Minimal Polynomial and Diagonalizability

Can you tell whether a linear operator is diagonalizable without ever computing an eigenvector? Remarkably, yes. A single polynomial attached to the operator carries the complete answer. This article proves one of the most elegant facts in Eigenvalues and Eigenvectors theory. An operator on a finite-dimensional vector space is diagonalizable exactly when its minimal polynomial factors into distinct linear terms. No repeated roots, no irreducible quadratics over the reals, nothing extra. Just a product of first-degree factors, each appearing once. This matters because the minimal polynomial is computable in principle from the matrix alone. This works via the sequence of powers I,T,T2,…I, T, T^2, \dots and linear dependence. It requires no appeal to eigenvector bases or geometric multiplicities. The criterion converts a geometric question ("does a basis of eigenvectors exist?") into an algebraic one. That question becomes: "does this specific polynomial split with no repeated roots?" It unifies and sharpens the diagonalizability criteria you have already seen. Those criteria were phrased in terms of algebraic and geometric multiplicities. This theorem explains why those criteria work. It does so by exhibiting the precise algebraic obstruction to diagonalizability. The big picture is that the minimal polynomial governs the finest possible decomposition of a vector space into TT-invariant pieces. When it splits with distinct roots, the space decomposes as a direct sum of eigenspaces. That is the cleanest possible picture. When roots repeat, the space cannot be so decomposed. Instead, one is forced toward generalized eigenspaces and Jordan blocks. This theorem is thus the exact dividing line between the diagonalizable world and the world where Jordan Canonical Form is needed. That structure is needed to organize the operator's behavior. This article states the precise definitions. It proves the theorem in full using the primary decomposition method. It works two nontrivial examples, one diagonalizable and one not, both with asymmetric matrices. It shows exactly which hypothesis fails, and how, when repeated roots appear.

Key takeaways
  • The minimal polynomial mTm_T of an operator TT is the monic polynomial of least degree with mT(T)=0m_T(T) = 0.
  • Main theorem: TT is diagonalizable   ⟺  \iff mT(x)=(x−λ1)(x−λ2)⋯(x−λk)m_T(x) = (x - \lambda_1)(x-\lambda_2)\cdots(x-\lambda_k) with the λi\lambda_i distinct scalars.
  • The proof direction "distinct linear factors ⇒\Rightarrow diagonalizable" uses partial fractions / Lagrange-style idempotents built from mTm_T.
  • The converse direction uses the fact that eigenspaces already give a diagonal basis, from which mTm_T is read off directly.
  • A repeated root in mTm_T is not automatically fatal to diagonalizability of eigenvalues in general, but is fatal precisely because mTm_T having a repeated factor is equivalent to non-diagonalizability — this article proves that equivalence exactly.
  • Key idea: Diagonalizability is not about eigenvalues having high enough algebraic multiplicity — it is about the minimal polynomial having no repeated roots at all.

Where this lives today

The minimal-polynomial diagonalizability test underlies numerical linear algebra routines that decide whether an iterative eigenvalue algorithm can safely assume a simple spectral structure, since diagonalizable matrices admit far better-conditioned eigenvector bases. In quantum mechanics, observables are represented by self-adjoint (Hermitian) operators, which the Spectral Theorem guarantees are always diagonalizable — a special case where the minimal polynomial automatically has distinct real roots. In Markov chain theory and graph theory, the transition/adjacency matrix's minimal polynomial determines whether long-run behavior converges cleanly to an eigenvector decomposition or requires more delicate asymptotic (Jordan-block) analysis. In control theory, the minimal polynomial of a system matrix governs the structure of state-space realizations and stability analysis via modal decomposition. In cryptography and coding theory, minimal polynomials of linear-feedback-shift-register matrices over finite fields determine periodicity and structure of pseudorandom sequences, with the distinct-roots criterion again separating "clean" diagonalizable dynamics from more complex repeated-root behavior. Finally, this theorem is the essential stepping stone to the Jordan Canonical Form, which handles precisely the case this theorem identifies as failing: when mTm_T has repeated linear factors (or the field is not algebraically closed), Jordan form supplies the finest possible substitute for a diagonal matrix.

Ask the tutor: How does the primary decomposition argument change if the minimal polynomial has a repeated factor — what replaces the eigenspaces EλiE_{\lambda_i}?

DefinitionMinimal polynomial and diagonalizability

Let VV be a finite-dimensional vector space over a field F\mathbb{F}, with dim⁡V=n≥1\dim V = n \geq 1, and let T:V→VT : V \to V be a linear operator.

