The Characteristic Polynomial

How can we find every eigenvalue of a linear operator without guessing? This question sits at the heart of Eigenvalues and Eigenvectors. Its answer is one of the cleanest constructions in linear algebra. We build a single polynomial whose roots are exactly the eigenvalues. That polynomial, the characteristic polynomial, converts a geometric search into an algebraic one. The geometric question is: "for which scalars does T−λIT - \lambda I fail to be injective?" The algebraic question is: "where does a determinant vanish?" The idea matters because it makes eigenvalues computable and countable. A linear operator on an nn-dimensional space has at most nn eigenvalues. This fact is not obvious from the definition of eigenvalue alone. It falls out immediately once we know the eigenvalues are roots of a degree-nn polynomial. The characteristic polynomial also carries the trace and determinant of the operator as its coefficients. This links three seemingly different invariants into one algebraic object. The big picture is that diagonalization, Jordan form, and the entire theory of canonical forms all begin with one question. That question is: "what is the characteristic polynomial, and what are its roots?" Multiplicities of roots foreshadow the distinction between geometric and algebraic multiplicity. The polynomial itself is the star of the Cayley–Hamilton theorem. That theorem says every operator satisfies its own characteristic equation. This article gives the precise definition of the characteristic polynomial via the determinant. It proves that its roots are exactly the eigenvalues. It works two concrete non-trivial examples by hand. It shows exactly which hypotheses (finite dimension, a field, a fixed basis) are load-bearing.

Key takeaways
  • The characteristic polynomial of TT is χT(λ)=det⁡(λI−T)\chi_T(\lambda) = \det(\lambda I - T), a degree-nn monic polynomial when dim⁡V=n\dim V = n.
  • Its roots (in the field of scalars, or in an algebraic closure) are exactly the eigenvalues of TT.
  • It is basis-independent: similar matrices have the same characteristic polynomial.
  • The coefficient of λn−1\lambda^{n-1} is −tr⁡(T)-\operatorname{tr}(T); the constant term is (−1)ndet⁡(T)(-1)^n \det(T).
  • An operator on an nn-dimensional space has at most nn distinct eigenvalues, since χT\chi_T has degree nn.
  • Key idea: turning the eigenvalue equation Tv=λvTv = \lambda v into a determinant equation converts an existence question into a root-finding problem.

Where this lives today

The characteristic polynomial is the computational engine behind eigenvalue theory across mathematics and its applications. In differential equations, the characteristic polynomial of the coefficient matrix of a linear system x˙=Ax\dot{x} = Ax determines the stability of equilibria: eigenvalues with negative real part correspond to decaying modes, and the whole classification of phase portraits (node, saddle, spiral, center) is read off from the roots of χA\chi_A. In physics, characteristic polynomials of Hamiltonian or moment-of-inertia matrices give quantized energy levels and principal axes of rotation. In graph theory and network science, the characteristic polynomial of a graph's adjacency matrix (spectral graph theory) encodes structural information — the number of closed walks, bipartiteness, and connectivity properties — used in algorithms for clustering and network analysis. In statistics, the characteristic polynomial of a covariance matrix underlies principal component analysis, since PCA extracts eigenvectors ranked by eigenvalue. In cryptography and coding theory, characteristic polynomials of matrices over finite fields determine the periods of linear feedback shift registers used in stream ciphers. In numerical linear algebra, while directly computing χT\chi_T and factoring it is numerically unstable for large matrices (small perturbations in coefficients can hugely perturb roots), the theoretical existence of the characteristic polynomial still underlies the correctness of the QR algorithm and other modern eigenvalue solvers, and the polynomial itself remains the conceptual bridge from "operator" to "root-finding problem" used throughout the theory of Diagonalization.

Ask the tutor: How does the multiplicity of a root of the characteristic polynomial relate to the dimension of its eigenspace, and when does that gap prevent diagonalization?

