Characteristic Polynomial Calculator

Enter Calculation

Matrix size 2×2 3×3

a11

a12

a13

a21

a22

a23

a31

a32

a33

For a 2×2 matrix only the top-left four entries (a11, a12, a21, a22) are used.

Results

Characteristic Polynomial p(λ)

1λ² -4λ +3

= 0

λ³ coefficient	0
λ² coefficient	1
λ¹ coefficient	-4
λ⁰ (constant)	3
Trace tr(A)	4
Determinant det(A)	3

What is the characteristic polynomial?

The characteristic polynomial of a square matrix A is defined as \(p(\lambda) = \det(A - \lambda I)\), where I is the identity matrix and \(\lambda\) is a scalar variable. Its roots are exactly the eigenvalues of A, which makes it a cornerstone of linear algebra, differential equations, stability analysis, and quantum mechanics. This calculator handles both 2×2 and 3×3 matrices and returns the polynomial coefficients along with the trace and determinant.

The characteristic polynomial comes from \(\det(A - \lambda I)\), subtracting \(\lambda\) along the diagonal.

How to use this calculator

Pick the matrix size (2×2 or 3×3), then type each entry into its labelled cell. For a 2×2 matrix only a11, a12, a21 and a22 are used; the other cells are ignored. Press calculate to see the polynomial written in standard form together with each coefficient.

The formulas

For a 2×2 matrix the result is compact: $$p(\lambda) = \lambda^{2} - \operatorname{tr}(A)\lambda + \det(A),$$ where \(\operatorname{tr}(A) = \text{a11} + \text{a22}\) and \(\det(A) = \text{a11}\cdot\text{a22} - \text{a12}\cdot\text{a21}\).

For a 3×3 matrix: $$p(\lambda) = -\lambda^{3} + \operatorname{tr}(A)\lambda^{2} - m\cdot\lambda + \det(A),$$ where \(\operatorname{tr}(A)\) is the sum of the diagonal entries and \(m\) is the sum of the three principal 2×2 minors (the minors obtained by deleting one matching row and column through the diagonal).

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For a 2×2 matrix the polynomial is \(\lambda^{2} - (\text{trace})\lambda + (\text{determinant})\).

Worked example

Take the 2×2 matrix \(\begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}\). The trace is \(2 + 2 = 4\) and the determinant is \(2\cdot 2 - 1\cdot 1 = 3\). So $$p(\lambda) = \lambda^{2} - 4\lambda + 3,$$ which factors as \((\lambda - 1)(\lambda - 3)\), giving eigenvalues 1 and 3.

FAQ

What are the roots of the characteristic polynomial? They are the eigenvalues of the matrix.

Why is the leading coefficient −1 for 3×3? Because expanding \(\det(A - \lambda I)\) for an odd-sized matrix introduces a factor of \((-\lambda)^{3} = -\lambda^{3}\). Many texts multiply through by −1 to make it monic; both forms have the same roots.

Does it work for non-symmetric matrices? Yes — the formula uses every entry, so any real 2×2 or 3×3 matrix works.