Minimal polynomial. The set of polynomials p∈F[x]p \in \mathbb{F}[x] with p(T)=0p(T) = 0 (the zero operator) is a nonzero ideal of F[x]\mathbb{F}[x] (nonzero because I,T,T2,…,Tn2I, T, T^2, \dots, T^{n^2} are n2+1n^2+1 operators in the n2n^2-dimensional space L(V,V)\mathcal{L}(V,V), hence linearly dependent, giving a nonzero annihilating polynomial of degree at most n2n^2). Since F[x]\mathbb{F}[x] is a principal ideal domain, this ideal is generated by a unique monic polynomial of minimal degree, denoted mT(x)m_T(x), called the minimal polynomial of TT. It is characterized by: 1. mT(T)=0m_T(T) = 0; 2. mTm_T is monic; 3. if p(T)=0p(T) = 0 for any polynomial pp, then mT∣pm_T \mid p.

Diagonalizable. TT is diagonalizable if there exists a basis of VV consisting entirely of eigenvectors of TT; equivalently, the matrix of TT with respect to some basis is diagonal.

Splits with distinct linear factors. We say mTm_T splits into distinct linear factors over F\mathbb{F} if there exist distinct scalars λ1,…,λk∈F\lambda_1, \dots, \lambda_k \in \mathbb{F} such that

mT(x)=(x−λ1)(x−λ2)⋯(x−λk),m_T(x) = (x - \lambda_1)(x-\lambda_2)\cdots(x-\lambda_k),

with each factor appearing to the first power exactly once (no repeated λi\lambda_i, and no irreducible factor of degree ≥2\geq 2).

Intuition

Think of TT acting on VV and ask: what is the smallest polynomial relation that TT itself must obey? Every eigenvalue λ\lambda of TT must be a root of mTm_T (proved below), so mTm_T always "knows about" every eigenvalue. The subtlety is multiplicity: does mTm_T need a repeated factor (x−λ)2(x-\lambda)^2 to kill off vectors associated to λ\lambda, or does a single factor (x−λ)(x - \lambda) suffice? If a single power of (x−λ)(x-\lambda) always suffices for every eigenvalue — that is precisely when (T−λI)(T - \lambda I) already sends the whole "λ\lambda-part" of VV to zero, meaning every generalized eigenvector for λ\lambda is already an honest eigenvector. That is exactly the diagonalizable situation: no operator is needed beyond the direct sum of eigenspaces. If some eigenvalue needs (x−λ)2(x - \lambda)^2 or higher to be annihilated, there must be vectors that are "almost" eigenvectors — killed by (T−λI)2(T-\lambda I)^2 but not by (T−λI)(T - \lambda I) — and such vectors obstruct any eigenvector basis. The picture below shows the two possible worlds for a 3×33\times 3 operator with eigenvalues 22 and 55: one where mTm_T has simple roots (space splits cleanly into eigenlines/eigenplanes) and one where mTm_T has a repeated root (a Jordan block appears, and no eigenbasis exists).

Figure
Two operators with the same characteristic polynomial but different minimal polynomials
Char. poly = (x−2)(x−5)² in both cases Case A: diagonalizable m_T(x) = (x−2)(x−5) distinct linear factors λ=2 eigenline λ=5 eigenPLANE ⊕ Case B: not diagonalizable m_T(x) = (x−2)(x−5)² repeated linear factor λ=2 eigenline λ=5: only 1-dim eigenspace Jordan block needed

Central theorem

TheoremMinimal Polynomial Criterion for Diagonalizability

Let VV be a finite-dimensional vector space over a field F\mathbb{F} with dim⁡V=n≥1\dim V = n \geq 1, and let T:V→VT: V \to V be a linear operator with minimal polynomial mT∈F[x]m_T \in \mathbb{F}[x]. Then:

T is diagonalizable  ⟺  mT(x)=(x−λ1)(x−λ2)⋯(x−λk)T \text{ is diagonalizable} \iff m_T(x) = (x-\lambda_1)(x - \lambda_2)\cdots(x-\lambda_k)

for some distinct scalars λ1,…,λk∈F\lambda_1, \dots, \lambda_k \in \mathbb{F} (i.e. mTm_T splits over F\mathbb{F} into linear factors, each occurring exactly once).

LemmaRoots of the minimal polynomial are exactly the eigenvalues

Let T:V→VT: V \to V be a linear operator on a finite-dimensional space V≠{0}V \neq \{0\} over F\mathbb{F}, with minimal polynomial mTm_T. A scalar λ∈F\lambda \in \mathbb{F} satisfies mT(λ)=0m_T(\lambda) = 0 if and only if λ\lambda is an eigenvalue of TT.