DefinitionCharacteristic Polynomial

Let VV be a finite-dimensional vector space over a field F\mathbf{F} with dim⁡V=n≥1\dim V = n \geq 1, and let T:V→VT: V \to V be a linear operator. Fix any basis β\beta of VV and let A=[T]βA = [T]_\beta be the matrix of TT with respect to β\beta. The characteristic polynomial of TT is

χT(λ)  =  det⁡(λIn−A)  ∈  F[λ],\chi_T(\lambda) \;=\; \det(\lambda I_n - A) \;\in\; \mathbf{F}[\lambda],

where InI_n is the n×nn \times n identity matrix and λ\lambda is an indeterminate. Equivalently, for a square matrix A∈Fn×nA \in \mathbf{F}^{n \times n} directly (with no reference to an operator), define χA(λ)=det⁡(λIn−A)\chi_A(\lambda) = \det(\lambda I_n - A). A scalar λ0∈F\lambda_0 \in \mathbf{F} is a root of χT\chi_T if χT(λ0)=0\chi_T(\lambda_0) = 0. The definition is well-posed independent of the choice of basis β\beta, a fact proved below as part of the central theorem.

Intuition

The eigenvalue equation Tv=λvTv = \lambda v for a nonzero vector vv can be rewritten as (λI−T)v=0(\lambda I - T)v = 0. This says vv lies in the kernel of the operator λI−T\lambda I - T. A linear operator on a finite-dimensional space has nontrivial kernel if and only if it fails to be invertible, and invertibility of a matrix is detected exactly by whether its determinant is zero. So the search for eigenvalues — a search over an infinite field for special scalars — collapses to finding the roots of one polynomial, det⁡(λI−T)\det(\lambda I - T), in the single variable λ\lambda. Picture the determinant as a function of λ\lambda: it is a polynomial curve, and the eigenvalues are precisely where that curve crosses (or touches) zero. Since determinants are alternating, multilinear functions of the rows or columns of a matrix built from λI−A\lambda I - A, expanding along any row produces a genuine polynomial in λ\lambda of degree exactly nn, with leading coefficient 11 — hence "monic" of degree nn. The diagram below shows this for a 2×22\times 2 matrix: the graph of χA(λ)\chi_A(\lambda) is a upward parabola crossing the λ\lambda-axis at the two eigenvalues.

Figure
The characteristic polynomial as a curve crossing zero at the eigenvalues
λ χ(λ) λ₁ = 1 λ₂ = 4 roots of χ(λ) = det(λI − A) are the eigenvalues

Crucially, the recipe det⁡(λI−A)\det(\lambda I - A) does not care which basis produced the matrix AA: change of basis replaces AA by P−1APP^{-1}AP, and the determinant construction is engineered precisely so that this substitution leaves χT\chi_T unchanged. This is what allows us to speak of the characteristic polynomial of the operator TT, not merely of a matrix representing it.

TheoremThe Characteristic Polynomial Theorem

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

(1) Well-definedness. If β\beta and γ\gamma are two bases of VV, with A=[T]βA = [T]_\beta and B=[T]γB = [T]_\gamma, then

det⁡(λI−A)=det⁡(λI−B).\det(\lambda I - A) = \det(\lambda I - B).

Hence χT(λ):=det⁡(λI−[T]β)\chi_T(\lambda) := \det(\lambda I - [T]_\beta) does not depend on the choice of basis β\beta.

(2) Degree and leading behavior. χT(λ)\chi_T(\lambda) is a monic polynomial in λ\lambda of degree exactly nn. Writing

χT(λ)=λn−c1λn−1+c2λn−2−⋯+(−1)ncn,\chi_T(\lambda) = \lambda^n - c_1 \lambda^{n-1} + c_2 \lambda^{n-2} - \cdots + (-1)^n c_n,

we have c1=tr⁡(T)c_1 = \operatorname{tr}(T) and cn=det⁡(T)c_n = \det(T).

(3) Roots are eigenvalues. For λ0∈F\lambda_0 \in \mathbf{F}, the scalar λ0\lambda_0 is a root of χT\chi_T (that is, χT(λ0)=0\chi_T(\lambda_0) = 0) if and only if λ0\lambda_0 is an eigenvalue of TT.

(4) Bound on eigenvalues. TT has at most nn distinct eigenvalues in F\mathbf{F}.

