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Eigenvalues and Eigenvectors


Form A − λI, expand the characteristic polynomial, find its roots, and read an eigenvector from each null space.

Preset?An eigenvector of A is a non-zero vector that A merely scales: A v = λ v. The scale factor λ is the eigenvalue. Rewriting as (A − λI) v = 0, a non-zero v exists exactly when det(A − λI) = 0, so the eigenvalues are the roots of the characteristic polynomial det(A − λI), and each eigenvector is a null-space vector of A − λI for its λ. The roots may be complex; symmetric matrices always have real ones.
Size (square)
A3× 3
Entries of A
A3×3
2
0
0
0
3
4
0
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9
Step 1 / 8

Step explanations

1Eigenvalues and eigenvectors of a 3×3 matrix
An eigenvector is a direction A does not turn: A v = λ v, with the eigenvalue λ as the scale factor. Rewritten as (A − λI) v = 0, a non-zero v exists only when A − λI is singular, so the plan is: subtract λ from the diagonal, expand the determinant into a polynomial in λ, find its roots, and for each root read an eigenvector from the null space of A − λI.
Shift, expand, find roots, reduce - the same four moves for every matrix. Learn more about the first move · what they are








Key Terms

Eigenvector — a non-zero vector v\mathbf{v} with Av=λvA\mathbf{v} = \lambda\mathbf{v}: a direction that AA scales but does not turn.

Eigenvalue — the scale factor λ\lambda in Av=λvA\mathbf{v} = \lambda\mathbf{v}.

Characteristic polynomialp(λ)=det(AλI)p(\lambda) = \det(A - \lambda I), whose roots are the eigenvalues; shown monic in the tool.

Characteristic equationdet(AλI)=0\det(A - \lambda I) = 0.

Eigenspace — the null space of AλIA - \lambda I, all eigenvectors for λ\lambda together with 0\mathbf{0}.

Algebraic multiplicity — how many times λ\lambda is a root of pp.

Geometric multiplicity — the dimension of the eigenspace, the number of free columns of AλIA - \lambda I; never exceeds the algebraic multiplicity.

Defective — a matrix with some eigenvalue whose geometric multiplicity is smaller than its algebraic one; such a matrix cannot be diagonalized.

Trace and determinant — the eigenvalues sum to trA\operatorname{tr} A and multiply to detA\det A, counting multiplicity.

Getting Started with the Visualizer

Choose a matrix, then watch its eigenvalues and eigenvectors emerge the way they are found by hand.

• Use the Preset pills for six matrices: three distinct eigenvalues, a symmetric 2×22 \times 2, a repeated eigenvalue with a plane of eigenvectors, a defective matrix, a triangular one, and a rotation with complex eigenvalues
• Use the Size stepper for 2×22 \times 2 or 3×33 \times 3
• Edit any entry directly, or press Shuffle for a random matrix built to have small integer eigenvalues
• Hover the ? icon for a reminder of the definition and why the eigenvalues are roots of a polynomial
• Press play or step manually; the step log on the right keeps every stage

The run is always the same four moves: form AλIA - \lambda I, expand its determinant, find the roots, and row reduce once per real eigenvalue. If the roots are complex the run stops after finding them, with an explanation, since there are then no real eigenvectors to show.

The Four Moves

The visualizer follows the textbook procedure exactly.

ShiftAλIA - \lambda I is written with λ\lambda subtracted from each diagonal entry; the eigenvalues are the λ\lambda that make it singular
Expanddet(AλI)\det(A - \lambda I) is expanded into the characteristic polynomial and its coefficients are laid out in a row; for a 2×22 \times 2 they are 11, trA-\operatorname{tr} A, detA\det A; for a 3×33 \times 3, the tool multiplies by 1-1 so the polynomial is monic
Roots — the eigenvalues, found by the quadratic formula or, for a cubic, by spotting a rational root or numerically; the caption shows the factored form and checks the trace and determinant
Eigenvectors — for each real eigenvalue, AλIA - \lambda I is row reduced, the free columns are counted, and one eigenvector per free column is read off with the free variable set to 11, then scaled to small integers
Done — the eigenvalues and eigenvectors side by side, each checked by multiplying, and a verdict on whether AA is diagonalizable

When an eigenvalue is repeated, the eigenvector scene reports both multiplicities. A full set of eigenvectors means AA can be diagonalized; a shortfall means it cannot.

