2D Eigenvector Explorer

matrix action on basis vectors and transformed unit circle

Worked example — 2D Eigenvector Explorer

Eigenvector rule	0we solve it for λ
Area scale \|det\|	3

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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trace and determinant fix the eigenvalues between them 🖖

Two of the four numbers you type decide the eigenvalues. The characteristic equation is λ² − (trace)λ + det = 0, so trace and determinant fix both roots between them. The sign of tr² − 4det then says which of three worlds the map lives in: two real eigendirections, one, or none. The panel prints all three numbers. Reflection and Swap axes share a trace of 0 and a determinant of −1, so they share the eigenvalues 1 and −1 — and they are still different maps, because everything else went into the eigenvectors.

the arrows that refuse to turn 🖖

Apply the matrix to most vectors and they swing to a brand-new direction. Eigenvectors are the special ones that stay on their own line — the transform only stretches or shrinks them by a number, the eigenvalue, and may flip them 180°. Put a vector on a green eigenvector line and this matrix keeps it on that line however many times you apply it, so those lines are the map's fixed skeleton.

Google was built on a giant eigenvector 🖖

The same idea powers Google's original PageRank. Treat the entire web as one enormous matrix of links, and the vector of every page's rank is the dominant eigenvector — the single invariant direction that survives endless random clicking. What you nudge here in 2D is the toy version of a calculation once run across billions of pages.

EIGENVECTORS — WHICH DIRECTIONS SURVIVE THE TRANSFORMATION?

Which Eigenvector Case Are You In?

An eigenvector is a direction the matrix refuses to turn: it only stretches it, by the eigenvalue λ. Whether such directions exist at all is settled by one number, the discriminant Δ = tr(A)² − 4·det(A). Positive gives two of them, zero gives one, negative gives none in the real plane. Orthogonality, flips and rotation all follow from where your matrix lands in that classification.

Δ > 0 — two real directions, and symmetry makes them perpendicular Δ > 0 ⇒ λ₁, λ₂ ∈ ℝ

A negative eigenvalue — the direction survives but points the other way det < 0 ⇒ λ = ±1

Δ = 0 — one eigenvalue twice, and only one direction to show for it Δ = 0 ⇒ λ₁ = λ₂

Δ < 0 — a rotation, and no real eigenvector at all Δ < 0 ⇒ λ ∈ ℂ

Δ > 0 — two real directions, and symmetry makes them perpendicular

What you know: The discriminant is positive, so there are two distinct real eigenvalues and two independent eigendirections. When the matrix is symmetric, meaning a₁₂ = a₂₁, those directions always meet at a right angle.

Test: Δ > 0 ⇒ λ₁, λ₂ ∈ ℝ

Worked example: A = [[2, 1], [1, 2]]: tr = 4, det = 3, Δ = 16 − 12 = 4 → λ = 3 along (1, 1) and λ = 1 along (1, −1)

Open this case: symmetric stretch

Δ > 0 — two real directions, and symmetry makes them perpendicular. Two perpendicular directions survive, and everything else is dragged towards the stronger one. The discriminant is positive, so there are two distinct real eigenvalues and two independent eigendirections. When the matrix is symmetric, meaning a₁₂ = a₂₁, those directions always meet at a right angle. — Two perpendicular directions survive, and everything else is dragged towards the stronger one.

A negative eigenvalue — the direction survives but points the other way

What you know: Two real eigenvalues again, but one of them is negative, so det < 0. Vectors along that direction come back reversed and the whole plane is flipped.

Test: det < 0 ⇒ λ = ±1

Worked example: A = [[1, 0], [0, −1]]: tr = 0, det = −1, Δ = 4 → λ = 1 along (1, 0) and λ = −1 along (0, 1)

Open this case: reflection

A negative eigenvalue — the direction survives but points the other way. One axis is left alone, the other is sent back along itself: a mirror. Two real eigenvalues again, but one of them is negative, so det < 0. Vectors along that direction come back reversed and the whole plane is flipped. — One axis is left alone, the other is sent back along itself: a mirror.

Δ = 0 — one eigenvalue twice, and only one direction to show for it

What you know: The discriminant is exactly zero, so the two eigenvalues coincide. A repeated eigenvalue does not promise two directions: here there is only one, and the matrix cannot be diagonalised.

Test: Δ = 0 ⇒ λ₁ = λ₂

Worked example: A = [[1, 1], [0, 1]]: tr = 2, det = 1, Δ = 4 − 4 = 0 → λ = 1 twice, with (1, 0) the only eigendirection

Open this case: shear only

Δ = 0 — one eigenvalue twice, and only one direction to show for it. The horizontal axis is fixed while every other direction is tilted, so no second eigenvector exists. The discriminant is exactly zero, so the two eigenvalues coincide. A repeated eigenvalue does not promise two directions: here there is only one, and the matrix cannot be diagonalised. — The horizontal axis is fixed while every other direction is tilted, so no second eigenvector exists.

Δ < 0 — a rotation, and no real eigenvector at all

What you know: A negative discriminant means the eigenvalues are a complex conjugate pair. Geometrically the matrix turns the plane, and a turn leaves no direction pointing where it began.