Proof

We prove each part in turn, using Determinants (multiplicativity, and that a square matrix is invertible iff its determinant is nonzero) and Eigenvalues and Eigenvectors (the definition of eigenvalue via nontrivial kernel of T−λIT - \lambda I).

(1) Well-definedness. Let β,γ\beta, \gamma be bases of VV and let QQ be the change-of-basis matrix from γ\gamma-coordinates to β\beta-coordinates, so that QQ is invertible and

B=[T]γ=Q−1AQ,A=[T]β.B = [T]_\gamma = Q^{-1} A Q, \qquad A = [T]_\beta.

This is the standard change-of-basis formula for the matrix of a linear operator. Then, treating λ\lambda as an indeterminate (equivalently, working over the field of rational functions F(λ)\mathbf{F}(\lambda), or simply substituting any fixed scalar value for λ\lambda and using that the identity of polynomials follows once it holds for infinitely many field elements — but here we argue directly and formally),

λI−B=λI−Q−1AQ=Q−1(λI)Q−Q−1AQ=Q−1(λI−A)Q,\lambda I - B = \lambda I - Q^{-1}AQ = Q^{-1}(\lambda I)Q - Q^{-1}AQ = Q^{-1}(\lambda I - A)Q,

using that Q−1(λI)Q=λQ−1Q=λIQ^{-1}(\lambda I)Q = \lambda Q^{-1}Q = \lambda I. Taking determinants and using multiplicativity of the determinant together with det⁡(Q−1)=det⁡(Q)−1\det(Q^{-1}) = \det(Q)^{-1},

det⁡(λI−B)=det⁡(Q−1)det⁡(λI−A)det⁡(Q)=det⁡(λI−A)⋅det⁡(Q)−1det⁡(Q)=det⁡(λI−A).\det(\lambda I - B) = \det(Q^{-1})\det(\lambda I - A)\det(Q) = \det(\lambda I - A) \cdot \det(Q)^{-1}\det(Q) = \det(\lambda I - A).

This equality holds identically as polynomials in λ\lambda (the computation is valid entry-by-entry in the matrix λI−A\lambda I - A before any determinant is expanded, so it is an identity of polynomial expressions, not merely a numerical coincidence at each substituted value). Hence χT\chi_T is independent of the basis chosen to compute it, and the definition of χT(λ)\chi_T(\lambda) for the operator TT is legitimate.

(2) Degree and leading behavior. Fix a basis β\beta and let A=[T]β=(aij)A = [T]_\beta = (a_{ij}). By the Leibniz formula for the determinant,

det⁡(λI−A)=∑σ∈Snsgn⁡(σ)∏i=1n(λI−A)i,σ(i).\det(\lambda I - A) = \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \prod_{i=1}^n (\lambda I - A)_{i,\sigma(i)}.

Each diagonal entry of λI−A\lambda I - A is λ−aii\lambda - a_{ii}, a degree-11 polynomial in λ\lambda, and each off-diagonal entry (λI−A)ij=−aij(\lambda I - A)_{ij} = -a_{ij} (for i≠ji \neq j) is a constant (degree 00) in λ\lambda. The term corresponding to the identity permutation σ=id\sigma = \mathrm{id} is

∏i=1n(λ−aii),\prod_{i=1}^n (\lambda - a_{ii}),

which is a polynomial of degree exactly nn with leading term λn\lambda^n. Every other permutation σ≠id\sigma \neq \mathrm{id} moves at least two indices (since a permutation fixing all but one index must fix all indices), so its product includes at most n−2n-2 diagonal factors and at least 22 off-diagonal (constant) factors, giving a term of degree at most n−2<n−1n - 2 < n-1. Hence the only contributions to the coefficients of λn\lambda^n and λn−1\lambda^{n-1} come from σ=id\sigma = \mathrm{id}:

χT(λ)=∏i=1n(λ−aii)+(terms of degree≤n−2).\chi_T(\lambda) = \prod_{i=1}^n (\lambda - a_{ii}) + (\text{terms of degree} \le n-2).

