over ℚ compare T=C((t²+1)²) and S=C(t²+1) ⊕ C(t²+1). Write both matrices, compute the two minimal polynomials, and decide whether similarity is possible. Then test the changed pair C(t²) ⊕ C(t−1): can its two blocks merge into one cyclic block?
Check the mechanism and the changed case
Since (t²+1)²=t⁴+2t²+1, T has rows (0,0,0,−1), (1,0,0,0), (0,1,0,−2), (0,0,1,0). Each block of S has rows (0,−1),(1,0). Both characteristic polynomials are (t²+1)², but their minimal polynomials are respectively (t²+1)² and t²+1. Similarity preserves annihilating polynomials, so they are not similar. Neither has rational eigenvalues, since their characteristic polynomials have no rational roots. The absence of rational eigenvalues does not prevent classification.
For the changed pair, t² and t−1 are coprime because the latter does not divide t². CRT therefore merges the blocks into C(t³−t²), whose last column is (0,0,1). Its minimal and characteristic polynomials are both t³−t². This cyclic operator is not diagonalizable: its minimal polynomial contains the repeated factor t². Cyclic and diagonalizable are different properties.
Continue
Return to primary decomposition to split coprime factors of each invariant factor. When the characteristic polynomial splits over F, the existing Jordan chain construction refines these primary pieces into powers of linear factors. Without that splitting hypothesis, companion blocks still work over F. Record a similarity certificate by stating its field, its invariant factors, and either an intertwining basis matrix or a proved module isomorphism; a list of approximate eigenvalues is not that certificate.
