Every matrix has special directions it cannot rotate — only stretch. Those directions and their stretch factors are the matrix's fingerprint.
The idea, before the algebra
Picture a matrix transforming the plane: most vectors get knocked off their original line — rotated, sheared, redirected. But a few stubborn vectors stay on their own line. The matrix can stretch them, shrink them, or flip them, but it cannot turn them.
Those are the eigenvectors ("eigen" is German for own or characteristic). The factor by which each one stretches is its eigenvalue, written \(\lambda\). The defining equation:
$$Av = \lambda v$$
In words: applying the matrix \(A\) to the vector \(v\) does exactly the same thing as multiplying \(v\) by a plain number \(\lambda\). The matrix, in that direction, behaves like simple scalar arithmetic.
Finding eigenvalues: the characteristic equation
Rearrange \(Av = \lambda v\) to \((A - \lambda I)v = 0\). A nonzero solution \(v\) exists only when \(A - \lambda I\) is singular — that is, when its determinant vanishes:
$$\det(A - \lambda I) = 0$$
This is the characteristic equation. For an \(n \times n\) matrix it is a degree-\(n\) polynomial in \(\lambda\), so an \(n \times n\) matrix has up to \(n\) eigenvalues.
Worked example
Take the matrix from the widget above:
$$A = \begin{bmatrix} 3 & 1 \\ 0 & 2 \end{bmatrix}$$
Step 1 — eigenvalues. Set up the characteristic equation:
$$\det\begin{bmatrix} 3-\lambda & 1 \\ 0 & 2-\lambda \end{bmatrix} = (3-\lambda)(2-\lambda) - 0 = 0$$
So \(\lambda_1 = 3\) and \(\lambda_2 = 2\).
Step 2 — eigenvectors. For \(\lambda_1 = 3\), solve \((A - 3I)v = 0\):
$$\begin{bmatrix} 0 & 1 \\ 0 & -1 \end{bmatrix}\begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \;\Rightarrow\; v_2 = 0 \;\Rightarrow\; v = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$$
For \(\lambda_2 = 2\), solving \((A - 2I)v = 0\) gives \(v_1 + v_2 = 0\), so \(v = \begin{bmatrix} 1 \\ -1 \end{bmatrix}\).
Check: \(A\begin{bmatrix}1\\0\end{bmatrix} = \begin{bmatrix}3\\0\end{bmatrix} = 3\begin{bmatrix}1\\0\end{bmatrix}\). The matrix stretched the vector by exactly 3 and didn't turn it. ✓
Diagonalization: a matrix in its own coordinates
If an \(n \times n\) matrix has \(n\) independent eigenvectors, pack them into a matrix \(P\) and the eigenvalues into a diagonal matrix \(D\). Then:
$$A = PDP^{-1}$$
This says something profound: viewed in its eigenvector coordinate system, the matrix is just a set of independent stretches. All the apparent complexity was a change of perspective. The practical payoff is enormous — powers become trivial:
$$A^{100} = PD^{100}P^{-1} \qquad D^{100} = \begin{bmatrix} \lambda_1^{100} & 0 \\ 0 & \lambda_2^{100} \end{bmatrix}$$
Raising a diagonal matrix to a power just raises each diagonal entry. This is how systems that evolve step by step — populations, Markov chains, Google's PageRank — are analyzed over long horizons.
Where eigen-analysis shows up
| Application | What the eigenvectors mean |
|---|
| PCA (machine learning) | Directions of maximum variance in data — the features that matter |
| Google PageRank | The dominant eigenvector of the web's link matrix ranks every page |
| Vibration analysis | Natural resonant modes of bridges and buildings |
| Quantum mechanics | Eigenvalues of operators are the only measurable energy levels |
| Stability of systems | Eigenvalues beyond magnitude 1 mean the system blows up |
The machine learning case — PCA and beyond — is covered in depth in Matrices in AI.