Expanding ∏i=1n(λ−aii)=λn−(∑iaii)λn−1+⋯\prod_{i=1}^n(\lambda - a_{ii}) = \lambda^n - \left(\sum_i a_{ii}\right)\lambda^{n-1} + \cdots, we read off:

  • The polynomial is monic of degree nn (leading coefficient 11), since no other term contributes a λn\lambda^n term and the identity-permutation term supplies exactly λn\lambda^n.
  • The coefficient of λn−1\lambda^{n-1} is −∑iaii=−tr⁡(A)=−tr⁡(T)-\sum_i a_{ii} = -\operatorname{tr}(A) = -\operatorname{tr}(T) (trace is basis-independent, consistent with part (1)). So c1=tr⁡(T)c_1 = \operatorname{tr}(T).
  • The constant term, obtained by setting λ=0\lambda = 0, is det⁡(0⋅I−A)=det⁡(−A)=(−1)ndet⁡(A)=(−1)ndet⁡(T)\det(0 \cdot I - A) = \det(-A) = (-1)^n \det(A) = (-1)^n\det(T). Comparing with the stated form χT(λ)=λn−c1λn−1+⋯+(−1)ncn\chi_T(\lambda) = \lambda^n - c_1\lambda^{n-1} + \cdots + (-1)^n c_n, the constant term is (−1)ncn(-1)^n c_n, so cn=det⁡(T)c_n = \det(T).

(3) Roots are eigenvalues. Fix λ0∈F\lambda_0 \in \mathbf{F}. By definition, λ0\lambda_0 is an eigenvalue of TT if and only if there exists a nonzero v∈Vv \in V with Tv=λ0vTv = \lambda_0 v, equivalently (λ0I−T)v=0(\lambda_0 I - T)v = 0 for some v≠0v \neq 0, equivalently the linear operator λ0I−T\lambda_0 I - T has nontrivial kernel, equivalently λ0I−T\lambda_0 I - T is not injective. Since VV is finite-dimensional, a linear operator on VV is injective if and only if it is surjective, if and only if it is invertible (rank–nullity theorem applied to the operator λ0I−T:V→V\lambda_0 I - T : V \to V). And a linear operator on a finite-dimensional space is invertible if and only if its matrix (with respect to any basis) is invertible, if and only if that matrix has nonzero determinant. Taking β\beta our fixed basis and A=[T]βA = [T]_\beta, the matrix of λ0I−T\lambda_0 I - T is λ0I−A\lambda_0 I - A, so:

λ0 is an eigenvalue of T  ⟺  λ0I−T not invertible  ⟺  det⁡(λ0I−A)=0  ⟺  χT(λ0)=0.\lambda_0 \text{ is an eigenvalue of } T \iff \lambda_0 I - T \text{ not invertible} \iff \det(\lambda_0 I - A) = 0 \iff \chi_T(\lambda_0) = 0.

This chain of equivalences is exactly the claim.

(4) Bound on eigenvalues. By part (2), χT\chi_T is a nonzero polynomial of degree nn over the field F\mathbf{F}. A nonzero polynomial of degree nn over a field has at most nn distinct roots in that field — this is a standard fact from polynomial algebra (proved by induction on degree using polynomial division: if λ0\lambda_0 is a root, factor χT(λ)=(λ−λ0)q(λ)\chi_T(\lambda) = (\lambda - \lambda_0)q(\lambda) with deg⁡q=n−1\deg q = n-1, and any other root of χT\chi_T distinct from λ0\lambda_0 must be a root of qq since F\mathbf{F} has no zero divisors). By part (3), the distinct roots of χT\chi_T in F\mathbf{F} are exactly the distinct eigenvalues of TT in F\mathbf{F}. Hence TT has at most nn distinct eigenvalues. ■\blacksquare

∎

Worked examples

Example 1: A 3×33\times 3 matrix with an asymmetric structure (not symmetric, not triangular). Let

A=(2−10031−204).A = \begin{pmatrix} 2 & -1 & 0 \\ 0 & 3 & 1 \\ -2 & 0 & 4 \end{pmatrix}.