Proof

(⇐\Leftarrow) Suppose λ\lambda is an eigenvalue of TT, so Tv=λvTv = \lambda v for some v≠0v \neq 0. By induction, Tjv=λjvT^j v = \lambda^j v for all j≥0j \geq 0, so for any polynomial pp, p(T)v=p(λ)vp(T)v = p(\lambda) v. Taking p=mTp = m_T: since mT(T)=0m_T(T) = 0, we get 0=mT(T)v=mT(λ)v0 = m_T(T)v = m_T(\lambda) v. As v≠0v \neq 0, this forces mT(λ)=0m_T(\lambda) = 0. (⇒\Rightarrow) Suppose mT(λ)=0m_T(\lambda) = 0. By the factor theorem in F[x]\mathbb{F}[x], we may write mT(x)=(x−λ)q(x)m_T(x) = (x - \lambda) q(x) for some polynomial qq with deg⁡q=deg⁡mT−1<deg⁡mT\deg q = \deg m_T - 1 < \deg m_T. Since mTm_T has minimal degree among nonzero annihilators of TT, q(T)≠0q(T) \neq 0 (as qq has strictly smaller degree than mTm_T, so if q(T)=0q(T)=0 this would contradict minimality of deg⁡mT\deg m_T, unless qq is the zero polynomial — but deg⁡q=deg⁡mT−1≥0\deg q = \deg m_T - 1 \geq 0 since mTm_T is nonconstant, as mT(T)=0m_T(T)=0 and TT acts on a nonzero space, so mTm_T cannot be the nonzero constant polynomial). Hence there exists w∈Vw \in V with q(T)w≠0q(T)w \neq 0. Set v:=q(T)w≠0v := q(T) w \neq 0. Then

(T−λI)v=(T−λI)q(T)w=mT(T)w=0,(T - \lambda I) v = (T-\lambda I) q(T) w = m_T(T) w = 0,

so Tv=λvTv = \lambda v with v≠0v \neq 0: λ\lambda is an eigenvalue of TT.

∎

Now we prove the main theorem, using the primary decomposition technique.

Proof

(⇐\Leftarrow) Suppose mT(x)=(x−λ1)⋯(x−λk)m_T(x) = (x-\lambda_1)\cdots(x-\lambda_k) with λ1,…,λk∈F\lambda_1, \dots, \lambda_k \in \mathbb{F} pairwise distinct. We show VV is spanned by eigenvectors of TT. For each i=1,…,ki = 1, \dots, k define the polynomial

pi(x):=∏j≠i(x−λj).p_i(x) := \prod_{j \neq i} (x - \lambda_j).

Since the λj\lambda_j are distinct, the polynomials p1,…,pkp_1, \dots, p_k have no common root, hence (as polynomials over a field) gcd⁡(p1,…,pk)=1\gcd(p_1, \dots, p_k) = 1. By repeated application of Bézout's identity in F[x]\mathbb{F}[x] (a PID), there exist polynomials q1,…,qk∈F[x]q_1, \dots, q_k \in \mathbb{F}[x] such that

q1(x)p1(x)+q2(x)p2(x)+⋯+qk(x)pk(x)=1.q_1(x) p_1(x) + q_2(x) p_2(x) + \cdots + q_k(x) p_k(x) = 1.

Substituting TT for xx (a valid operation since polynomial identities are preserved under the ring homomorphism F[x]→L(V,V)\mathbb{F}[x] \to \mathcal{L}(V,V), p↦p(T)p \mapsto p(T)), we get the operator identity

∑i=1kqi(T) pi(T)=I.(∗)\sum_{i=1}^k q_i(T)\, p_i(T) = I. \tag{$\ast$}

Define Ei:=qi(T)pi(T)∈L(V,V)E_i := q_i(T) p_i(T) \in \mathcal{L}(V,V) for each ii. Then (∗)(\ast) says

E1+E2+⋯+Ek=I.E_1 + E_2 + \cdots + E_k = I.

Claim 1: (T−λiI)Ei=0(T - \lambda_i I) E_i = 0 for each ii. Indeed, (x−λi)pi(x)(x - \lambda_i) p_i(x) is, up to reordering factors, exactly mT(x)m_T(x) times qi(x)q_i(x)... more directly: (x−λi) pi(x)=mT(x)(x-\lambda_i)\, p_i(x) = m_T(x) since pi(x)=∏j≠i(x−λj)p_i(x) = \prod_{j\neq i}(x - \lambda_j) and multiplying by the missing factor (x−λi)(x-\lambda_i) recovers mTm_T. Hence

(T−λiI)Ei=(T−λiI) qi(T) pi(T)=qi(T) (T−λiI)pi(T)=qi(T) mT(T)=qi(T)⋅0=0,(T - \lambda_i I) E_i = (T-\lambda_i I)\, q_i(T)\, p_i(T) = q_i(T)\, (T - \lambda_i I) p_i(T) = q_i(T)\, m_T(T) = q_i(T) \cdot 0 = 0,

using that all these operators are polynomials in TT and hence commute with one another (polynomials in a fixed operator TT always commute, since F[T]\mathbb{F}[T] is a commutative subring of L(V,V)\mathcal{L}(V,V)). Claim 1 shows: for every v∈Vv \in V, the vector EivE_i v satisfies (T−λiI)(Eiv)=0(T - \lambda_i I)(E_i v) = 0, so Eiv∈ker⁡(T−λiI)=EλiE_i v \in \ker(T - \lambda_i I) = E_{\lambda_i}, the eigenspace of λi\lambda_i (or Eiv=0E_i v = 0).