Forming A Minus Lambda I

The first move subtracts the unknown λ\lambda from each diagonal entry, leaving the rest of AA as it is. The frozen picture below shows the default preset's AλIA - \lambda I between determinant bars, its diagonal reading 2λ2 - \lambda, 3λ3 - \lambda, 9λ9 - \lambda.

This matrix is singular for exactly the right values of λ\lambda, and nothing else about it matters yet.
A − λI3×32 − λ0003 − λ4049 − λ
A − λI, frozen

The default preset with λ subtracted from each diagonal entry and nothing else touched. The eigenvalues are the values of λ that make this matrix singular.

The shift is where the definition becomes computation. Av=λvA\mathbf{v} = \lambda\mathbf{v} has an unknown on both sides; moving everything to the left gives (AλI)v=0(A - \lambda I)\mathbf{v} = \mathbf{0}, a homogeneous system with a parameter in it. A homogeneous system has a non-zero solution exactly when its matrix is singular, so the eigenvalues are the values of λ\lambda at which AλIA - \lambda I loses rank.

Only the diagonal carries λ\lambda because λI\lambda I is diagonal. This is the entry-level mistake most worth avoiding: the off-diagonal entries of AλIA - \lambda I are those of AA, unchanged.

The Characteristic Polynomial

Expanding det(AλI)\det(A - \lambda I) produces a polynomial in λ\lambda of degree nn. The frozen picture below shows the default preset's coefficients laid out beside the shifted matrix: 11, 14-14, 3535, 22-22, for λ314λ2+35λ22\lambda^3 - 14\lambda^2 + 35\lambda - 22.

The second coefficient is minus the trace and the last is minus the determinant, which are the two checks the tool applies throughout.
A − λI3×32 − λ0003 − λ4049 − λλ³ · λ² · λ · 11−1435−22
Characteristic polynomial, frozen

Coefficients 1, −14, 35, −22: λ³ − 14λ² + 35λ − 22. Minus the trace in second place, minus the determinant in last - the two standing checks.

For a 3×33 \times 3 matrix the raw expansion has leading term λ3-\lambda^3; the tool multiplies through by 1-1 so the polynomial is monic and its roots are easier to read. The middle coefficient is the sum of the three principal 2×22 \times 2 minors, a quantity with no common name but a definite meaning: the second elementary symmetric function of the eigenvalues.

The polynomial is the whole eigenvalue problem in one object. Two matrices with the same characteristic polynomial share their eigenvalues, trace and determinant; similar matrices always do, which is what makes eigenvalues a property of the transformation rather than of the particular matrix representing it.

Finding the Roots

The eigenvalues are the roots of the characteristic polynomial. The frozen picture below shows the default preset's three roots, 1111, 22 and 11, beside the coefficient row.

For a 2×22 \times 2 the quadratic formula does it; for a 3×33 \times 3 the tool uses the cubic formula, then polishes the roots and snaps them to simple values when they are within rounding error of one.
λ³ · λ² · λ · 11−1435−22eigenvalues λ1121
Eigenvalues, frozen

The roots 11, 2, 1 beside the coefficient row. They sum to 14, the trace, and multiply to 22, the determinant.

By hand the cubic is usually cracked by spotting a rational root among the divisors of the constant term, dividing it out, and solving the remaining quadratic; the factored form in the caption is what that produces. Triangular and block-triangular matrices short-circuit the whole step, since their eigenvalues sit on the diagonal.

The two checks are worth applying every time: the roots must sum to the trace and multiply to the determinant. A slip in the expansion almost always fails one of them.

Reading an Eigenvector

For each real eigenvalue, the tool substitutes it into AλIA - \lambda I, row reduces, and reads a null-space vector by setting the free variable to 11. The frozen picture below is the default preset's λ=11\lambda = 11: the shifted matrix, its reduced form with two pivots and one free column, and the eigenvector (0,1,2)(0, 1, 2).