We compute χA(λ)=det⁡(λI−A)\chi_A(\lambda) = \det(\lambda I - A) where

λI−A=(λ−2100λ−3−120λ−4).\lambda I - A = \begin{pmatrix} \lambda - 2 & 1 & 0 \\ 0 & \lambda - 3 & -1 \\ 2 & 0 & \lambda - 4 \end{pmatrix}.

Expand along the first row:

det⁡(λI−A)=(λ−2)det⁡(λ−3−10λ−4)−1⋅det⁡(0−12λ−4)+0.\det(\lambda I - A) = (\lambda - 2)\det\begin{pmatrix}\lambda-3 & -1 \\ 0 & \lambda - 4\end{pmatrix} - 1 \cdot \det\begin{pmatrix}0 & -1 \\ 2 & \lambda-4\end{pmatrix} + 0.

The first 2×22\times 2 minor: (λ−3)(λ−4)−(−1)(0)=(λ−3)(λ−4)(\lambda - 3)(\lambda-4) - (-1)(0) = (\lambda-3)(\lambda-4). The second 2×22\times 2 minor: 0⋅(λ−4)−(−1)(2)=20\cdot(\lambda - 4) - (-1)(2) = 2. So

χA(λ)=(λ−2)(λ−3)(λ−4)−1⋅2.\chi_A(\lambda) = (\lambda - 2)(\lambda-3)(\lambda-4) - 1\cdot 2.

Expand (λ−2)(λ−3)(λ−4)(\lambda-2)(\lambda-3)(\lambda-4): first (λ−2)(λ−3)=λ2−5λ+6(\lambda-2)(\lambda-3) = \lambda^2 - 5\lambda + 6, then

(λ2−5λ+6)(λ−4)=λ3−4λ2−5λ2+20λ+6λ−24=λ3−9λ2+26λ−24.(\lambda^2 - 5\lambda + 6)(\lambda - 4) = \lambda^3 - 4\lambda^2 - 5\lambda^2 + 20\lambda + 6\lambda - 24 = \lambda^3 - 9\lambda^2 + 26\lambda - 24.

So

χA(λ)=λ3−9λ2+26λ−24−2=λ3−9λ2+26λ−26.\chi_A(\lambda) = \lambda^3 - 9\lambda^2 + 26\lambda - 24 - 2 = \lambda^3 - 9\lambda^2 + 26\lambda - 26.

Sanity checks using part (2) of the theorem. The trace of AA is 2+3+4=92 + 3 + 4 = 9, matching the coefficient −c1=−9-c_1 = -9 of λ2\lambda^2, i.e. c1=9=tr⁡(A)c_1 = 9 = \operatorname{tr}(A). Good. The determinant of AA: expand along the first row of AA itself,

det⁡A=2det⁡(3104)−(−1)det⁡(01−24)+0=2(12)+1(0⋅4−1⋅(−2))=24+2=26.\det A = 2\det\begin{pmatrix}3&1\\0&4\end{pmatrix} - (-1)\det\begin{pmatrix}0&1\\-2&4\end{pmatrix} + 0 = 2(12) + 1(0\cdot 4 - 1\cdot(-2)) = 24 + 2 = 26.

Part (2) predicts constant term (−1)3det⁡A=−26(-1)^3 \det A = -26, matching the constant term −26-26 found above. Both checks confirm the computation. Does χA\chi_A have nice rational roots? By the rational root theorem, any rational root divides 26=2⋅1326 = 2\cdot 13, so candidates are ±1,±2,±13,±26\pm 1, \pm 2, \pm 13, \pm 26. Testing λ=1\lambda = 1: 1−9+26−26=−8≠01 - 9 + 26 - 26 = -8 \neq 0. Testing λ=2\lambda=2: 8−36+52−26=−2≠08 - 36 + 52 - 26 = -2 \neq 0. None of the small candidates work cleanly, illustrating a genuine and important point: characteristic polynomials of general matrices need not factor nicely over Q\mathbf{Q}, and finding exact eigenvalues may require numerical methods or working in an extension field. The eigenvalues of AA exist (as complex numbers, by the fundamental theorem of algebra) but are irrational/non-obvious real or complex numbers here — this is realistic and expected for a "generic" matrix.