Claim 2: every v∈Vv \in V is a sum of eigenvectors (or zero) for λ1,…,λk\lambda_1, \dots, \lambda_k. Take any v∈Vv \in V. By (∗)(\ast),

v=Iv=∑i=1kEiv,v = I v = \sum_{i=1}^k E_i v,

and each EivE_i v lies in EλiE_{\lambda_i} by Claim 1. So vv is a sum of vectors, one from each eigenspace EλiE_{\lambda_i} (possibly zero). This shows

V=Eλ1+Eλ2+⋯+Eλk.V = E_{\lambda_1} + E_{\lambda_2} + \cdots + E_{\lambda_k}.

Since eigenvectors for distinct eigenvalues are always linearly independent (a standard fact used freely from Diagonalizability and Its Criteria: eigenspaces for distinct eigenvalues form a direct sum), this sum is in fact a direct sum:

V=Eλ1⊕Eλ2⊕⋯⊕Eλk.V = E_{\lambda_1} \oplus E_{\lambda_2} \oplus \cdots \oplus E_{\lambda_k}.

Choosing a basis for each eigenspace EλiE_{\lambda_i} and concatenating these bases yields a basis of VV consisting entirely of eigenvectors of TT. Hence TT is diagonalizable.

(⇒\Rightarrow) Suppose TT is diagonalizable. We show mTm_T splits into distinct linear factors. Since TT is diagonalizable, VV has a basis v1,…,vnv_1, \dots, v_n of eigenvectors, with Tvi=μiviTv_i = \mu_i v_i for scalars μi\mu_i (not necessarily distinct as listed; let λ1,…,λk\lambda_1, \dots, \lambda_k be the distinct values occurring among μ1,…,μn\mu_1, \dots, \mu_n, i.e. the distinct eigenvalues of TT, with 1≤k≤n1 \le k \le n). Define

g(x):=(x−λ1)(x−λ2)⋯(x−λk),g(x) := (x - \lambda_1)(x-\lambda_2) \cdots (x - \lambda_k),

a monic polynomial with distinct roots λ1,…,λk∈F\lambda_1,\dots,\lambda_k \in \mathbb F. We claim g(T)=0g(T) = 0. It suffices to check g(T)vi=0g(T) v_i = 0 for each basis vector viv_i, since a linear operator vanishing on a basis is the zero operator. Fix viv_i; then Tvi=λjviTv_i = \lambda_j v_i for the unique jj with μi=λj\mu_i = \lambda_j. Since g(x)=(x−λj)h(x)g(x) = (x - \lambda_j) h(x) for some polynomial hh (as (x−λj)(x-\lambda_j) is one of the factors of gg), and all factors are polynomials in TT hence commute,

g(T)vi=h(T) (T−λjI)vi=h(T)(λjvi−λjvi)=h(T)⋅0=0.g(T) v_i = h(T)\,(T - \lambda_j I) v_i = h(T)\big(\lambda_j v_i - \lambda_j v_i\big) = h(T)\cdot 0 = 0.

So g(T)=0g(T) = 0 on the whole basis, hence g(T)=0g(T) = 0 as an operator. By the defining property of the minimal polynomial, mT∣gm_T \mid g. On the other hand, by the Lemma above, every root of mTm_T is an eigenvalue of TT, hence lies in {λ1,…,λk}\{\lambda_1, \dots, \lambda_k\}; and conversely each λi\lambda_i is an eigenvalue of TT, so by the same Lemma (the ⇐\Leftarrow direction argument reversed — actually we use the Lemma's ⇒\Rightarrow-direction contrapositive is not needed; instead) each λi\lambda_i is a root of mTm_T directly: since λi\lambda_i is an eigenvalue, the Lemma's proof shows mT(λi)=0m_T(\lambda_i) = 0. So every λi\lambda_i, i=1,…,ki=1,\dots,k, is a root of mTm_T, meaning (x−λi)∣mT(x)(x - \lambda_i) \mid m_T(x) for each ii. As the λi\lambda_i are pairwise distinct, the linear factors (x−λ1),…,(x−λk)(x-\lambda_1), \dots, (x - \lambda_k) are pairwise coprime, so their product divides mTm_T:

g(x)=(x−λ1)⋯(x−λk)∣mT(x).g(x) = (x-\lambda_1)\cdots(x-\lambda_k) \mid m_T(x).

Combined with mT∣gm_T \mid g from above, and both being monic, we conclude mT=gm_T = g. Thus

mT(x)=(x−λ1)(x−λ2)⋯(x−λk)m_T(x) = (x - \lambda_1)(x-\lambda_2)\cdots(x-\lambda_k)

with the λi\lambda_i distinct scalars in F\mathbb{F}, as required. ■\blacksquare

∎
CorollaryMultiplicity-free eigenvalue equation test

TT is diagonalizable if and only if there exists some polynomial (not necessarily mTm_T itself) that splits into distinct linear factors over F\mathbb F and annihilates TT. Equivalently, it suffices to exhibit any polynomial with distinct roots killing TT; one need not compute mTm_T exactly to certify diagonalizability.

