Compute p(λ) = det(A − λI) for any 2×2 or 3×3 matrix — the polynomial in λ, its coefficients, trace, determinant, and eigenvalues (the roots of the characteristic equation), with step-by-step working.
For a 2×2 matrix: p(λ) = λ² − tr(A)·λ + det(A). For 3×3: p(λ) = −λ³ + tr(A)·λ² − s₂·λ + det(A), where s₂ is the sum of the three principal 2×2 minors. Quadratic roots are exact; cubic roots use Cardano's formula.
Trace: tr(A) = 4 + 3 = 7. Determinant: det(A) = 4·3 − 2·1 = 12 − 2 = 10.
Solve λ² − 7λ + 10 = 0. Discriminant D = 49 − 40 = 9, so λ = (7 ± 3)/2.
Trace: tr(A) = 0 + 0 = 0. Determinant: det(A) = 0·0 − (−1)·1 = 1.
A real matrix can have nonreal eigenvalues — here a 90° rotation has no real eigenvector.
Trace: tr(A) = 2 + 3 + 4 = 9. Principal 2×2 minors: M₁ = 3·4 − 1·0 = 12, M₂ = 2·4 − 0·0 = 8, M₃ = 2·3 − 1·0 = 6, so s₂ = 12 + 8 + 6 = 26. Determinant (triangular): det(A) = 2·3·4 = 24.
Check: p(2) = −8 + 36 − 52 + 24 = 0 ✓, p(3) = −27 + 81 − 78 + 24 = 0 ✓, p(4) = −64 + 144 − 104 + 24 = 0 ✓.
Trace: tr(A) = 1 + 4 + 0 = 5. Principal 2×2 minors: M₁ = 4·0 − 5·0 = 0, M₂ = 1·0 − 3·0 = 0, M₃ = 1·4 − 2·0 = 4, so s₂ = 4. Determinant: det(A) = 0 (zero row).
The zero constant term (det(A) = 0) forces λ = 0 to be an eigenvalue — the matrix is singular. Check: p(1) = −1 + 5 − 4 = 0 ✓, p(4) = −64 + 80 − 16 = 0 ✓.
The characteristic polynomial of a square matrix A is p(λ) = det(A − λI), where λ is a scalar variable and I is the identity matrix of the same size. It is a polynomial of degree n for an n×n matrix, and its coefficients are built from the trace, determinant, and principal minors of A.
For A = [[a, b], [c, d]], it is p(λ) = λ² − tr(A)·λ + det(A), where tr(A) = a + d and det(A) = ad − bc. Example: A = [[4, 2], [1, 3]] has tr(A) = 7 and det(A) = 10, so p(λ) = λ² − 7λ + 10 = (λ − 2)(λ − 5).
p(λ) = −λ³ + tr(A)·λ² − s₂·λ + det(A), where s₂ is the sum of the three principal 2×2 minors of A. Example: the upper-triangular matrix [[2, 1, 0], [0, 3, 1], [0, 0, 4]] has tr(A) = 9, s₂ = 12 + 8 + 6 = 26 and det(A) = 24, so p(λ) = −λ³ + 9λ² − 26λ + 24.
The eigenvalues of A are exactly the roots of its characteristic polynomial: the values λ with p(λ) = 0, equivalently det(A − λI) = 0. Example: the rotation matrix [[0, −1], [1, 0]] has p(λ) = λ² + 1, whose roots are λ = i and λ = −i — so a real matrix can have nonreal eigenvalues.
The constant term of p(λ) = det(A − λI) equals det(A). If it is zero, then p(0) = 0, so λ = 0 is an eigenvalue and A is singular (not invertible). Example: [[1, 2, 3], [0, 4, 5], [0, 0, 0]] gives p(λ) = −λ³ + 5λ² − 4λ = −λ(λ − 1)(λ − 4), with eigenvalues 0, 1, 4.
A repeated root means an eigenvalue with algebraic multiplicity greater than 1. For example, [[2, 1, 0], [1, 2, 0], [0, 0, 3]] has p(λ) = −λ³ + 7λ² − 15λ + 9 = −(λ − 1)(λ − 3)², so the eigenvalues are 1, 3, 3. Nonreal eigenvalues of a real matrix always come in conjugate pairs, and every cubic characteristic polynomial has at least one real root.
No. The characteristic polynomial always has degree n, while the minimal polynomial is the lowest-degree monic polynomial that annihilates A; it divides the characteristic polynomial. Example: the 2×2 identity matrix has characteristic polynomial (λ − 1)² but minimal polynomial (λ − 1).
No. This calculator writes p(λ) = det(A − λI), whose leading coefficient is (−1)ⁿ, while many textbooks use det(λI − A) = (−1)ⁿ·p(λ). The two forms differ only by an overall sign, so they have exactly the same roots and the same eigenvalues. For 2×2 matrices both conventions coincide because (−1)² = 1.
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