Eigenvalues and Eigenvectors

Most vectors get both stretched and rotated when a matrix acts on them. A rare few only get stretched: the matrix leaves their direction alone and just scales them. Those special directions are the eigenvectors, and the scale factors are the eigenvalues. They are the axes along which a linear map is simplest.¹

The defining equation

A nonzero vector $v$ is an eigenvector of the square matrix $A$ with eigenvalue $\lambda$ when

$$A v = \lambda v .$$

The map only scales $v$; it does not rotate it. Rewrite this as $(A - \lambda I)\,v = 0$. A nonzero $v$ solves it only when $A - \lambda I$ is singular, which means

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

This is the characteristic equation. Its roots are the eigenvalues, and for each $\lambda$ the null space of $A - \lambda I$ holds the matching eigenvectors.²

The 2x2 formula

For $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ write the trace $T = a + d$ and the determinant $D = ad - bc$. Then

$$\det(A - \lambda I) = \lambda^{2} - T\lambda + D = 0 \quad\Longrightarrow\quad \lambda = \frac{T \pm \sqrt{T^{2} - 4D}}{2} .$$

The two roots satisfy $\lambda_1 + \lambda_2 = T$ and $\lambda_1 \lambda_2 = D$, a fast sanity check. The discriminant $T^{2} - 4D$ decides everything: positive gives two real eigenvalues, zero a repeated one, and negative a complex pair. In the last case there is no real direction that $A$ merely scales, because $A$ is acting as a rotation.

A worked example

Take

$$A = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}, \qquad T = 4, \qquad D = 4 - 1 = 3 .$$

The discriminant is $T^{2} - 4D = 16 - 12 = 4 > 0$, so

$$\lambda = \frac{4 \pm \sqrt{4}}{2} = \frac{4 \pm 2}{2} \;\Longrightarrow\; \lambda_1 = 3,\; \lambda_2 = 1 .$$

For $\lambda_1 = 3$ solve $(A - 3I)v = 0$, that is $\begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix} v = 0$, giving $v_1 = (1, 1)$. For $\lambda_2 = 1$ the same step gives $v_2 = (1, -1)$. Verify directly:

$$A \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \end{bmatrix} = 3 \begin{bmatrix} 1 \\ 1 \end{bmatrix}. \checkmark$$

Feel the eigen-directions

The matrix below starts at the worked example. Sweep the test vector $v$ by its angle: almost everywhere $A v$ points off in a new direction, but when $v$ lands on a green ray, $A v$ becomes parallel to $v$ and the label appears. Those rays are the eigenvectors solved from the formula above. Push $b$ and $c$ to opposite signs to make $T^{2} - 4D < 0$ and watch the real eigenvectors vanish.

Eigen-explorer · A v = λ vinteractive

a = 2.0b = 1.0c = 1.0d = 2.0

angle θ = 20°

trace T

4.00

det D

3.00

T² − 4D

4.00

λ₁ = 3.000
λ₂ = 1.000

Sweep θ until the blue v lands on a green ray: there A v is parallel to v, so A only scales it.

Diagonalization

When an $n\times n$ matrix has $n$ independent eigenvectors, stack them as the columns of $P$ and put the eigenvalues on the diagonal of $\Lambda$. Then $A P = P \Lambda$, so

$$A = P \Lambda P^{-1} .$$

This is the payoff. Powers telescope because the inner factors cancel:

$$A^{k} = P \Lambda^{k} P^{-1}, \qquad \Lambda^{k} = \operatorname{diag}(\lambda_1^{k}, \ldots, \lambda_n^{k}) .$$

Raising $A$ to a power reduces to raising each scalar $\lambda_i$ to that power, which is why eigenvalues govern long-run behavior: the largest $|\lambda_i|$ dominates as $k$ grows. The same identity powers matrix exponentials and the stability analysis of linear dynamics.¹

Footnotes

1. Strang, Introduction to Linear Algebra (5th ed., 2016). ↩ ↩²

2. Axler, Linear Algebra Done Right (3rd ed., 2015). ↩