Last updated: 2026-07-01

What this calculator does

This calculator finds the eigenvalues and eigenvectors of a 2×2 matrix. Enter the four entries of A = [[a, b], [c, d]] and it returns the trace, the determinant, the characteristic equation, both eigenvalues — real or complex — and an eigenvector for each real eigenvalue, along with a step-by-step derivation.

An eigenvector of A is a non-zero vector whose direction is left unchanged when the matrix acts on it. The amount by which it is scaled is the eigenvalue. In symbols:

$A\mathbf{v} = \lambda\mathbf{v}, \qquad \mathbf{v} \neq \mathbf{0}$

The characteristic equation

Rewrite $A\mathbf{v} = \lambda\mathbf{v}$ as $(A - \lambda I)\mathbf{v} = \mathbf{0}$. A non-zero solution exists only when the matrix $A - \lambda I$ is singular, that is when its determinant is zero:

$\det(A - \lambda I) = 0$

For a 2×2 matrix this expands to a quadratic in $\lambda$:

$\lambda^2 - (\operatorname{tr}A)\,\lambda + \det A = 0$

where the trace is $\operatorname{tr}A = a + d$ and the determinant is $\det A = ad - bc$. These two invariants alone determine both eigenvalues.

Solving for the eigenvalues

Apply the quadratic formula to the characteristic equation:

$\lambda = \frac{\operatorname{tr}A \pm \sqrt{(\operatorname{tr}A)^2 - 4\det A}}{2}$

The quantity under the root, $\Delta = (\operatorname{tr}A)^2 - 4\det A$, is the discriminant. Its sign decides the nature of the eigenvalues:

• $\Delta > 0$: two distinct real eigenvalues.
• $\Delta = 0$: one repeated real eigenvalue.
• $\Delta < 0$: a complex conjugate pair $\alpha \pm \beta i$.

Worked example

Take the symmetric matrix [[2, 1], [1, 2]].

First the trace and determinant:

$\operatorname{tr}A = 2 + 2 = 4, \qquad \det A = (2)(2) - (1)(1) = 3$

The characteristic equation and its discriminant:

$\lambda^2 - 4\lambda + 3 = 0, \qquad \Delta = 4^2 - 4(3) = 4$

Since $\Delta > 0$, the eigenvalues are real:

$\lambda_1 = \frac{4 + \sqrt{4}}{2} = 3, \qquad \lambda_2 = \frac{4 - \sqrt{4}}{2} = 1$

Finding the eigenvectors

For each eigenvalue $\lambda$, solve $(A - \lambda I)\mathbf{v} = \mathbf{0}$. For a 2×2 matrix the two rows are proportional, so a single row fixes the direction. A reliable choice is $\mathbf{v} = (b,\ \lambda - a)$ whenever $b \neq 0$.

For $\lambda_1 = 3$:

For $\lambda_2 = 1$:

Any non-zero scalar multiple of these vectors is also an eigenvector, since scaling a vector does not change its direction. A quick check confirms $A\mathbf{v}_1 = 3\mathbf{v}_1$ and $A\mathbf{v}_2 = \mathbf{v}_2$.

The complex case

When $\Delta < 0$ the square root is imaginary and the eigenvalues become a complex conjugate pair:

$\lambda = \frac{\operatorname{tr}A}{2} \pm \frac{\sqrt{|\Delta|}}{2}\,i = \alpha \pm \beta i$

Rotation matrices are the classic example. The 90° rotation [[0, −1], [1, 0]] has trace 0 and determinant 1, so $\Delta = 0 - 4 = -4$ and the eigenvalues are $\pm i$. A rotation leaves no real direction fixed, which is exactly why there are no real eigenvectors — the eigenvectors live in the complex plane instead.

Trace, determinant, and the eigenvalues

The two invariants relate directly to the eigenvalues, which gives a fast sanity check:

$\lambda_1 + \lambda_2 = \operatorname{tr}A, \qquad \lambda_1 \lambda_2 = \det A$

In the worked example, $3 + 1 = 4 = \operatorname{tr}A$ and $3 \cdot 1 = 3 = \det A$. If your eigenvalues do not reproduce the trace and determinant, an arithmetic slip has crept in.

Where eigenvalues are used

• Diagonalization: a matrix with distinct eigenvalues can be written as $A = PDP^{-1}$, where $D$ holds the eigenvalues, which makes powers $A^n$ easy to compute.
• Stability: for the system $\mathbf{x}' = A\mathbf{x}$, the signs of the real parts of the eigenvalues determine whether solutions grow or decay.
• Principal directions: for symmetric matrices the eigenvectors are orthogonal and mark the axes of stretching, which underlies principal component analysis and the geometry of quadratic forms.