Eigenvalues of a 2×2 matrix from det(A − λI) = 0
The two scalars that survive when a matrix acts on its eigenvectors — found by solving one quadratic equation.
What this shows
An eigenvalue of a square matrix A is a scalar λ such that there exists a non-zero vector v with
A · v = λ · v
Geometrically: when A acts on v it just stretches v by the factor λ, without rotating it. The vector v is called an eigenvector for λ. Eigenvalues exist for any square matrix (complex ones in general, real ones for symmetric matrices).
The equation Av = λv rearranges to (A − λI)v = 0, and that has a non-zero solution v exactly when (A − λI) is singular — i.e. when det(A − λI) = 0. Expanding the determinant for a 2×2 matrix gives the characteristic polynomial:
det(A − λI) = λ² − (trace A)·λ + (det A)
For A = [[2, 1], [1, 2]]:
trace A = 2 + 2 = 4
det A = 2·2 − 1·1 = 3
poly = λ² − 4λ + 3 = (λ − 1)(λ − 3)
roots = λ = 1, λ = 3Plugging each back: (A − 1·I)v = 0 forces v in the direction (1, −1); (A − 3·I)v = 0 forces v in the direction (1, 1). The two eigenvectors are perpendicular, which is no accident — it always happens for a symmetric matrix.
Where it shows up
Eigenvalues are the diagonal entries of a matrix once it's been rotated into its eigenbasis. Whenever you see a fast power iteration, a Markov-chain stationary distribution, a principal component analysis, or a normal-mode decomposition of a vibrating system — eigenvalues are doing the work.
In differential equations, the eigenvalues of the system's coefficient matrix decide whether a steady state is stable (negative real parts), unstable (positive real parts), or oscillating (purely imaginary). In statistics, the eigenvalues of a covariance matrix tell you the variance along each principal axis of the data cloud.
Frequently asked questions
What if the discriminant of the characteristic polynomial is negative?
Then the eigenvalues are a complex-conjugate pair. Real matrices can still have complex eigenvalues — a 90° rotation matrix, for instance, has eigenvalues i and −i. Its eigenvectors are complex, but they still satisfy Av = λv.
Why does the trace appear in the formula?
Because for any n×n matrix, the sum of the eigenvalues equals the trace, and the product equals the determinant. For 2×2 those two facts uniquely determine the characteristic polynomial, which is why it's λ² − (trace)·λ + det.
Are eigenvectors unique?
Only up to a scalar. If v is an eigenvector for λ, so is c·v for any non-zero scalar c. Conventionally you normalise eigenvectors to have length 1, but the direction is the only thing that matters.
What if both eigenvalues are equal?
You get a repeated eigenvalue. The matrix may still be diagonalisable (if there are two independent eigenvectors) or it may be defective (if there's only one), in which case you need a Jordan-form generalisation. The simplest defective example is [[1, 1], [0, 1]], with the single eigenvalue λ = 1.