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Characteristic Equation



From Eigenvectors to the Determinant Condition
Eigenvalue Notation
The Characteristic Polynomial
Computing the Characteristic Polynomial: 2×2
Computing the Characteristic Polynomial: 3×3
Larger Matrices
Algebraic Multiplicity
Finding Eigenvectors After Finding Eigenvalues
The Cayley-Hamilton Theorem
Characteristic Polynomial and Similarity
Summary: The Characteristic Polynomial at a Glance
Common Pitfalls
Characteristic Equation FAQ



The Polynomial Whose Roots Are the Eigenvalues

The eigenvalue problem Av = λv converts into a determinant condition: det(A − λI) = 0. This determinant is a polynomial in λ whose roots are the eigenvalues. Computing the characteristic polynomial and factoring it is the standard method for finding eigenvalues of small matrices — and the polynomial's coefficients encode the trace, determinant, and other invariants of the matrix.



From Eigenvectors to the Determinant Condition

The equation Av=λvA\mathbf{v} = \lambda\mathbf{v} rearranges to (A−λI)v=0(A - \lambda I)\mathbf{v} = \mathbf{0}. This is a homogeneous system, and eigenvectors are its nontrivial solutions. Nontrivial solutions exist if and only if the coefficient matrix A−λIA - \lambda I is singular:

Characteristic Equation
det⁡(A−λI)=0\det(A - \lambda I) = 0

This is the characteristic equation. It holds for exactly those values of λ\lambda that are eigenvalues of AA. Every other value of λ\lambda makes A−λIA - \lambda I invertible, the system has only the trivial solution, and no eigenvector exists for that λ\lambda.

The characteristic equation transforms the geometric question "which directions does AA preserve?" into the algebraic question "for which λ\lambda is this determinant zero?"

Eigenvalue Notation

Notation

Eigenvalue Notation

The Greek letter no one dares rename, the identity matrix that must not be dropped, and the subscript that names a space after a number.
det⁡\det and its bars — determinant notation; II and matrix algebra — matrix notation; Null\text{Null} — the four subspaces.
Av=λvA\mathbf{v} = \lambda\mathbf{v}
A v equals lambda v
λ\lambda is the subject's reserved letter — eigenvalues are λ\lambda in every textbook on earth, enumerated λ1,λ2,…\lambda_1, \lambda_2, \ldots when there are several. The defining equation reads as a balance: matrix action on the left, plain scaling on the right, same vector both sides.
CasesThe full set of eigenvalues has its own mark in advanced texts: the spectrum σ(A)\sigma(A) — a set-valued operator, sibling to Col\text{Col} and Null\text{Null}.
Also writtenμ\mu when λ\lambda is spoken for, in perturbation and two-matrix arguments; German texts historically used λ\lambda from the start — the letter is Hilbert-era and stuck.
Do not confuseA matrix. λ\lambda is a scalar — which is exactly why A−λA - \lambda is illegal and the next entry's II exists.
det⁡(A−λI)=0\det(A - \lambda I) = 0
The determinant of A minus lambda I, set to zero
The II is load-bearing: a scalar cannot be subtracted from a matrix, so λ\lambda must first become λI\lambda I — a scaling of the identity. Writing A−λA - \lambda is the classic species error; the notation A−λIA - \lambda I is its permanent correction. The equation itself is derived in From Eigenvectors to the Determinant Condition above.
CasesTwo sign conventions circulate: det⁡(A−λI)\det(A - \lambda I) — this site's, keeping AA first — and det⁡(λI−A)\det(\lambda I - A), which makes the polynomial monic. They differ by (−1)n(-1)^n; the roots, being what matters, agree.
Also writtenp(λ)p(\lambda), or pA(λ)p_A(\lambda) with the matrix as subscript — the characteristic polynomial named as a function, per The Characteristic Polynomial below.
Do not confuseAn equation to solve for AA. The unknown is λ\lambda — the matrix is data, the scalar is the variable, an inversion of the usual roles that the notation quietly performs.
Eλ=Null(A−λI)E_{\lambda} = \text{Null}(A - \lambda I)
The eigenspace for lambda
A space named after a number: the subscript on EλE_{\lambda} is the eigenvalue whose eigenvectors (plus 0\mathbf{0}) fill it. One subscripted letter per eigenvalue — E3E_3 is the eigenspace for λ=3\lambda = 3, not a third eigenspace.
CasesEach λ\lambda carries two counted quantities: algebraic multiplicity — its exponent as a root of p(λ)p(\lambda), the multiplicity notation of polynomials — and geometric multiplicity, dim⁡Eλ\dim E_{\lambda}. The pair drives Algebraic Multiplicity below and the whole of diagonalization.
Also writtenker⁡(A−λI)\ker(A - \lambda I) — the map-dialect spelling, per kernel notation.
Do not confuseThe standard-basis letter. ei\mathbf{e}_i is a basis vector, EijE_{ij} a matrix unit, EλE_{\lambda} an eigenspace — one letter, three species, separated by bold, double subscripts, and a Greek subscript respectively.