Example 2: A non-diagonalizable-looking 2×22\times 2 matrix over R\mathbf{R}, illustrating a repeated root. Let

B=(5−141).B = \begin{pmatrix} 5 & -1 \\ 4 & 1 \end{pmatrix}.

Then

λI−B=(λ−51−4λ−1),\lambda I - B = \begin{pmatrix} \lambda - 5 & 1 \\ -4 & \lambda - 1 \end{pmatrix},

so

χB(λ)=(λ−5)(λ−1)−(1)(−4)=λ2−6λ+5+4=λ2−6λ+9=(λ−3)2.\chi_B(\lambda) = (\lambda-5)(\lambda - 1) - (1)(-4) = \lambda^2 - 6\lambda + 5 + 4 = \lambda^2 - 6\lambda + 9 = (\lambda - 3)^2.

Check with part (2): tr⁡(B)=5+1=6\operatorname{tr}(B) = 5 + 1 = 6, matching c1=6c_1 = 6. det⁡(B)=5(1)−(−1)(4)=5+4=9\det(B) = 5(1) - (-1)(4) = 5+4=9, matching constant term c2=9c_2 = 9 (with (−1)2c2=9(-1)^2 c_2 = 9). Both match. Here χB(λ)=(λ−3)2\chi_B(\lambda) = (\lambda-3)^2 has a repeated root λ0=3\lambda_0 = 3 with multiplicity 22 as a root of the polynomial. By part (3) of the theorem, λ0=3\lambda_0 = 3 is the only eigenvalue of BB. This is a first glimpse of the distinction to come between the algebraic multiplicity of an eigenvalue (its multiplicity as a root of χT\chi_T, here 22) and its geometric multiplicity (the dimension of the eigenspace ker⁡(B−3I)\ker(B - 3I)), a distinction explored fully under Geometric vs Algebraic Multiplicity. Direct computation shows

B−3I=(2−14−2),B - 3I = \begin{pmatrix}2 & -1\\4 & -2\end{pmatrix},

whose rows are proportional (4=2⋅24 = 2\cdot 2, −2=2⋅(−1)-2 = 2\cdot(-1)), so this matrix has rank 11 and nullity 11: the eigenspace is 11-dimensional, strictly smaller than the algebraic multiplicity 22. This exact phenomenon is why the characteristic polynomial alone does not determine whether TT is diagonalizable — it only bounds the possibilities, setting up the need for the deeper multiplicity theory and eventually the The Cayley-Hamilton Theorem.

CounterexampleWhy finite dimension is needed

The entire construction of χT(λ)=det⁡(λI−T)\chi_T(\lambda) = \det(\lambda I - T) requires VV to be finite-dimensional, because the determinant is only defined for matrices — that is, for linear maps between finite-dimensional spaces expressed in finite bases. Drop this hypothesis and the theorem's proof breaks down immediately at the step "let A=[T]βA = [T]_\beta." Consider the infinite-dimensional space V=F[x]V = \mathbf{F}[x] of all polynomials over a field F\mathbf{F}, and the linear operator

T:V→V,T(p(x))=x⋅p(x)T : V \to V, \qquad T(p(x)) = x\cdot p(x)

(multiplication by xx). This TT is injective (if xp(x)=0xp(x) = 0 then p(x)=0p(x)=0) but not surjective (the constant polynomial 11 is not in the image, since every element of the image has xx as a factor). So TT is injective but not invertible — already a phenomenon impossible in finite dimensions, where injective implies invertible (as used in part (3) of the proof). Does TT have eigenvalues? Suppose Tp=λpTp = \lambda p for some scalar λ\lambda and nonzero polynomial pp. Then xp(x)−λp(x)=0xp(x) - \lambda p(x) = 0, i.e., (x−λ)p(x)=0(x - \lambda)p(x) = 0 in F[x]\mathbf{F}[x], forcing p(x)=0p(x) = 0 since F[x]\mathbf{F}[x] has no zero divisors and x−λ≠0x - \lambda \neq 0. So TT has no eigenvalues at all. There is no finite-degree "characteristic polynomial" to write down: VV has no finite basis, no matrix [T]β[T]_\beta exists, and the very expression det⁡(λI−T)\det(\lambda I - T) is meaningless. The bound "at most nn eigenvalues" from part (4) also has no analogue — the correct statement in infinite dimensions is that TT can have zero eigenvalues even though intuitively "λI−T\lambda I - T failing to be invertible" is still a meaningful (spectral-theoretic) question, just not one answered by a polynomial. This shows finite-dimensionality is not a technical nicety but load-bearing: without it, there is no characteristic polynomial to construct, and the eigenvalue-counting theorem fails outright.

