The Payoff: Diagonalizability is a Polynomial Test

The diagonalizability criterion

The capstone theorem: T is diagonalizable if and only if its minimal polynomial splits into DISTINCT linear factors, m_T(x) = (x - lambda_1)...(x - lambda_r) with the lambda_i all different. This is the minimal polynomial diagonalizability test. No counting of eigenvectors, no comparing geometric to algebraic multiplicities — one factorization decides it.

1. (=>) If T is diagonalizable with distinct eigenvalues lambda_1,...,lambda_r, then prod (T - lambda_i I) kills every eigenvector, hence the whole space. So prod (x - lambda_i) annihilates T and m_T divides it — m_T is squarefree.
2. (<=) If m_T = prod (x - lambda_i) with distinct roots, the factors are pairwise coprime. By guide 4, V = (+) ker(T - lambda_i I), a direct sum of eigenspaces. A basis of eigenvectors exists, so T is diagonalizable.

Worked diagnosis: same char poly, different fate

Both matrices have char poly p(x) = (x - 5)^2.

A = [5, 0;     B = [5, 1;
     0, 5]          0, 5]

A - 5I = 0          B - 5I = [0,1; 0,0] != 0
=> m_A(x) = x - 5    => (B-5I)^2 = 0, (B-5I) != 0
   (distinct, deg 1)    so m_B(x) = (x - 5)^2 (repeated)

Verdict via the test:
  m_A squarefree   -> A IS diagonalizable (already diagonal)
  m_B has (x-5)^2  -> B is NOT diagonalizable (Jordan block)

Caveat on the field: 'splits into linear factors' presupposes the eigenvalues live in your field. If m_T is irreducible (e.g. x^2 + 1 over the reals, a rotation by 90 degrees), there are no real eigenvalues and T is not diagonalizable over R — but it IS over the splitting field C, where m_T = (x - i)(x + i) is squarefree. Diagonalizability is field-dependent; the polynomial test makes the dependence explicit.

Two coda: coefficients as power sums, and commuting

The characteristic polynomial's coefficients are symmetric functions of the eigenvalues, and Newton's identities express them through power sums p_k = sum lambda_i^k = trace(T^k). You compute these power sums by tracing matrix powers — no eigenvalues needed — then recover all coefficients of p_T algebraically. The trace and determinant are just the first and last of this family.

Finally, the structural reason all of this hangs together: any two polynomials in the same operator commute, f(T) g(T) = g(T) f(T), since they are sums of powers of one T. That is why the projections of guide 4, the inverse of guide 2, and the functions of the calculus all live in one commutative algebra F[T] — the algebra m_T governs. Knowing m_T means knowing the entire algebra T generates. By the companion matrix construction, every monic polynomial is realized as some operator's characteristic AND minimal polynomial — so this theory is not a special case, it is the whole landscape of operators up to similarity.