The Characteristic Polynomial

The characteristic polynomial of AA is the determinant of A−λIA - \lambda I viewed as a function of λ\lambda:

Characteristic Polynomial
p(λ)=det⁡(A−λI)p(\lambda) = \det(A - \lambda I)

It is a polynomial of degree nn in the variable λ\lambda.

For an n×nn \times n matrix, p(λ)p(\lambda) has degree nn with leading term (−1)nλn(-1)^n \lambda^n. The constant term is p(0)=det⁡(A)p(0) = \det(A) — the determinant of the matrix itself. The coefficient of λn−1\lambda^{n-1} is (−1)n−1tr(A)(-1)^{n-1}\text{tr}(A), connecting the next-to-leading term to the trace.

THEOREM
The eigenvalues are precisely the roots of p(λ)=0p(\lambda) = 0.

Every root is an eigenvalue, and every eigenvalue is a root. The characteristic polynomial packages the entire eigenvalue structure of the matrix into a single algebraic expression.
A − λI3×32 − λ0003 − λ4049 − λ→λ³ · λ² · λ · 11−1435−22
The determinant expanded into a polynomial

Taking the determinant of the shifted matrix has produced a polynomial in λ whose degree equals the size of the matrix. Its roots are the eigenvalues, so the question has been converted from a matrix problem into a root-finding problem. Expand it for a matrix of your own on the eigenvalues and eigenvectors visualizer.

Because the degree matches the size, an n×n matrix has exactly n eigenvalues once multiplicity and complex roots are counted.

Computing the Characteristic Polynomial: 2×2

For A=(abcd)A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}, the characteristic polynomial is

Computing the Characteristic Polynomial: 2×2
p(λ)=det⁡(a−λbcd−λ)=(a−λ)(d−λ)−bc=λ2−(a+d)λ+(ad−bc)p(\lambda) = \det\begin{pmatrix} a - \lambda & b \\ c & d - \lambda \end{pmatrix} = (a - \lambda)(d - \lambda) - bc = \lambda^2 - (a + d)\lambda + (ad - bc)

Rewritten in terms of trace and determinant:

Characteristic Polynomial 2x2
p(λ)=λ2−tr(A) λ+det⁡(A)p(\lambda) = \lambda^2 - \text{tr}(A)\,\lambda + \det(A)

The eigenvalues follow from the quadratic formula:

Computing the Characteristic Polynomial: 2×2
λ=tr(A)±tr(A)2−4det⁡(A)2\lambda = \frac{\text{tr}(A) \pm \sqrt{\text{tr}(A)^2 - 4\det(A)}}{2}

The discriminant Δ=tr(A)2−4det⁡(A)\Delta = \text{tr}(A)^2 - 4\det(A) classifies the eigenvalue type. When Δ>0\Delta > 0, there are two distinct real eigenvalues. When Δ=0\Delta = 0, there is one repeated real eigenvalue. When Δ<0\Delta < 0, the eigenvalues are a complex conjugate pair.

Worked Example
For A=(5234)A = \begin{pmatrix} 5 & 2 \\ 3 & 4 \end{pmatrix}: tr(A)=9\text{tr}(A) = 9, det⁡(A)=14\det(A) = 14, Δ=81−56=25\Delta = 81 - 56 = 25. The eigenvalues are λ=9±52\lambda = \frac{9 \pm 5}{2}, giving λ1=7\lambda_1 = 7 and λ2=2\lambda_2 = 2.