Proof

If such a polynomial gg exists, the argument in the (⇐\Leftarrow) direction above used only that g(T)=0g(T) = 0 and gg has distinct roots λ1,…,λk\lambda_1,\dots,\lambda_k — it never required g=mTg = m_T. So the same construction of idempotents EiE_i from gg's factors produces the eigenspace decomposition V=Eλ1⊕⋯⊕EλkV = E_{\lambda_1}\oplus\cdots\oplus E_{\lambda_k} verbatim. Conversely if TT is diagonalizable then mTm_T itself is such a polynomial, by the (⇒\Rightarrow) direction. ■\blacksquare

∎

Worked examples

Example 1 (Diagonalizable, verified via minimal polynomial). Let T:R3→R3T : \mathbb{R}^3 \to \mathbb{R}^3 be given, with respect to the standard basis, by

A=(41−125−2112).A = \begin{pmatrix} 4 & 1 & -1 \\ 2 & 5 & -2 \\ 1 & 1 & 2 \end{pmatrix}.

Step 1: characteristic polynomial. Compute

det⁡(A−xI)=det⁡(4−x1−125−x−2112−x).\det(A - xI) = \det\begin{pmatrix} 4-x & 1 & -1 \\ 2 & 5-x & -2 \\ 1 & 1 & 2 - x\end{pmatrix}.

Expand along the first row:

(4−x)[(5−x)(2−x)−(−2)(1)]−1[2(2−x)−(−2)(1)]+(−1)[2(1)−(5−x)(1)].(4-x)\big[(5-x)(2-x) - (-2)(1)\big] - 1\big[2(2-x) - (-2)(1)\big] + (-1)\big[2(1) - (5-x)(1)\big].

Compute the bracketed terms:

  • (5−x)(2−x)+2=(10−7x+x2)+2=x2−7x+12(5-x)(2-x) + 2 = (10 - 7x + x^2) + 2 = x^2 - 7x + 12.
  • 2(2−x)+2=4−2x+2=6−2x2(2-x) + 2 = 4 - 2x + 2 = 6 - 2x.
  • 2−(5−x)=x−32 - (5-x) = x - 3.

So the determinant is

(4−x)(x2−7x+12)−(6−2x)−(x−3).(4-x)(x^2 - 7x + 12) - (6-2x) - (x-3).

Expand (4−x)(x2−7x+12)=4x2−28x+48−x3+7x2−12x=−x3+11x2−40x+48(4-x)(x^2-7x+12) = 4x^2 - 28x + 48 - x^3 + 7x^2 - 12x = -x^3 + 11x^2 - 40x + 48. Then subtract (6−2x)+(x−3)=3−x(6-2x) + (x - 3) = 3 - x:

−x3+11x2−40x+48−3+x=−x3+11x2−39x+45.-x^3 + 11x^2 - 40x + 48 - 3 + x = -x^3 + 11x^2 - 39x + 45.

So det⁡(A−xI)=−x3+11x2−39x+45\det(A-xI) = -x^3 + 11x^2 - 39x + 45, i.e. the characteristic polynomial (monic, using det⁡(xI−A)=−det⁡(A−xI)\det(xI - A) = -\det(A-xI)) is

χA(x)=x3−11x2+39x−45.\chi_A(x) = x^3 - 11x^2 + 39x - 45.

Test x=3x = 3: 27−99+117−45=027 - 99 + 117 - 45 = 0. ✓ So (x−3)(x-3) divides χA\chi_A. Divide:

x3−11x2+39x−45=(x−3)(x2−8x+15)=(x−3)(x−3)(x−5)=(x−3)2(x−5).x^3 - 11x^2 + 39x - 45 = (x-3)(x^2 - 8x + 15) = (x-3)(x-3)(x-5) = (x-3)^2(x-5).

Check: (x−3)2(x−5)=(x2−6x+9)(x−5)=x3−5x2−6x2+30x+9x−45=x3−11x2+39x−45(x-3)^2(x-5) = (x^2-6x+9)(x-5) = x^3 -5x^2 -6x^2+30x+9x-45 = x^3 -11x^2+39x-45. ✓ Matches. So χA(x)=(x−3)2(x−5)\chi_A(x) = (x-3)^2(x-5), eigenvalues 33 (algebraic multiplicity 2) and 55 (algebraic multiplicity 1). Step 2: geometric multiplicity of λ=3\lambda = 3. Compute A−3IA - 3I:

A−3I=(11−122−211−1).A - 3I = \begin{pmatrix} 1 & 1 & -1 \\ 2 & 2 & -2 \\ 1 & 1 & -1\end{pmatrix}.

