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Characteristic Polynomial Calculator

Calculate the characteristic polynomial of a matrix effortlessly with our step-by-step calculator. Get instant solutions and detailed explanations for eigenvalue calculations.

Characteristic Polynomial Tool

Enter your square matrix and get the characteristic polynomial with eigenvalues

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Master Characteristic Polynomials with Our Advanced Calculator

Our characteristic polynomial calculator is designed to help students, researchers, and engineers understand the fundamental properties of matrices through their characteristic equations. Whether you're working onlinear algebra homework, preparing for university mathematics exams, or conductingeigenvalue analysis in engineering applications, this tool provides comprehensive step-by-step solutions.

The characteristic polynomial calculator computes p(λ)=det⁡(A−λI)p(\lambda) = \det(A - \lambda I) for any square matrix AA. This eigenvalue calculator reveals the roots of the characteristic equation, which are the eigenvalues of the matrix. Our linear algebra calculator is particularly useful for university mathematics courses where understanding the relationship between matrices and their eigenvalues is crucial.

Perfect for linear algebra students learning about eigenvalues and eigenvectors,engineering students working with system dynamics, and researchersanalyzing matrix properties. The characteristic polynomial tool provides not just the polynomial expression, but also the corresponding eigenvalues and detailed explanations of the mathematical process.

Types of Characteristic Polynomials

Matrix SizePolynomial FormDegreeComplexity
2×2

p(λ)=λ2−tr(A)λ+det⁡(A)p(\lambda) = \lambda^2 - \text{tr}(A)\lambda + \det(A)

2Simple
3×3

p(λ)=−λ3+tr(A)λ2−c1λ+det⁡(A)p(\lambda) = -\lambda^3 + \text{tr}(A)\lambda^2 - c_1\lambda + \det(A)

3Intermediate
n×n

p(λ)=(−1)nλn+cn−1λn−1+...+c0p(\lambda) = (-1)^n\lambda^n + c_{n-1}\lambda^{n-1} + ... + c_0

nAdvanced
Symmetric

Real eigenvalues, orthogonal eigenvectors

nIntermediate
Diagonal

p(λ)=(λ−d1)(λ−d2)...(λ−dn)p(\lambda) = (\lambda - d_1)(\lambda - d_2)...(\lambda - d_n)

nSimple

Common Mistakes to Avoid

Non-Square Matrices

Characteristic polynomials are only defined for square matrices. For an m×nm \times n matrix where m≠nm \neq n, the characteristic polynomial does not exist.

Sign Errors

Remember that p(λ)=det⁡(A−λI)p(\lambda) = \det(A - \lambda I), not det⁡(A+λI)\det(A + \lambda I). The negative sign is crucial for correct eigenvalue calculation.

Complex Eigenvalues

Don't be surprised by complex eigenvalues. Many matrices have complex eigenvalues, especially non-symmetric ones. This is mathematically valid and important.

How to Calculate Characteristic Polynomials

The characteristic polynomial of a square matrix A is a fundamental concept in linear algebra that reveals the eigenvalues of the matrix. It's defined as the determinant of the matrix A minus λ times the identity matrix.

p(λ)=det⁡(A−λI)p(\lambda) = \det(A - \lambda I)

where II is the identity matrix of the same size as AA, and λ\lambda is a scalar variable. The roots of this polynomial are the eigenvalues of matrix AA.

Step-by-Step Method

1

Verify square matrix

Ensure the matrix is square (same number of rows and columns)

2

Form A - λI

Subtract λ times the identity matrix from the original matrix

3

Calculate determinant

Compute the determinant of the resulting matrix

4

Find eigenvalues

Set the polynomial equal to zero and solve for λ

Key Properties

Eigenvalue Connection

The roots of p(λ)=0p(\lambda) = 0 are the eigenvalues of matrix AA

Degree and Size

For an n×nn \times n matrix, the characteristic polynomial has degree nn

Examples

2×2 Matrix

A = [[1,2],[3,4]]

Solution:

  1. A−λI=(1−λ234−λ)A - \lambda I = \begin{pmatrix} 1-\lambda & 2 \\ 3 & 4-\lambda \end{pmatrix}

  2. det⁡(A−λI)=(1−λ)(4−λ)−2×3\det(A - \lambda I) = (1-\lambda)(4-\lambda) - 2×3

  3. p(λ)=λ2−5λ−2p(\lambda) = \lambda^2 - 5\lambda - 2

λ₁ ≈ 5.37, λ₂ ≈ -0.37

Symmetric 2×2 Matrix

A = [[2,1],[1,2]]

Solution:

  1. A−λI=(2−λ112−λ)A - \lambda I = \begin{pmatrix} 2-\lambda & 1 \\ 1 & 2-\lambda \end{pmatrix}

  2. det⁡(A−λI)=(2−λ)2−1\det(A - \lambda I) = (2-\lambda)^2 - 1

  3. p(λ)=λ2−4λ+3p(\lambda) = \lambda^2 - 4\lambda + 3

λ₁ = 3, λ₂ = 1

3×3 Matrix

A = [[1,2,3],[4,5,6],[7,8,9]]

Solution:

  1. A−λI=(1−λ2345−λ6789−λ)A - \lambda I = \begin{pmatrix} 1-\lambda & 2 & 3 \\ 4 & 5-\lambda & 6 \\ 7 & 8 & 9-\lambda \end{pmatrix}

  2. Calculate determinant of 3×3 matrix

  3. p(λ)=−λ3+15λ2−c1λ+0p(\lambda) = -\lambda^3 + 15\lambda^2 - c_1\lambda + 0

λ₁ ≈ 16.12, λ₂ ≈ -1.12, λ₃ = 0

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Frequently Asked Questions

What is a characteristic polynomial?

A characteristic polynomial is a polynomial equation associated with a square matrix. It is defined as p(λ)=det⁡(A−λI)p(\lambda) = \det(A - \lambda I), where A is the matrix, I is the identity matrix, and λ is a variable. The roots of this polynomial are the eigenvalues of the matrix.

Why do we need characteristic polynomials?

Characteristic polynomials are essential for finding eigenvalues, which are crucial in many applications including diagonalization, solving differential equations, analyzing system stability, and understanding matrix transformations.

Can non-square matrices have characteristic polynomials?

No, characteristic polynomials are only defined for square matrices. For rectangular matrices, the concept of eigenvalues and characteristic polynomials does not apply.

What is the relationship between characteristic polynomials and eigenvalues?

The eigenvalues of a matrix are exactly the roots of its characteristic polynomial. If p(λ)=0p(\lambda) = 0 has solutions λ1,λ2,...,λn\lambda_1, \lambda_2, ..., \lambda_n, then these are the eigenvalues of the matrix.

Can characteristic polynomials have complex roots?

Yes, characteristic polynomials can have complex roots, especially for non-symmetric matrices. Complex eigenvalues are mathematically valid and important in many applications.

How do I find eigenvectors from the characteristic polynomial?

First find the eigenvalues by solving the characteristic equation, then for each eigenvalue λ, solve the system (A−λI)v=0(A - \lambda I)\mathbf{v} = \mathbf{0} to find the corresponding eigenvectors.

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Last updated: 24/08/2025 — Written by the AskMathAI team