Pitfall: Flipping the sign of the trace term
Writing p(λ)=λ2+tr(A)λ+det⁡(A)p(\lambda) = \lambda^2 + \text{tr}(A)\lambda + \det(A) for A=(5234)A = \begin{pmatrix} 5 & 2 \\ 3 & 4 \end{pmatrix} gives λ2+9λ+14=(λ+7)(λ+2)\lambda^2 + 9\lambda + 14 = (\lambda + 7)(\lambda + 2) and the eigenvalues −7-7 and −2-2. With the middle term −tr(A)λ-\text{tr}(A)\lambda, the roots are 77 and 22.

In the worked example, both inputs to the formula are read straight off the matrix: the trace 5+4=95 + 4 = 9 from the diagonal, and the determinant 5⋅4−2⋅3=145 \cdot 4 - 2 \cdot 3 = 14 from the product of the diagonal entries minus the product of the off-diagonal ones. Nothing else about the matrix is needed. The discriminant, computed from the same two numbers, already showed that the two eigenvalues would be real and distinct before either was found.
Discriminant Δ = tr(A)² − 4·det(A) Eigenvalue type Independent real eigenvectors
Δ > 0 two distinct real eigenvalues 2 — always diagonalizable
Δ = 0 one repeated real eigenvalue 1 or 2 — diagonalizable only when A = λ · I (otherwise defective)
Δ < 0 complex conjugate pair a ± bi 0 real (one conjugate pair of complex eigenvectors over ℂ)
λ³ · λ² · λ · 11−1435−22→eigenvalues λ1121
Roots extracted from the characteristic polynomial

The polynomial has been solved and its roots listed — these are the eigenvalues, and a repeated root is recorded with its multiplicity rather than collapsed into one. That bookkeeping matters later, because algebraic multiplicity and the number of independent eigenvectors need not agree. See the repeated and complex cases on the eigenvalues and eigenvectors visualizer.

Beyond 4×4 no formula exists for these roots, which is why large eigenvalue problems are solved iteratively rather than algebraically.

Computing the Characteristic Polynomial: 3×3

For a 3×33 \times 3 matrix, expanding det⁡(A−λI)\det(A - \lambda I) using cofactor expansion produces a cubic polynomial:

Computing the Characteristic Polynomial: 3×3
p(λ)=−λ3+tr(A)λ2−(sum of 2×2 principal minors)λ+det⁡(A)p(\lambda) = -\lambda^3 + \text{tr}(A)\lambda^2 - (\text{sum of } 2 \times 2 \text{ principal minors})\lambda + \det(A)

The computation is lengthier but follows the same cofactor mechanics as any 3×33 \times 3 determinant.

Worked Example


For A=(210031001)A = \begin{pmatrix} 2 & 1 & 0 \\ 0 & 3 & 1 \\ 0 & 0 & 1 \end{pmatrix}, this is upper triangular, so A−λIA - \lambda I is also upper triangular with diagonal entries 2−λ2 - \lambda, 3−λ3 - \lambda, 1−λ1 - \lambda:

Computing the Characteristic Polynomial: 3×3
p(λ)=(2−λ)(3−λ)(1−λ)p(\lambda) = (2 - \lambda)(3 - \lambda)(1 - \lambda)

The eigenvalues are λ=1,2,3\lambda = 1, 2, 3 — readable directly from the diagonal.

RULE
For triangular matrices, the characteristic polynomial always factors as the product of the diagonal terms, making the eigenvalues visible by inspection.

For non-triangular 3×33 \times 3 matrices, the cubic must be factored by finding rational roots (testing factors of the constant term), by inspection, or by the cubic formula.

Larger Matrices

For an n×nn \times n matrix, the characteristic polynomial has degree nn, and finding its roots becomes increasingly difficult as nn grows.

THEOREMAbel–Ruffini theorem
There is no general closed-form formula for roots of polynomials of degree 55 or higher (Abel-Ruffini theorem), so explicit factoring is limited to small matrices or matrices with special structure.

Fact: Root finders compute eigenvalues
Every polynomial λn+cn−1λn−1+⋯+c0\lambda^n + c_{n-1}\lambda^{n-1} + \cdots + c_0 is, up to sign, the characteristic polynomial of its companion matrix: 11s just below the diagonal and −c0,…,−cn−1-c_0, \ldots, -c_{n-1} down the last column. The roots functions of MATLAB and NumPy run the idea backwards, finding a polynomial's roots as that matrix's eigenvalues.

Diagonal and triangular matrices are immediate: the eigenvalues are the diagonal entries. Block triangular matrices factor block by block: the characteristic polynomial is the product of the characteristic polynomials of the diagonal blocks.