Rows 1 and 3 are identical, row 2 is twice row 1: rank 11. So nullity =3−1=2= 3 - 1 = 2: geometric multiplicity of λ=3\lambda=3 is 22, matching its algebraic multiplicity. Step 3: minimal polynomial. Since the eigenvalue 33 already achieves geometric multiplicity equal to algebraic multiplicity, TT is diagonalizable by the standard rank criterion, and consequently mT(x)=(x−3)(x−5)m_T(x) = (x-3)(x-5) (no repeated factor needed) — let us verify directly that (A−3I)(A−5I)=0(A - 3I)(A-5I) = 0:

A−5I=(−11−120−211−3).A - 5I = \begin{pmatrix}-1 & 1 & -1\\ 2 & 0 & -2\\ 1 & 1 & -3\end{pmatrix}.

Compute (A−3I)(A−5I)(A-3I)(A-5I): Row 1 of A−3IA-3I is (1,1,−1)(1,1,-1). Dot with columns of A−5IA - 5I:

  • col 1 (−1,2,1)(-1,2,1): 1(−1)+1(2)+(−1)(1)=−1+2−1=01(-1)+1(2)+(-1)(1) = -1+2-1=0.
  • col 2 (1,0,1)(1,0,1): 1(1)+1(0)+(−1)(1)=1+0−1=01(1)+1(0)+(-1)(1) = 1+0-1 = 0.
  • col 3 (−1,−2,−3)(-1,-2,-3): 1(−1)+1(−2)+(−1)(−3)=−1−2+3=01(-1)+1(-2)+(-1)(-3) = -1-2+3=0.

Row 2 of A−3IA - 3I is (2,2,−2)=2×(2,2,-2) = 2\times row 1, so its product is also (0,0,0)(0,0,0). Row 3 of A−3IA-3I is (1,1,−1)(1,1,-1), identical to row 1, giving (0,0,0)(0,0,0) again. So (A−3I)(A−5I)=0(A-3I)(A-5I) = 0, confirming mT∣(x−3)(x−5)m_T \mid (x-3)(x-5). Since TT has two distinct eigenvalues 3,53,5, by the Lemma both must be roots of mTm_T, so mTm_T is divisible by (x−3)(x−5)(x-3)(x-5); combined with the above, mT(x)=(x−3)(x−5)m_T(x) = (x-3)(x-5) exactly — distinct linear factors. By the Main Theorem, TT is diagonalizable, confirming Step 2's conclusion by an entirely independent, purely algebraic computation.

Example 2 (Not diagonalizable, minimal polynomial has a repeated factor). Let S:R3→R3S : \mathbb{R}^3 \to \mathbb{R}^3 be given by

B=(210021003).B = \begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 3 \end{pmatrix}.

Step 1: characteristic polynomial. BB is upper triangular, so

χB(x)=(x−2)2(x−3).\chi_B(x) = (x-2)^2(x-3).

Step 2: check geometric multiplicity of λ=2\lambda = 2. Compute

B−2I=(010001001).B - 2I = \begin{pmatrix}0&1&0\\0&0&1\\0&0&1\end{pmatrix}.

Row reduce: rows 2 and 3 are identical, giving rank 22 (rows (0,1,0)(0,1,0) and (0,0,1)(0,0,1) independent). So nullity =3−2=1= 3-2=1. Geometric multiplicity of λ=2\lambda=2 is 11, strictly less than its algebraic multiplicity 22. By the rank criterion for diagonalizability, SS is not diagonalizable, since geometric multiplicity fails to match algebraic multiplicity for λ=2\lambda = 2. Step 3: confirm via the minimal polynomial directly. We show mS(x)=(x−2)2(x−3)m_S(x) = (x-2)^2(x-3), i.e. mS=χBm_S = \chi_B, so the factor (x−2)(x-2) is genuinely repeated in mSm_S. First, (x−2)(x−3)(x-2)(x-3) does not annihilate BB: compute

(B−2I)(B−3I).(B-2I)(B-3I).
B−3I=(−1100−11000).B - 3I = \begin{pmatrix}-1&1&0\\0&-1&1\\0&0&0\end{pmatrix}.

Multiply (B−2I)(B−3I)(B-2I)(B-3I): Row 1 of B−2IB-2I is (0,1,0)(0,1,0). Dotted with columns of B−3IB-3I:

  • col1 (−1,0,0)(-1,0,0): 00.
  • col2 (1,−1,0)(1,-1,0): 0⋅1+1⋅(−1)+0⋅0=−10\cdot1 + 1\cdot(-1) + 0\cdot 0 = -1.
  • col3 (0,1,0)(0,1,0): 0⋅0+1⋅1+0⋅0=10\cdot0+1\cdot1+0\cdot0=1.