Scope, edge cases & common traps

SituationWhat happensWhy
F=R\mathbf{F} = \mathbf{R}, roots of χT\chi_T are complex, non-realTT has no real eigenvalues even though χT\chi_T has degree nnPart (3) only detects eigenvalues in the field F\mathbf{F}; a root of χT\chi_T in C∖R\mathbf{C}\setminus\mathbf{R} is not an eigenvalue of TT viewed as a real operator
Repeated root of χT\chi_TAlgebraic multiplicity >1>1 does not imply the eigenspace has the same dimensionSee Example 2; geometric multiplicity ≤\le algebraic multiplicity always, but can be strictly less
χT\chi_T has degree <n< nNever happens for finite-dimensional VVPart (2) proves degree is exactly nn, always, for any operator on an nn-dimensional space
Convention det⁡(T−λI)\det(T - \lambda I) vs det⁡(λI−T)\det(\lambda I - T)Differ by a factor of (−1)n(-1)^nBoth are used in textbooks; roots are identical either way, only overall sign (and monic-ness) differs
Infinite-dimensional VVNo characteristic polynomial existsDeterminants and finite matrices are undefined; see counterexample above

The mathematician behind it

Origins · 1 minAugustin-Louis Cauchy and how this was discovered
AC
Augustin-Louis Cauchy
1789–1857 · Early 19th-century France, École Polytechnique and the Paris Academy of Sciences

Cauchy was one of the most prolific mathematicians in history, laying rigorous foundations for real and complex analysis while also making foundational contributions to determinant theory and the study of what would become eigenvalues — which he called "characteristic roots" in his work on quadratic forms and mechanical systems (principal axes of inertia, vibration frequencies of coupled systems). Cauchy's 1829 work on the diagonalization of quadratic forms associated with symmetric systems is among the earliest systematic uses of the determinant det⁡(A−λI)\det(A - \lambda I) to find the special scalars that made a linear system degenerate — precisely the characteristic polynomial construction formalized in this article. His insistence on rigor (epsilon-delta arguments, careful convergence proofs) set the standard that made this kind of "define an object, then prove it well-behaved" theorem-and-proof style, used throughout this article, the norm in mathematics.

History
- 1829 — Cauchy studies the characteristic equation of symmetric matrices in connection with quadratic forms and principal axes.
- 1858 — Arthur Cayley publishes A Memoir on the Theory of Matrices, developing matrix algebra and stating what becomes the Cayley–Hamilton theorem.
- 1878 — Ferdinand Georg Frobenius gives a rigorous general proof of the Cayley–Hamilton theorem for arbitrary square matrices.
- 1904 — David Hilbert's work on integral equations popularizes the term "eigenvalue" (from German eigen, "own" or "characteristic") in the broader operator-theoretic setting.
- 20th century — The characteristic polynomial becomes a standard tool in linear algebra curricula, unifying trace, determinant, and eigenvalue theory in the language of this article.
Why do we use λI−T\lambda I - T instead of T−λIT - \lambda I?

Both conventions appear in textbooks, and they differ only by an overall sign: det⁡(λI−T)=(−1)ndet⁡(T−λI)\det(\lambda I - T) = (-1)^n \det(T - \lambda I) for an n×nn \times n operator. The roots are identical either way, since a product of (−1)n(-1)^n (a nonzero scalar) does not change where a polynomial equals zero. The convention det⁡(λI−T)\det(\lambda I - T) is preferred because it produces a monic polynomial (leading coefficient +1+1), which is more convenient for stating results like Cayley–Hamilton and for uniquely factoring the polynomial over an algebraically closed field. Some authors, especially in physics-oriented texts, use det⁡(T−λI)\det(T - \lambda I) instead; just be consistent within a single computation.