For general large matrices, eigenvalues are computed numerically by iterative algorithms — most importantly the QR algorithm, which repeatedly applies QR decompositions to converge on the eigenvalues without ever forming the characteristic polynomial explicitly. Computing the polynomial and then finding its roots is numerically unstable for large nn and is never used in practice.
Matrix structure How to find the eigenvalues
Diagonal eigenvalues are the diagonal entries — read off by inspection
Triangular (upper or lower) eigenvalues are the diagonal entries (det of a triangular matrix is the product of its diagonal)
Block triangular multiply the characteristic polynomials of each diagonal block; eigenvalues are the union of block eigenvalues
General 2×2 quadratic formula on λ² − tr(A)·λ + det(A) = 0
General 3×3 or 4×4 expand det(A − λI) and factor (rational-root testing, cubic/quartic formulas, or inspection)
General n ≥ 5 (numerical) iterative algorithms such as QR iteration — the polynomial is never formed explicitly
The characteristic polynomial defines the eigenvalues, but beyond small or structured matrices it is not how they are computed.

Algebraic Multiplicity

If λ0\lambda_0 is a root of the characteristic polynomial p(λ)p(\lambda), its algebraic multiplicity is the largest power kk such that (λ−λ0)k(\lambda - \lambda_0)^k divides p(λ)p(\lambda). Equivalently, it is the multiplicity of λ0\lambda_0 as a root.

If p(λ)=(λ−2)3(λ+1)p(\lambda) = (\lambda - 2)^3(\lambda + 1), then λ=2\lambda = 2 has algebraic multiplicity 33 and λ=−1\lambda = -1 has algebraic multiplicity 11. The algebraic multiplicities of all eigenvalues sum to nn — the degree of the polynomial — when complex roots are included.

THEOREM
The algebraic multiplicity is an upper bound for the geometric multiplicity: 1≤mg(λ)≤ma(λ)1 \leq m_g(\lambda) \leq m_a(\lambda).

Pitfall: Assuming a double root gives two eigenvectors
Counting a double root as two independent eigenvectors fails for A=(2102)A = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}: p(λ)=(λ−2)2p(\lambda) = (\lambda - 2)^2, but A−2I=(0100)A - 2I = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} has only one free variable. The eigenspace E2E_2 is the line through (1,0)T(1, 0)^T, so ma=2m_a = 2 while mg=1m_g = 1.

The geometric multiplicity is the dimension of the eigenspace, and it can be strictly smaller than the algebraic multiplicity. When this gap occurs for any eigenvalue, the matrix is not diagonalizable.
A − (1)I3×3111111111→rref3×3111000000→eigenvectors3×211−100−1
A repeated eigenvalue, λ = 1

Here λ = 1 is a double root of the characteristic polynomial, so its algebraic multiplicity is 2. Substituting it back, A − I has two free columns, and the solutions of (A − I)v = 0 fill a whole plane — for this matrix the eigenspace is as large as the multiplicity allows. That is not guaranteed, which is what the geometric multiplicity measures. Work through the repeated preset on the eigenvalues and eigenvectors visualizer.

Algebraic multiplicity counts the root; how many independent eigenvectors it brings is a separate question.

Finding Eigenvectors After Finding Eigenvalues

Once the eigenvalues are known, the eigenvectors for each λi\lambda_i are found by solving the homogeneous system (A−λiI)v=0(A - \lambda_i I)\mathbf{v} = \mathbf{0}.

Row reduce A−λiIA - \lambda_i I and express the general solution in parametric form. Each free variable contributes one basis vector for the eigenspace EλiE_{\lambda_i}.