So row 1 of the product is (0,−1,1)≠(0,0,0)(0,-1,1) \neq (0,0,0). Hence (B−2I)(B−3I)≠0(B-2I)(B-3I) \neq 0, so (x−2)(x−3)(x-2)(x-3) does not annihilate BB, confirming that the minimal polynomial cannot be this degree-2, distinct-root polynomial. Since χB(x)=(x−2)2(x−3)\chi_B(x) = (x-2)^2(x-3) annihilates BB by the Cayley–Hamilton theorem (a fact usable here as background, though we can also verify directly), and we've just shown no polynomial of the form (x−2)(x−3)(x-2)(x-3) (its only proper monic divisor with both roots) works, and any other divisor would omit a needed root (violating the Lemma, since both 2,32,3 are eigenvalues and so must be roots of mSm_S), we conclude the minimal polynomial must be exactly

mS(x)=(x−2)2(x−3),m_S(x) = (x-2)^2(x-3),

which has a repeated linear factor (x−2)2(x-2)^2. By the Main Theorem, this repeated factor is precisely what obstructs diagonalizability — matching Step 2's conclusion via an entirely different, purely polynomial-algebraic route.

CounterexampleDropping "distinct" — repeated linear factors still block diagonalizability even though every root is real and the field is algebraically closed

The hypothesis "distinct" in the theorem cannot be weakened to merely "mTm_T splits into linear factors" (i.e. dropping distinctness while keeping everything linear). Consider again

B=(210021003)B = \begin{pmatrix} 2&1&0\\0&2&1\\0&0&3\end{pmatrix}

from Example 2, over F=R\mathbb{F} = \mathbb{R} (or even C\mathbb{C} — the field is algebraically closed, so every operator's characteristic polynomial splits into linear factors here; there is no obstruction from missing roots). We showed mB(x)=(x−2)2(x−3)m_B(x) = (x-2)^2(x-3): this does split into linear factors over R\mathbb{R}, satisfying every hypothesis of the theorem except distinctness of the roots. Yet SS (equivalently BB) is demonstrably not diagonalizable (Step 2 of Example 2: geometric multiplicity 1<2=1 < 2 = algebraic multiplicity for λ=2\lambda=2; equivalently no basis of R3\mathbb R^3 of eigenvectors of BB exists, since the λ=2\lambda=2-eigenspace is only 11-dimensional and cannot supply the two linearly independent eigenvectors needed together with the λ=3\lambda=3 eigenvector to reach dimension 3). This shows that "splits into linear factors" alone is not sufficient; the distinctness of the roots of mTm_T is doing indispensable work in the proof — specifically, distinctness is exactly what permits construction of coprime polynomials p1,…,pkp_1,\dots,p_k satisfying Bézout's identity ∑qipi=1\sum q_i p_i = 1 in the forward direction of the proof. If (x−λ1)(x-\lambda_1) appeared with multiplicity ≥2\geq 2 in mTm_T, then p1=∏j≠1(x−λj)p_1 = \prod_{j\ne 1}(x-\lambda_j) and the other pip_i would no longer be coprime as a full collection matching mTm_T's exact factorization, and the identity (∗)(\ast) used to build the eigenspace decomposition breaks down — reflecting the genuine geometric fact that generalized eigenvectors beyond honest eigenvectors then exist.

Scope, edge cases & common traps

SituationWhat happens
mTm_T splits, roots distinctTT diagonalizable (Main Theorem)
mTm_T splits, some root repeatedTT not diagonalizable (Counterexample)
mTm_T has an irreducible factor of degree ≥2\geq 2 over F\mathbb F (e.g. rotation by 90°90° over R\mathbb R)TT not diagonalizable over F\mathbb F; may become diagonalizable over an extension field where the factor splits
χT\chi_T (characteristic polynomial) has repeated rootsSays nothing by itself — must check whether mTm_T also repeats that root; Example 1 shows χ\chi can repeat a root while mTm_T does not
V={0}V = \{0\}Degenerate case: mT=1m_T = 1 (empty product), vacuously "distinct linear factors" (none), and TT is vacuously diagonalizable

The mathematician behind it

Origins · 1 minCamille Jordan and how this was discovered
CJ
Camille Jordan
1838–1922 · French mathematician, École Polytechnique, foundational work on linear operators and group theory

Camille Jordan's 1870 treatise Traité des substitutions et des équations algébriques systematized the study of linear substitutions (what we now call linear operators and matrices) and their canonical forms, laying groundwork that directly led to the classification of operators by minimal and characteristic polynomials. The decomposition into eigenspaces described by this article's theorem is the simplest possible case of the general canonical form theory that bears Jordan's name — every operator, diagonalizable or not, admits a canonical block decomposition, and diagonalizability is exactly the case where every Jordan block has size 11.