The reduced form must have at least one free column; that is what being singular means, and it is the source of the eigenvector.
A − (11)I3×3−9000−8404−2rref3×310001−1/2000v3×1012
Eigenvector for λ = 11, frozen

A − 11I row reduced to two pivots and one free column, and the eigenvector (0, 1, 2) read off with the free variable set to 1 and scaled to integers. A v = 11 v.

An eigenvector is never unique. Any non-zero multiple of (0,1,2)(0, 1, 2) is also an eigenvector for 1111, and the whole line through it is the eigenspace. The tool scales to small integers for readability; a unit vector, or the one with free variable 11, would be equally valid.

When the reduced form has more than one free column, the eigenspace is a plane or larger, with one eigenvector per free column. That happens only for repeated eigenvalues, and whether it happens is the question of diagonalizability.

A Repeated Eigenvalue with a Plane of Eigenvectors

On the repeated preset, λ=1\lambda = 1 is a double root, and AIA - I is the all-ones matrix, which reduces to a single non-zero row. Two free columns, so two independent eigenvectors: the eigenspace is a plane.

The frozen picture below shows that scene: one pivot, two free columns, and both eigenvectors read off.
A − (1)I3×3111111111rref3×3111000000eigenvectors3×211−100−1
Repeated eigenvalue, frozen

The repeated preset at λ = 1: A − I reduces to a single row, two free columns, two eigenvectors. Geometric multiplicity 2 matches algebraic multiplicity 2 - a full set.

Here the geometric multiplicity, 22, equals the algebraic multiplicity, and the matrix has a full set of three eigenvectors: it is diagonalizable, with the eigenvalue 11 appearing twice on the diagonal. Symmetric matrices, like this one, always achieve this.

The defective preset shows the other possibility. Its λ=1\lambda = 1 is also a double root, but AIA - I has only one free column, one eigenvector, geometric multiplicity 11 against algebraic multiplicity 22. No basis of eigenvectors exists, and the matrix cannot be diagonalized; the closest it comes is the Jordan form.

Complex Eigenvalues

On the rotation preset the characteristic polynomial is λ2+1\lambda^2 + 1, with roots ±i\pm i. The frozen picture below shows the run stopping there: the eigenvalues are complex, and no real vector keeps its direction under a quarter turn.

The tool stops because there are no real eigenvectors to reduce for; the matrix is not defective, just not real-diagonalizable.
A2×20−110eigenvalues λi−i
Complex eigenvalues, frozen

The quarter-turn rotation: λ² + 1 = 0, eigenvalues ±i, and the run stops. No real direction survives a rotation; the eigenvectors live in ℂ².

Complex eigenvalues of a real matrix always come in conjugate pairs, and for a 2×22 \times 2 they mean the map is a rotation combined with a scaling by λ|\lambda|. Over the complex numbers the eigenvectors exist, and in the real plane the pair corresponds to a rotation-scaling block rather than to two stretch directions.

This is also the reason symmetric matrices are special. Their eigenvalues are always real, so this case never arises for them, and the spectral theorem guarantees a full orthogonal set of real eigenvectors.

Reading the Scene Player

Each scene shows the matrices for one move with the arithmetic in the caption.

• In the shift and expand scenes, the diagonal of AλIA - \lambda I is highlighted and the diagonal entries read ai,iλa_{i,i} - \lambda; the coefficient row is accent
• In the roots scene, the eigenvalue row is accent when the roots are real and secondary when they are complex
• In an eigenvector scene, the diagonal of AλIA - \lambda I is highlighted, the pivots of its reduced form are accent and the free columns secondary, and the eigenvector column is accent, with an arrow from the free column that generated it
• In the done scene, each eigenvector column is accent with its eigenvalue above it
• Eigenvectors are scaled to small integers whenever a multiple by 11 to 1212 makes every entry integral; otherwise they are shown to three decimals with the first non-zero entry positive

Choosing a Matrix

The six presets each make a different point.

