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Why Jordan Form Isn't Enough

Vol I gave you diagonalization and Jordan form — but both lean on roots of the characteristic polynomial. Here is what breaks when the field is too small.

Where Vol I left off

In a first course you learned diagonalization: if a matrix A has a full set of independent eigenvectors, you can write A = P D P^-1 with D diagonal. Not every matrix is diagonalizable, so Jordan form patched the gap by allowing 1's just above the diagonal. Both tools answer the same question — when are two matrices similar? — but both have a hidden cost.

The hidden cost is eigenvalues. Jordan blocks are built from the roots lambda of the characteristic polynomial. If those roots do not live in your field, there is no Jordan form over that field at all.

A matrix with no eigenvalues

Work over the rationals Q. Take the rotation-by-90-degrees matrix. Its characteristic polynomial is x^2 + 1, which has no root in Q (or even in R). So over Q this matrix is not diagonalizable and has no Jordan form — yet it is a perfectly ordinary linear operator that we ought to be able to classify.

A = [ 0  -1 ]     over the field Q
    [ 1   0 ]

char poly  det(xI - A) = x^2 + 1
roots in Q?  none      (x^2 + 1 is irreducible over Q)

=>  A is NOT diagonalizable over Q
=>  A has NO Jordan form over Q
but A is a genuine operator -- we still want a canonical form for it.
Rotation by 90 degrees: a real operator with no rational (or real) eigenvalues.

The form that needs no roots

The rational canonical form (RCF) solves exactly this. It is built not from eigenvalues but from polynomials with coefficients in your field — so it exists for every operator over every field, no algebraic closure required. The price is that its blocks look different: instead of near-diagonal Jordan blocks, you get companion matrices of polynomials.