History
- 1870 — Camille Jordan publishes Traité des substitutions, introducing canonical forms for linear substitutions over finite fields and C\mathbb{C}.
- 1878 — Georg Frobenius develops the theory of minimal polynomials and rational canonical forms independently, emphasizing polynomial-algebraic invariants of matrices.
- 1898 — Kurt Hensel and others refine the ideal-theoretic viewpoint of F[x]\mathbb F[x]-modules underlying the primary decomposition used in this article's proof.
- 1930s — The module-theoretic formulation (via F[x]\mathbb F[x]-modules and the structure theorem for finitely generated modules over a PID) becomes the standard modern lens for minimal-polynomial arguments, as used in Hoffman–Kunze.
- Present — The distinct-linear-factors criterion is standard in every linear algebra curriculum as the sharp algebraic test preceding the general Jordan Canonical Form.
Why is it the minimal polynomial and not the characteristic polynomial that governs diagonalizability?

The characteristic polynomial always has degree n=dim⁡Vn = \dim V and encodes algebraic multiplicities, but it can have repeated roots even when the operator is perfectly diagonalizable — Example 1 in this article shows χA(x)=(x−3)2(x−5)\chi_A(x) = (x-3)^2(x-5) for a diagonalizable operator. The minimal polynomial, by contrast, encodes exactly the smallest annihilating relation, which is precisely sensitive to whether generalized eigenvectors beyond honest eigenvectors are needed. A repeated root in χT\chi_T is compatible with diagonalizability (if the eigenspace is "full-sized"), whereas a repeated root in mTm_T is never compatible with diagonalizability. This is why the sharp criterion must be phrased in terms of mTm_T.

Does every operator have a minimal polynomial that splits into linear factors, over some field?

Not over every field, but yes over an algebraically closed field such as C\mathbb C: the fundamental theorem of algebra guarantees every nonconstant polynomial over C\mathbb C factors completely into linear factors. Over R\mathbb R, a minimal polynomial can contain an irreducible quadratic factor like x2+1x^2+1 (rotation by 90°90°), and such an operator is not diagonalizable over R\mathbb R even though it becomes diagonalizable after extending scalars to C\mathbb C. So "splits into linear factors" is itself a real hypothesis depending on the field, separate from the distinctness condition.

How does this theorem relate to the rank-based criterion (geometric multiplicity equals algebraic multiplicity for every eigenvalue)?

They are equivalent, but the minimal-polynomial criterion is more economical to state and often faster to verify computationally, since it requires finding one polynomial rather than checking every eigenspace's dimension individually. Internally, the rank criterion is essentially unpacking what "distinct linear factors of mTm_T" means eigenspace by eigenspace: a repeated factor (x−λ)2(x-\lambda)^2 in mTm_T occurs exactly when the geometric multiplicity of λ\lambda falls short of its algebraic multiplicity. Both criteria are proved from the same underlying primary decomposition machinery.

Can the minimal polynomial have distinct factors but the operator still fail to be diagonalizable, for some exotic reason?

No — this article's Main Theorem is a genuine if-and-only-if statement, proved in both directions from first principles using only the definition of minimal polynomial, Bézout's identity in F[x]\mathbb F[x], and basic facts about eigenspaces. There is no additional hidden hypothesis; distinct linear factors of mTm_T are both necessary and sufficient for diagonalizability, over any field F\mathbb F, for any finite-dimensional VV.

What happens in infinite dimensions — does this theorem still hold?

This article assumes dim⁡V<∞\dim V < \infty throughout, which is essential: the existence of a minimal polynomial itself relies on the pigeonhole argument that I,T,…,Tn2I, T, \dots, T^{n^2} must be linearly dependent in the finite-dimensional space L(V,V)\mathcal L(V,V). In infinite dimensions, an operator need not satisfy any polynomial relation at all (e.g. the shift operator on sequence spaces), so "the minimal polynomial" may simply not exist, and the theorem has no direct analogue without additional spectral-theoretic machinery.

How does this connect to the Jordan Canonical Form?

The Jordan Canonical Form generalizes this theorem to handle exactly the failure case: when mTm_T has a repeated linear factor (x−λ)k(x-\lambda)^k with k≥2k \geq 2, the operator cannot be diagonalized, but it can still be put into a near-diagonal block form using generalized eigenvectors, with Jordan blocks of size up to kk attached to λ\lambda. Diagonalizable operators are exactly those whose Jordan form consists entirely of 1×11\times1 blocks — the theorem in this article is the k=1k=1 special case of the general Jordan theory.

Is checking mTm_T for repeated roots computationally easier than checking eigenspace dimensions directly?

Often yes: one can compute mTm_T via the Euclidean algorithm / polynomial gcd applied to χT\chi_T and χT′\chi_T' (its derivative) to detect repeated roots symbolically, without explicitly solving for eigenvectors, and this scales well symbolically over exact fields such as Q\mathbb Q. In floating-point numerical practice, however, eigenvalue algorithms (QR algorithm, etc.) are typically preferred for stability, and diagonalizability is often assessed via condition numbers of the eigenvector matrix rather than exact polynomial factorization.

Before this

Where this leads