Three distinct — eigenvalues 11, 22 and 1111, each with a one-dimensional eigenspace; the general case, and the default
Symmetric 2×22 \times 2 — eigenvalues 11 and 33 with perpendicular eigenvectors (1,1)(1, -1) and (1,1)(1, 1); symmetric matrices always have real eigenvalues and orthogonal eigenvectors
Repeated eigenvalueλ=1\lambda = 1 twice and λ=4\lambda = 4; the reduced form of AIA - I has two free columns, so the eigenspace is a plane and the matrix is still diagonalizable
Defective — the shear with λ=1\lambda = 1 twice but a single eigenvector; algebraic multiplicity 22, geometric multiplicity 11, not diagonalizable
Triangular — the eigenvalues are the diagonal entries, because det(AλI)\det(A - \lambda I) of a triangular matrix is the product of the diagonal
Rotation — the quarter turn, with characteristic polynomial λ2+1\lambda^2 + 1 and eigenvalues ±i\pm i; no real vector keeps its direction

Shuffle builds its matrices as PDP1PDP^{-1} with an integer unimodular PP, so the eigenvalues are small integers and the entries stay integers too.

What Eigenvalues and Eigenvectors Are

A non-zero vector v\mathbf{v} is an eigenvector of AA with eigenvalue λ\lambda when

Av=λvA\mathbf{v} = \lambda\mathbf{v}


Applying AA to v\mathbf{v} does nothing but stretch it, by the factor λ\lambda; a negative λ\lambda reverses it and λ=0\lambda = 0 collapses it. Rewriting as (AλI)v=0(A - \lambda I)\mathbf{v} = \mathbf{0} turns the definition into a solvable problem: a non-zero solution exists exactly when AλIA - \lambda I is singular, that is when

p(λ)=det(AλI)=0p(\lambda) = \det(A - \lambda I) = 0


This is the characteristic equation, and pp is a polynomial of degree nn whose roots are the eigenvalues. For a 2×22 \times 2 matrix it is λ2(trA)λ+detA\lambda^2 - (\operatorname{tr} A)\lambda + \det A; in general the coefficient of λn1\lambda^{n-1} is ±trA\pm\operatorname{tr} A and the constant is ±detA\pm\det A, which gives the two standard checks: the eigenvalues sum to the trace and multiply to the determinant.

Each eigenvalue then has an eigenspace, the null space of AλIA - \lambda I, found by row reduction. Its dimension, the geometric multiplicity, is at least 11 and at most the algebraic multiplicity. When every eigenvalue has a full set, the eigenvectors form a basis and AA is diagonalizable: A=PDP1A = PDP^{-1} with the eigenvectors as the columns of PP.

Real matrices can have complex eigenvalues, in conjugate pairs; a 2×22 \times 2 rotation is the standard example. Symmetric matrices never do: their eigenvalues are real and their eigenvectors can be chosen orthogonal. For the full theory, see the characteristic equation theory page and the eigenvalue properties page.

Key Properties

Facts that hold for every square matrix.

Trace and determinant: λi=trA\sum \lambda_i = \operatorname{tr} A and λi=detA\prod \lambda_i = \det A, with multiplicity
Singular matrices: λ=0\lambda = 0 is an eigenvalue exactly when detA=0\det A = 0
Triangular matrices: the eigenvalues are the diagonal entries
Powers and inverses: AkA^k has eigenvalues λk\lambda^k with the same eigenvectors; A1A^{-1} has 1/λ1/\lambda
Transpose: ATA^T has the same eigenvalues as AA, though not the same eigenvectors
Multiplicities: 11 \leq geometric \leq algebraic for every eigenvalue
Distinct eigenvalues: their eigenvectors are automatically linearly independent
Symmetric matrices: real eigenvalues, orthogonal eigenvectors, always diagonalizable
Complex eigenvalues of a real matrix come in conjugate pairs
Similar matrices: P1APP^{-1}AP has the same characteristic polynomial as AA

Why It Matters

Eigenvectors are the directions in which a linear map is simplest, and most of applied linear algebra is about finding them.

