Look for a direction the matrix preserves
An eigenvector is a nonzero vector whose image under a matrix is a scalar multiple of itself. The scalar is its eigenvalue. For a real eigenpair, a negative eigenvalue reverses the direction; zero sends that vector to zero.
Rearranging the defining equation gives a homogeneous system. It has a nonzero solution only when the shifted matrix is singular, which gives the characteristic equation. Finding its roots is the first step; finding the vectors is a separate calculation.
Start with a matrix we can check by hand
Enter the three displayed rows in the eigenvalue calculator, one row per line. The final coordinate is separate from the upper 2×2 block, so expanding det(λI − A) leaves one linear factor and one quadratic factor.
The quadratic is (λ − 2)² − 1. Factoring it gives the three eigenvalues 1, 3, and 4. Using det(A − λI) instead changes the sign of this cubic, but leaves its roots unchanged.
Check each vector with one multiplication
For λ = 1, the first two equations reduce to v₁ + v₂ = 0 and the last gives v₃ = 0. Taking v₁ = 1 gives [1, −1, 0]. For λ = 3, the first two components are equal. For λ = 4, only the third component is needed.
The calculator normalizes its vectors to unit length. These convenient unnormalized vectors describe the same directions: any nonzero scalar multiple remains an eigenvector. Ordering and signs can differ without changing the answer.
| Eigenvalue λ | One vector v | Both Av and λv |
|---|---|---|
| 1 | [1, −1, 0] | [1, −1, 0] |
| 3 | [1, 1, 0] | [3, 3, 0] |
| 4 | [0, 0, 1] | [0, 0, 4] |
A real matrix can have complex eigenvalues
This matrix rotates a real vector by a quarter turn, so it has no nonzero real eigenvector. Its characteristic equation is λ² + 1 = 0. Over the complex numbers, the eigenvalues are i and −i, with directions [1, −i] and [1, i], respectively.
Complex output is therefore a valid result for real input. A complex eigenvector can also be multiplied by any nonzero complex scalar; comparing components alone can make equivalent answers look different.
Count independent vectors for a repeated root
Here the characteristic polynomial is (λ − 2)². Solving (J − 2I)v = 0 forces the second component to zero, leaving only the direction [1, 0]. The root occurs twice but has one independent eigenvector, so this matrix cannot be diagonalized.
Compare that with 2I: it has the same repeated eigenvalue, but every nonzero vector is an eigenvector and two independent directions can be chosen. A repeated eigenvalue alone does not decide diagonalizability; the eigenspace dimension matters.
Read the residual alongside the answer
A residual measures how closely the computed pair satisfies Av = λv. It is a useful consistency check, but a small residual alone does not guarantee accurate eigenvalues for an ill-conditioned problem. Nearly repeated roots can make both the roots and the computed eigenspace sensitive to small changes.
This calculator accepts real 2×2 and 3×3 matrices. Results use floating-point arithmetic; the 3×3 eigenspace calculation uses a 10⁻⁹ tolerance after scaling the matrix. Review its warnings and computed eigenspace dimensions. Larger matrices, complex input matrices, and a general Jordan decomposition are outside this tool.
References & further reading
- MIT OpenCourseWare, 18.06SC — eigenvalues and eigenvectors
- MIT OpenCourseWare — diagonalizing a matrix
Equations describe the stated model. Worked values are rounded for reading; the calculator’s download retains its calculated precision. Read about our checks and their limits.