Worked Example
For A=(1243)A = \begin{pmatrix} 1 & 2 \\ 4 & 3 \end{pmatrix}, the characteristic polynomial is λ2−4λ−5=(λ−5)(λ+1)\lambda^2 - 4\lambda - 5 = (\lambda - 5)(\lambda + 1). Eigenvalues: λ1=5\lambda_1 = 5, λ2=−1\lambda_2 = -1.
For λ1=5\lambda_1 = 5: A−5I=(−424−2)A - 5I = \begin{pmatrix} -4 & 2 \\ 4 & -2 \end{pmatrix}. Row reducing: (1−1/200)\begin{pmatrix} 1 & -1/2 \\ 0 & 0 \end{pmatrix}. Free variable v2=tv_2 = t, so v1=t/2v_1 = t/2. Eigenvector: v1=(1,2)T\mathbf{v}_1 = (1, 2)^T.
For λ2=−1\lambda_2 = -1: A+I=(2244)A + I = \begin{pmatrix} 2 & 2 \\ 4 & 4 \end{pmatrix}. Row reducing: (1100)\begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}. Free variable v2=tv_2 = t, so v1=−tv_1 = -t. Eigenvector: v2=(−1,1)T\mathbf{v}_2 = (-1, 1)^T.
Verification: Av1=(510)=5v1A\mathbf{v}_1 = \begin{pmatrix} 5 \\ 10 \end{pmatrix} = 5\mathbf{v}_1 and Av2=(1−1)=−1⋅v2A\mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix} = -1 \cdot \mathbf{v}_2.

Pitfall: Taking a row as the eigenvector
Reading the row (1,−1/2)(1, -1/2) of the reduced A−5IA - 5I as the eigenvector fails: A(1,−1/2)T=(0,5/2)TA(1, -1/2)^T = (0, 5/2)^T, which is not a multiple of (1,−1/2)T(1, -1/2)^T. The eigenvector is a solution of that row's equation, (1,2)T(1, 2)^T.

In both cases the reduced matrix keeps a single nonzero row, which encodes one equation in the two unknowns: v1−12v2=0v_1 - \frac{1}{2}v_2 = 0 for λ1=5\lambda_1 = 5 and v1+v2=0v_1 + v_2 = 0 for λ2=−1\lambda_2 = -1. Setting the free variable v2=tv_2 = t turns each equation into a line of solutions through the origin, and any nonzero point on that line serves as the eigenvector. Choosing t=2t = 2 in the first case and t=1t = 1 in the second keeps the entries whole.
A − (11)I3×3−9000−8404−2→rref3×310001−1/2000→v3×1012
Solving the shifted system for one eigenvalue

With a specific eigenvalue substituted, the shifted matrix is being reduced and its null space read off. Any non-zero vector in that space is an eigenvector, and the whole space is the eigenspace — which is why eigenvectors are never unique and are usually quoted normalised. Run the solve for each eigenvalue in turn on the eigenvalues and eigenvectors visualizer.

Finding the eigenvalues is therefore only half the work; each one still has its own null space to compute.

The Cayley-Hamilton Theorem

Every square matrix satisfies its own characteristic polynomial. If p(λ)=det⁡(A−λI)p(\lambda) = \det(A - \lambda I) is the characteristic polynomial, then

Cayley-Hamilton
p(A)=Op(A) = O

where OO is the zero matrix and λ\lambda is replaced by AA (with constant terms multiplied by II).

Fact: Cayley checked only two sizes
Cayley published the theorem in 1858 after checking it for 2×22 \times 2 and 3×33 \times 3 matrices, and wrote that a general proof did not seem necessary. Hamilton had met the quaternion case earlier; the first proof for every size came from Frobenius in 1878.

For example, if p(λ)=λ2−5λ+6p(\lambda) = \lambda^2 - 5\lambda + 6, then A2−5A+6I=OA^2 - 5A + 6I = O. This can be rearranged to express A−1A^{-1} as a polynomial in AA: A−1=16(5I−A)A^{-1} = \frac{1}{6}(5I - A) (provided det⁡(A)=6≠0\det(A) = 6 \neq 0). More generally, the Cayley-Hamilton theorem guarantees that, whenever AA is invertible, A−1A^{-1} can be written as a polynomial in AA of degree at most n−1n - 1.

The theorem also shows that any power AkA^k with k≥nk \geq n can be reduced to a polynomial in AA of degree at most n−1n - 1 — the characteristic polynomial provides a recurrence that expresses higher powers in terms of lower ones.

Because an n×nn \times n matrix satisfies its own characteristic polynomial, every power of it reduces to a combination of I,A,…,An−1I, A, \ldots, A^{n-1}.

Characteristic Polynomial and Similarity

Similar matrices have the same characteristic polynomial:

Characteristic Polynomial and Similarity
det⁡(P−1AP−λI)=det⁡(P−1(A−λI)P)=det⁡(A−λI)\det(P^{-1}AP - \lambda I) = \det(P^{-1}(A - \lambda I)P) = \det(A - \lambda I)

The second equality uses the multiplicative property of the determinant: det⁡(P−1)det⁡(A−λI)det⁡(P)=det⁡(A−λI)\det(P^{-1})\det(A - \lambda I)\det(P) = \det(A - \lambda I), since det⁡(P−1)det⁡(P)=1\det(P^{-1})\det(P) = 1.