Diagonalization: in the eigenvector basis AA is diagonal, and powers, exponentials and functions of AA become trivial
Dynamical systems: xt+1=Axt\mathbf{x}_{t+1} = A\mathbf{x}_t grows, decays or oscillates along each eigenvector according to its eigenvalue; the dominant eigenvalue decides the long-run behaviour
Differential equations: x˙=Ax\dot{\mathbf{x}} = A\mathbf{x} has solutions eλtve^{\lambda t}\mathbf{v}, one per eigenpair
Vibrations and stability: natural frequencies are eigenvalues; a system is stable when every eigenvalue has negative real part
Principal components: the eigenvectors of a covariance matrix are the directions of greatest variance
Graphs and networks: PageRank is the dominant eigenvector of a link matrix, and spectral clustering uses eigenvectors of the graph Laplacian
Quantum mechanics: observables are operators whose eigenvalues are the possible measurements

Worked Example

Take the default preset,

A=(200034049)A = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 3 & 4 \\ 0 & 4 & 9 \end{pmatrix}


Characteristic polynomial. Expanding det(AλI)\det(A - \lambda I) along the first row, (2λ)[(3λ)(9λ)16]=(2λ)(λ212λ+11)(2 - \lambda)\left[(3 - \lambda)(9 - \lambda) - 16\right] = (2 - \lambda)(\lambda^2 - 12\lambda + 11). Made monic: p(λ)=λ314λ2+35λ22p(\lambda) = \lambda^3 - 14\lambda^2 + 35\lambda - 22. Check: the trace is 1414 and the determinant is 211=222 \cdot 11 = 22.

Roots. λ212λ+11=(λ1)(λ11)\lambda^2 - 12\lambda + 11 = (\lambda - 1)(\lambda - 11), so p(λ)=(λ1)(λ2)(λ11)p(\lambda) = (\lambda - 1)(\lambda - 2)(\lambda - 11) and the eigenvalues are 11, 22 and 1111. They sum to 1414 and multiply to 2222.

Eigenvector for λ=11\lambda = 11. A11IA - 11I has rows (9,0,0)(-9, 0, 0), (0,8,4)(0, -8, 4), (0,4,2)(0, 4, -2). Row reducing: x1=0x_1 = 0 and x2=12x3x_2 = \tfrac{1}{2}x_3, with x3x_3 free. Set x3=2x_3 = 2: v=(0,1,2)\mathbf{v} = (0, 1, 2). Check: Av=(0,3+8,4+18)=(0,11,22)=11vA\mathbf{v} = (0, 3 + 8, 4 + 18) = (0, 11, 22) = 11\mathbf{v}.

Eigenvector for λ=2\lambda = 2. A2IA - 2I has rows (0,0,0)(0, 0, 0), (0,1,4)(0, 1, 4), (0,4,7)(0, 4, 7); the last two rows force x2=x3=0x_2 = x_3 = 0 and x1x_1 is free: v=(1,0,0)\mathbf{v} = (1, 0, 0).

Eigenvector for λ=1\lambda = 1. AIA - I has rows (1,0,0)(1, 0, 0), (0,2,4)(0, 2, 4), (0,4,8)(0, 4, 8): x1=0x_1 = 0, x2=2x3x_2 = -2x_3. Set x3=1x_3 = -1: v=(0,2,1)\mathbf{v} = (0, 2, -1). Check: Av=(0,64,89)=(0,2,1)A\mathbf{v} = (0, 6 - 4, 8 - 9) = (0, 2, -1).

Three eigenvalues, three independent eigenvectors: AA is diagonalizable. Note that the two eigenvectors of the symmetric 2×22 \times 2 block, (1,2)(1, 2) and (2,1)(2, -1), are perpendicular.

Common Mistakes

A few mistakes recur.

Subtracting λ\lambda from every entry — only the diagonal changes in AλIA - \lambda I; the off-diagonal entries are untouched
Solving Av=λvA\mathbf{v} = \lambda\mathbf{v} with v=0\mathbf{v} = \mathbf{0} — the zero vector satisfies it for every λ\lambda and is excluded by definition; eigenvectors are non-zero
Expecting AλIA - \lambda I to reduce to the identity — it must be singular, so its reduced form has at least one zero row; a full pivot set means the λ\lambda is wrong
Reading a repeated root as a repeated eigenvector — the multiplicity of the root says nothing by itself; the free columns of AλIA - \lambda I decide how many eigenvectors there are
Assuming real eigenvalues — a real matrix can have complex eigenvalues; only symmetric matrices are guaranteed real ones
Treating the eigenvector as unique — any non-zero multiple is also an eigenvector; the tool scales to small integers for readability, not because that scale is special
Forgetting the checks — the eigenvalues must sum to the trace and multiply to the determinant, and AvA\mathbf{v} must come out as λv\lambda\mathbf{v}