Does every matrix have a characteristic polynomial with real roots?

No. A real n×nn \times n matrix always has a characteristic polynomial with real coefficients, but the roots of that polynomial need not be real. For example, the rotation matrix (0−110)\begin{pmatrix}0&-1\\1&0\end{pmatrix} has characteristic polynomial λ2+1\lambda^2+1, whose roots are ±i\pm i, not real numbers. This reflects the geometric fact that a genuine rotation by 90° has no real eigenvector: no nonzero real vector is sent to a real scalar multiple of itself. Over C\mathbf{C}, however, the fundamental theorem of algebra guarantees nn roots counted with multiplicity, which is why eigenvalue theory is often developed over algebraically closed fields.

Is the characteristic polynomial the same as the minimal polynomial?

Not in general, though they are closely related. The characteristic polynomial always has degree exactly dim⁡V\dim V, while the minimal polynomial (the monic polynomial of least degree with p(T)=0p(T) = 0) can have strictly smaller degree. Both polynomials have exactly the same roots (the eigenvalues of TT), a consequence of the Cayley–Hamilton theorem together with basic facts about minimal polynomials, but their multiplicities can differ. For instance, the identity operator on Fn\mathbf{F}^n has characteristic polynomial (λ−1)n(\lambda - 1)^n but minimal polynomial simply λ−1\lambda - 1.

Can two non-similar matrices have the same characteristic polynomial?

Yes, this is an important subtlety. The characteristic polynomial is a similarity invariant (it is preserved by conjugation, as proved in part (1) of the theorem above), but it is not a complete invariant of similarity — two matrices can share a characteristic polynomial while genuinely not being similar to each other. A classic example: I2I_2 (the 2×22\times2 identity) and (1101)\begin{pmatrix}1&1\\0&1\end{pmatrix} both have characteristic polynomial (λ−1)2(\lambda-1)^2, but they are not similar, since I2I_2 is diagonalizable (indeed already diagonal) while the second matrix is not. Distinguishing such cases requires finer invariants like the minimal polynomial or the full Jordan canonical form.

How does the characteristic polynomial relate to eigenspaces and diagonalizability?

The characteristic polynomial tells you exactly which scalars are eigenvalues and with what algebraic multiplicity (multiplicity as a root), but it does not by itself tell you the dimension of each eigenspace. Diagonalizability requires, for every eigenvalue, that the eigenspace dimension (geometric multiplicity) equal the root multiplicity (algebraic multiplicity), and separately that χT\chi_T split completely into linear factors over the base field. The characteristic polynomial is thus a necessary first step and gives an upper bound on each eigenspace's dimension, but genuine diagonalizability analysis requires the extra work developed under Geometric vs Algebraic Multiplicity.

Why does the theorem require finite dimension so heavily?

Every step of the proof leans on finite dimension: the matrix [T]β[T]_\beta only exists relative to a finite basis, the determinant is only defined for finite square matrices, and the equivalence "injective iff surjective iff invertible" used in part (3) is a rank–nullity consequence that is specific to finite-dimensional spaces. In infinite dimensions, operators can be injective without being surjective (see the counterexample above with multiplication by xx on polynomials), so the entire logical chain collapses, and there is no analogous finite-degree polynomial to construct.

What's the fastest way to sanity-check a computed characteristic polynomial?

Compute the trace and determinant of the original matrix independently, then check that the coefficient of λn−1\lambda^{n-1} in χT\chi_T equals −tr⁡(T)-\operatorname{tr}(T) and the constant term equals (−1)ndet⁡(T)(-1)^n \det(T), exactly as done in both worked examples above. This catches the overwhelming majority of arithmetic slips in a cofactor expansion, since trace and determinant are each computed by an entirely independent, usually much shorter, calculation.

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