This means the characteristic polynomial is a property of the linear transformation itself, not of any particular matrix representation. Changing the basis changes the matrix but not the polynomial. Since the eigenvalues are the roots of the polynomial, similar matrices have the same eigenvalues with the same algebraic multiplicities.

The trace and determinant are just two of the nn coefficients of the characteristic polynomial. The polynomial carries more information than either one alone — it determines the complete multiset of eigenvalues, not just their sum and product.

Summary: The Characteristic Polynomial at a Glance

The characteristic polynomial p(λ) = det(A − λI) carries every spectral fact about A — its degree, individual coefficients, roots, and behavior under similarity all encode information about the matrix. The table below collects each structural property of the polynomial alongside its statement and the connection it makes to A.
The polynomial is usually formed in order to solve it, and the roots are indeed the point. But the coefficients are informative before any solving happens: the constant term is the determinant, the next coefficient is the trace, and both are readable off the matrix directly. The grouping below separates what the coefficients tell you from what the roots tell you, with the closed forms and their consequences last.
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Eigen · characteristic polynomial

What the polynomial encodes

The roots give the eigenvalues, which is the reason to form it. The coefficients give trace and determinant for free, which is the reason it is worth reading rather than only solving.

9readings
What the coefficients carry3
1
Degree and leading term§ 2
the sign alternates with nn
deg⁡p=n\deg p = n, leading term (−1)nλn(-1)^n\lambda^n
One root per dimension, counted with multiplicity, over C\mathbb{C}. Some texts negate the whole polynomial to make it monic — harmless, but it flips the sign of every coefficient below, so the convention has to be fixed before the formulas are quoted.
2
∑λi=tr⁡(A)\sum \lambda_i = \operatorname{tr}(A)
The trace is available without solving anything — it is the sum of the diagonal, and the polynomial says that sum is also the sum of the roots. A cheap consistency check on any eigenvalue computation.
3
p(0)=det⁡(A)=∏λip(0) = \det(A) = \prod \lambda_i
Setting λ=0\lambda = 0 in det⁡(A−λI)\det(A - \lambda I) leaves det⁡(A)\det(A). So AA is invertible exactly when p(0)≠0p(0) \neq 0, which is the same as saying no eigenvalue is zero — three statements that turn out to be one.
What the roots carry3
4
Roots are the eigenvalues§ 1
over C\mathbb{C}, with multiplicity
p(λ)=0  ⟺  λp(\lambda) = 0 \iff \lambda is an eigenvalue
The determinant condition is what turns "(A−λI)x=0(A - \lambda I)\mathbf{x} = \mathbf{0} has a nontrivial solution" into an equation that can be solved. Over R\mathbb{R} some roots may not exist — a rotation has none — which is why the spectrum is stated over C\mathbb{C}.
5
Algebraic multiplicity§ 6
mg≤mam_g \leq m_a, never the reverse
multiplicity of λ\lambda as a root
How many times a root repeats, which is not the same as how many independent eigenvectors it has. When the geometric count falls short the matrix is defective and cannot be diagonalized — the polynomial cannot detect this on its own.
6
pP−1AP(λ)=pA(λ)p_{P^{-1}AP}(\lambda) = p_A(\lambda)
Similar matrices share the polynomial, hence the eigenvalues, trace and determinant. Two matrices with different characteristic polynomials cannot represent the same transformation in different bases — which makes pp a usable test for similarity in one direction.
Closed forms and consequences3
7
2×2 form§ 3
worth memorising
p(λ)=λ2−tr⁡(A)λ+det⁡(A)p(\lambda) = \lambda^2 - \operatorname{tr}(A)\lambda + \det(A)
The discriminant tr⁡2−4det⁡\operatorname{tr}^2 - 4\det classifies the pair before any solving: positive gives two distinct reals, zero gives a repeated root, negative gives a complex conjugate pair.
8
3×3 form§ 4
M2M_2 = sum of principal 2×22 \times 2 minors
−λ3+tr⁡(A)λ2−M2λ+det⁡(A)-\lambda^3 + \operatorname{tr}(A)\lambda^2 - M_2\lambda + \det(A)
The pattern continues: each coefficient is a sum of principal minors of one size. Beyond 3×33 \times 3 the expansion is not worth doing by hand, and numerical work does not compute the polynomial at all.
9
Cayley–Hamilton§ 8
every square matrix satisfies its own polynomial
p(A)=Op(A) = O
Substituting the matrix into its own characteristic polynomial gives the zero matrix. The practical consequence is that AnA^n can be rewritten in terms of lower powers, so any power or the inverse is a polynomial in AA of degree below nn.
Every entry is a property of the transformation rather than of the matrix expressing it: similar matrices have the same polynomial, so all of trace, determinant and spectrum survive a change of basis. That invariance is what makes the polynomial a description of the map and not merely of the array.
What the polynomial encodes·/linear-algebra/eigen/characteristic-equationLearn Math Class
Two of these are worth using as checks rather than as facts. Since the roots must sum to the trace and multiply to the determinant, both are available from the matrix in seconds and can be compared against any computed spectrum — a mismatch means an arithmetic error, found without redoing the work. For a 2×22 \times 2 matrix the discriminant tr⁡2−4det⁡\operatorname{tr}^2 - 4\det goes further and classifies the eigenvalues before they are computed at all.
The one thing the polynomial cannot report is how many independent eigenvectors an eigenvalue has. Algebraic multiplicity counts root repetition; geometric multiplicity counts eigenvectors, and it can be strictly smaller. That gap is invisible in p(λ)p(\lambda) and only appears when the null space of A−λIA - \lambda I is actually computed — which is why diagonalizability is a separate question from finding the eigenvalues.

