Eigenvalue and Eigenvector Calculator
Calculate eigenvalue eigenvector instantly with our math tool. Shows detailed work, formulas used, and multiple solution methods.
Reviewed by Manoj Kumar, Mathematics Educator
Formula
det(A - lambda * I) = 0 (characteristic equation), then solve (A - lambda * I)v = 0 for eigenvectors
Eigenvalues are found by solving the characteristic equation, which sets the determinant of (A - lambda * I) to zero. For a 3x3 matrix, this produces a cubic polynomial. Each eigenvalue lambda has corresponding eigenvectors found by solving the homogeneous system (A - lambda * I)v = 0.
Worked Examples
Example 1: 3x3 Symmetric Matrix Eigenvalues
Problem:Find eigenvalues and eigenvectors of A = [[2, 1, 0], [1, 3, 1], [0, 1, 2]].
Solution:Characteristic polynomial: lambda^3 - 7*lambda^2 + 14*lambda - 8 = 0\nFactors: (lambda - 1)(lambda - 2)(lambda - 4) = 0\nEigenvalues: lambda = 1, 2, 4\n\nFor lambda = 1: [1, 1, 0; 1, 2, 1; 0, 1, 1]v = 0 -> v1 = [1, -1, 1]\nFor lambda = 2: [0, 1, 0; 1, 1, 1; 0, 1, 0]v = 0 -> v2 = [1, 0, -1]\nFor lambda = 4: [-2, 1, 0; 1, -1, 1; 0, 1, -2]v = 0 -> v3 = [1, 2, 1]
Result:Eigenvalues: 1, 2, 4 | Matrix is symmetric, so orthogonally diagonalizable
Example 2: Matrix with Repeated Eigenvalue
Problem:Find eigenvalues of A = [[3, 0, 0], [0, 3, 0], [0, 0, 5]].
Solution:This is a diagonal matrix, so eigenvalues are the diagonal entries.\nlambda1 = lambda2 = 3 (algebraic multiplicity 2)\nlambda3 = 5\nEigenvectors for lambda = 3: any vector in the xy-plane, v1 = [1,0,0], v2 = [0,1,0]\nEigenvector for lambda = 5: v3 = [0,0,1]\nGeometric multiplicity = algebraic multiplicity, so diagonalizable.
Result:Eigenvalues: 3, 3, 5 | Diagonalizable (geometric multiplicity = algebraic multiplicity)
References
Reviewed by Manoj Kumar, Mathematics Educator · Editorial policy