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CalcChef

Eigenvalue Calculator

Eigenvalues and eigenvectors of a 2 × 2 or 3 × 3 matrix from its characteristic equation, with the trace and determinant and complex pairs a ± bi. How it works

Use the Eigenvalue Calculator

Matrix [a b; c d], by row.

Eigenvalue λ₁
3
Eigenvalue λ₂ -1

Working

Trace = a + d = 2. Determinant = ad − bc = -3.

Characteristic equation: λ² - 2λ - 3 = 0

Discriminant = tr² − 4·det = 16. Two different real eigenvalues.

Shown to 8 significant digits.

Steps

  1. The eigenvalues are the roots of the characteristic equation det(A − λI) = 0, that is (a − λ)(d − λ) − bc = 0, which expands to λ² − (a + d)λ + (ad − bc) = 0.
  2. Trace: a + d = 1 + 1 = 2. Determinant: ad − bc = 1·1 − 2·2 = −3.
  3. The characteristic equation is λ² − 2λ − 3 = 0
  4. Discriminant: tr² − 4·det = 2² − 4·(−3) = 16.
  5. Square root of the discriminant: √16 = 4.
  6. Quadratic formula: λ = (tr ± √disc) / 2 = (2 ± 4) / 2.
  7. So λ₁ = 3 and λ₂ = −1.

Eigenvectors

  • λ = 3: v = (1, 1)
  • λ = -1: v = (1, -1)

Each vector satisfies (A − λI)v = 0. Any non-zero multiple of it is also an eigenvector. Rational eigenvalues give whole-number vectors; others give unit-length decimals to 6 significant digits.

About the Eigenvalue Calculator

The matrix [1 2; 2 1] has eigenvalues 3 and −1, the roots of its characteristic equation λ² − 2λ − 3 = 0. This eigenvalue calculator takes a 2 × 2 or 3 × 3 matrix, shows the trace, determinant and characteristic equation (with the working step by step for a 2 × 2), gives complex eigenvalues as a ± bi, and lists an eigenvector for each real eigenvalue: (1, 1) for 3 and (1, -1) for −1 here.

Eigenvalues of 2 × 2 matrices

Matrix [a b; c d]TraceDeterminantCharacteristic equationλ₁λ₂
[1 2; 2 1]2-3λ² - 2λ - 3 = 03-1
[2 0; 0 3]56λ² - 5λ + 6 = 032
[4 1; 2 3]710λ² - 7λ + 10 = 052
[2 1; 1 2]43λ² - 4λ + 3 = 031
[5 4; 1 2]76λ² - 7λ + 6 = 061
[1 2; 3 4]5-2λ² - 5λ - 2 = 05.3722813-0.37228132
[0.9 0.2; 0.1 0.8]1.70.7λ² - 1.7λ + 0.7 = 010.7
[1 1; 0 1]21λ² - 2λ + 1 = 011
[0 -1; 1 0]01λ² + 1 = 00 + 1i0 - 1i
[3 -2; 4 -1]25λ² - 2λ + 5 = 01 + 2i1 - 2i

Enter a and b in the top row of the tool and c and d in the bottom row. Values are shown to 8 significant digits.

Eigenvalues and eigenvectors of 3 × 3 matrices

Matrix [a b c; d e f; g h i]Characteristic equationEigenvaluesEigenvectors
[2 -1 0; -1 2 -1; 0 -1 2]λ³ - 6λ² + 10λ - 4 = 03.4142136, 2, 0.58578644(0.5, -0.707107, 0.5); (1, 0, -1); (0.5, 0.707107, 0.5)
[1 5 7; 0 2 9; 0 0 3]λ³ - 6λ² + 11λ - 6 = 03, 2, 1(26, 9, 1); (5, 1, 0); (1, 0, 0)
[2 0 0; 0 2 0; 0 0 3]λ³ - 7λ² + 16λ - 12 = 03, 2, 2(0, 0, 1); (1, 0, 0) and (0, 1, 0)
[0 -1 0; 1 0 0; 0 0 1]λ³ - λ² + λ - 1 = 01, 0 + 1i, 0 - 1i(0, 0, 1); none shown for the complex pair

Eigenvectors are listed in the order of the eigenvalues, a repeated eigenvalue once. Rational eigenvalues get exact whole-number vectors; 2 ± √2 in the first row gets unit-length decimal vectors to 6 significant digits.

Frequently asked questions

How do you find the eigenvalues of a 2x2 matrix step by step?
Solve det(A − λI) = 0, which for [a b; c d] is λ² − (a + d)λ + (ad − bc) = 0. For [4 1; 2 3] the trace is 7 and the determinant 10, so λ² − 7λ + 10 = 0, which factors as (λ − 5)(λ − 2) = 0: the eigenvalues are 5 and 2. The Steps panel under a 2 × 2 result shows the same working with your numbers: for [1 2; 2 1] the trace is 2, the determinant −3, the discriminant 2² − 4·(−3) = 16, its square root 4, and λ = (2 ± 4) / 2 gives 3 and −1.
Is there a shortcut formula for 2x2 eigenvalues?
λ = tr/2 ± √((tr/2)² − det). For [1 2; 3 4], tr/2 = 2.5 and det = -2, so λ = 2.5 ± √8.25, which the tool shows as 5.3722813 and -0.37228132.
When are the eigenvalues complex?
When the discriminant tr² − 4·det is negative. For [3 -2; 4 -1] it is 2² − 4×5 = -16, so the eigenvalues are 1 + 2i and 1 − 2i. The rotation matrix [0 -1; 1 0] gives ±i.
What does a repeated eigenvalue mean?
The discriminant is 0 and both roots are equal. [1 1; 0 1] has the eigenvalue 1 twice. The tool computes the discriminant exactly, so a repeated eigenvalue is shown as repeated, not as a complex pair.
Does it find eigenvectors?
Yes, one or more for each real eigenvalue, from solving (A − λI)v = 0. For [1 2; 2 1] and λ = 3, A − 3I = [-2 2; 2 -2], so v = (1, 1); for λ = -1, v = (1, -1). Check: [1 2; 2 1] times (1, 1) is (3, 3). A rational eigenvalue gets an exact whole-number vector, an irrational one a unit-length decimal vector, and complex eigenvalues get none.
Can it do 3x3 matrices?
Yes. Press 3 × 3 and fill in the nine entries. The characteristic equation is the cubic λ³ − tr·λ² + m·λ − det = 0, with m the sum of the principal 2 × 2 minors: [2 -1 0; -1 2 -1; 0 -1 2] gives λ³ - 6λ² + 10λ - 4 = 0 and the eigenvalues 3.4142136, 2 and 0.58578644, which are 2 + √2, 2 and 2 − √2.

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