Common Pitfalls

Three mistakes account for many wrong answers when the characteristic polynomial is used to find eigenvalues and eigenvectors. Each is marked with a box where it happens in the sections above. Here each is worked through in full, with the wrong and the right working side by side.

Flipping the sign of the trace term

The 2×22 \times 2 shortcut is remembered as λ2+tr(A)λ+det⁡(A)\lambda^2 + \text{tr}(A)\lambda + \det(A), with a plus sign in the middle. Every eigenvalue then comes out with the wrong sign.

Wrong working: for A=(5234)A = \begin{pmatrix} 5 & 2 \\ 3 & 4 \end{pmatrix}, with tr(A)=9\text{tr}(A) = 9 and det⁡(A)=14\det(A) = 14,
λ2+9λ+14=(λ+7)(λ+2)=0⇒λ=−7,  −2\lambda^2 + 9\lambda + 14 = (\lambda + 7)(\lambda + 2) = 0 \quad\Rightarrow\quad \lambda = -7, \; -2


Right working: expand the determinant.
(5−λ)(4−λ)−2⋅3=λ2−9λ+20−6=λ2−9λ+14=(λ−7)(λ−2)(5 - \lambda)(4 - \lambda) - 2 \cdot 3 = \lambda^2 - 9\lambda + 20 - 6 = \lambda^2 - 9\lambda + 14 = (\lambda - 7)(\lambda - 2)

The eigenvalues are λ1=7\lambda_1 = 7 and λ2=2\lambda_2 = 2.

Why it happens: in (a−λ)(d−λ)(a - \lambda)(d - \lambda) the two cross terms −aλ-a\lambda and −dλ-d\lambda both carry a minus sign. That sign is easy to lose when the result is memorised as a pattern rather than derived from the determinant.

How to check: the eigenvalues must add up to the trace. Here 7+2=9=tr(A)7 + 2 = 9 = \text{tr}(A), while −7+(−2)=−9-7 + (-2) = -9. The product check does not help with this slip, because (−7)(−2)=14=det⁡(A)(-7)(-2) = 14 = \det(A) as well.

Assuming a double root gives two eigenvectors

A root of multiplicity 22 is taken to supply two independent eigenvectors, and the matrix is declared diagonalizable as soon as its characteristic polynomial is factored.

Wrong working: for A=(2102)A = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}, the polynomial is p(λ)=(2−λ)2=(λ−2)2p(\lambda) = (2 - \lambda)^2 = (\lambda - 2)^2, so λ=2\lambda = 2 has ma=2m_a = 2. Two eigenvectors, such as (1,0)T(1, 0)^T and (0,1)T(0, 1)^T, are assumed, and AA is called diagonalizable.