Test: Δ < 0 ⇒ λ ∈ ℂ

Worked example: A = [[0, −1], [1, 0]]: tr = 0, det = 1, Δ = −4 → λ = ±i, a quarter turn with no real eigenvector

Open this case: rotation

$Δ < 0 — a rotation, and no real eigenvector at all. Every arrow moves off its own line, so there is nothing left for a real eigenvector to be. A negative discriminant means the eigenvalues are a complex conjugate pair. Geometrically the matrix turns the plane, and a turn leaves no direction pointing where it began.$

Every arrow moves off its own line, so there is nothing left for a real eigenvector to be.

References (1)

Insight block 3 — the dominant eigenvector Google was built on: S. Brin and L. Page, "The anatomy of a large-scale hypertextual Web search engine." Computer Networks and ISDN Systems 30(1–7), 107–117, 1998.

Problem solved in full

Applying a transformation ten times using the derived eigen vectors 7 steps
The panel derives the eigenvalues for you: trace, determinant, discriminant, and out come 3 and 1. Skip that route entirely. Find the eigenvectors first, by looking at the matrix — then use them for something the tool cannot do at all, and apply the transformation ten times.
1. The matrix is symmetric, and symmetry tells you where to look. Try the diagonal: A sends (1, 1) to (3, 3), which is the same arrow, three times as long. So (1, 1) is an eigendirection and 3 is its eigenvalue — one multiplication, no characteristic polynomial.
  
  $A (11) = (2 + 1 1 + 2) = (33) = 3 (11)$
2. Try the other diagonal. A leaves (1, −1) exactly where it was, so that is the second eigendirection, with eigenvalue 1. Two independent directions is all a 2×2 matrix has, so the search is over — and the pair agrees with what the panel prints.
  
  $A (1 - 1) = (2 - 1 1 - 2) = (1 - 1) = 1 \cdot (1 - 1)$
3. They are perpendicular, and not by luck: a symmetric matrix always has orthogonal eigendirections. That is the fact the next step spends, because it means the change of basis can be undone with a transpose instead of an inverse.
  
  $(11) \cdot (1 - 1) = 1 \cdot 1 + 1 \cdot (- 1) = 0$
4. Scale both to unit length and stand them in the columns of Q. Orthonormal columns make Qᵀ the inverse of Q, so A factorises into a rotation, a stretch along the axes, and the rotation undone.
  
  $Q = \frac{1}{2} (11 1 - 1), Λ = (3001), A = Q Λ Q^{T}, Q^{T} = Q^{- 1}$
5. Now the payoff. Multiply that factorisation by itself and every interior QᵀQ collapses to the identity, leaving Q, the n-th power of a diagonal matrix, and Qᵀ. Powering a diagonal matrix is just powering each entry, which is the whole reason anyone diagonalises anything.
  
  $A^{n} = (Q Λ Q^{T})^{n} = Q Λ^{n} Q^{T} = \frac{1}{2} (3^{n} + 1 3^{n} - 1 3^{n} - 1 3^{n} + 1)$
6. Put n = 10. Only 3¹⁰ does any work — the other eigenvalue is 1, and 1 to any power is still 1, so it contributes the same amount it did at the start.
  
  $n = 10 : 3^{10} = 59049 ⟹ A^{10} = \frac{1}{2} (59050590485904859050) = (29525295242952429525)$
7. The same factorisation says what happens to an arbitrary arrow, not just to the matrix. Split it along the two eigendirections: each application multiplies one part by 3 and leaves the other untouched, so the ratio between them changes by a factor of 3 every single time.
  
  $v = c_{1} (11) + c_{2} (1 - 1) ⟹ \frac{c _{2} \cdot 1 ^{n}}{c _{1} \cdot 3 ^{n}} = \frac{c _{2}}{c _{1}} 3^{- n}, 3^{- 10} = \frac{1}{59 049}$
Answer

A¹⁰ has 29,525 on the diagonal and 29,524 off it — a difference of 1 part in 29,525. That near-equality is the picture above, told forward. The (1, −1) component of any starting arrow is divided by 3 on every application, so after ten it is 1/59,049 of what it began as, and everything ends up lying along (1, 1). The tool shows you one application. The eigenvalues turn that into all of them, and tell you how fast.

$A^{n} = \frac{1}{2} (3^{n} + 1 3^{n} - 1 3^{n} - 1 3^{n} + 1); A^{10} = (29525295242952429525), 3^{- 10} = \frac{1}{59 049}$

Learning path

Matrices that move things

Leads to Singular value decomposition the directions the transform does not turn, and the factor it stretches each of them by.

By the 42Math team · Updated 2026-07-27

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

symmetric stretch - Eigenvalues 3 and 1, along (1, 1) and (1, -1). A symmetric matrix always has perpendicular eigendirections, and the worked problem below raises this one to the tenth power.
shear only - The discriminant is 0, so both eigenvalues are 1 and there is only one eigendirection, along the x-axis. A shear cannot be diagonalised.
rotation - A quarter turn: tr² - 4det = -4, the eigenvalues are ±i, and no real arrow comes out pointing where it started.
reflection - Trace 0, determinant -1, eigenvalues 1 and -1 along the two axes. The negative determinant is the mirror: area kept, orientation reversed.
anisotropic stretch - Eigenvalues 3 and 0.5 - six times more stretch along x than along y - and a determinant of 1.5, so every area grows by half.
swap axes - The same trace 0 and determinant -1 as Reflection, so the same eigenvalues 1 and -1. Swapping the axes is a reflection too, in the line y = x.

YouTube

3Blue1Brown·Eigenvectors and eigenvalues | Chapter 14, Essence of linear algebra