Right working: substitute the root back and count free variables.
A−2I=(0100)A - 2I = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}

The only equation is v2=0v_2 = 0, and v1=tv_1 = t is free, so E2E_2 is the line through (1,0)T(1, 0)^T and mg=1m_g = 1. The vector (0,1)T(0, 1)^T is not an eigenvector: A(0,1)T=(1,2)TA(0, 1)^T = (1, 2)^T, which is not a multiple of (0,1)T(0, 1)^T. With mg=1<ma=2m_g = 1 < m_a = 2, the matrix is not diagonalizable.

Why it happens: for distinct roots each root brings exactly one eigenvector direction, and for a matrix such as 2I2I a double root really does bring a whole plane. The polynomial cannot tell the two cases apart: 2I2I and this AA both have p(λ)=(λ−2)2p(\lambda) = (\lambda - 2)^2.

How to check: compute the geometric multiplicity directly, as n−rank(A−λI)n - \text{rank}(A - \lambda I). Here the rank of A−2IA - 2I is 11, so mg=2−1=1m_g = 2 - 1 = 1.

Taking a row as the eigenvector

After A−λIA - \lambda I is row reduced, its nonzero row is written down as the eigenvector. The row holds the coefficients of an equation that the eigenvector must satisfy; the eigenvector is a solution of that equation, not the row itself.

Wrong working: for A=(1243)A = \begin{pmatrix} 1 & 2 \\ 4 & 3 \end{pmatrix} and λ=5\lambda = 5, the reduced matrix is (1−1/200)\begin{pmatrix} 1 & -1/2 \\ 0 & 0 \end{pmatrix}, and (1,−1/2)T(1, -1/2)^T is taken as the eigenvector. But
A(1−1/2)=(05/2)≠5(1−1/2)=(5−5/2)A\begin{pmatrix} 1 \\ -1/2 \end{pmatrix} = \begin{pmatrix} 0 \\ 5/2 \end{pmatrix} \neq 5\begin{pmatrix} 1 \\ -1/2 \end{pmatrix} = \begin{pmatrix} 5 \\ -5/2 \end{pmatrix}


Right working: the row says v1−12v2=0v_1 - \frac{1}{2}v_2 = 0. With v2=tv_2 = t, v1=t/2v_1 = t/2, and t=2t = 2 gives v1=(1,2)T\mathbf{v}_1 = (1, 2)^T. Then
A(12)=(510)=5(12)A\begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 5 \\ 10 \end{pmatrix} = 5\begin{pmatrix} 1 \\ 2 \end{pmatrix}


Why it happens: the reduced matrix is the last thing written before the answer, and its nonzero row looks like a vector of the right size.

How to check: multiply AA by the candidate and compare with λ\lambda times the candidate. A quicker test: the eigenvector must give 00 when combined with each row of the reduced matrix. For (1,−1/2)(1, -1/2) this gives 1⋅1+(−1/2)(−1/2)=5/4≠01 \cdot 1 + (-1/2)(-1/2) = 5/4 \neq 0; for (1,2)(1, 2) it gives 1−1=01 - 1 = 0.

Characteristic Equation FAQ

Why is it det⁡(A−λI)\det(A - \lambda I) rather than det⁡(A−λ)\det(A - \lambda)?

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Because λ\lambda is a scalar and AA is a matrix, so A−λA - \lambda subtracts incompatible objects and is simply not defined. Multiplying by the identity turns the scalar into a matrix of the right shape, λI\lambda I, which subtracts λ\lambda from each diagonal entry and leaves everything else alone. The II is doing real work.Read more →

Is it det⁡(A−λI)\det(A - \lambda I) or det⁡(λI−A)\det(\lambda I - A)?

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Both conventions circulate and neither is wrong. They differ by a factor of (−1)n(-1)^n, so the polynomials are not identical, but their roots are the same and the roots are what you are after. The second form has the advantage of coming out monic. Pick one and stay with it inside a single calculation.Read more →

What is the difference between ei\mathbf{e}_i, EijE_{ij} and EλE_{\lambda}?

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Three different objects sharing one letter, separated only by their decoration. Bold ei\mathbf{e}_i is a standard basis vector, EijE_{ij} with two subscripts is a matrix unit, and EλE_{\lambda} with a Greek subscript is an eigenspace, a whole subspace rather than a single vector or array.Read more →