Eigenvectors: Directions a Matrix Only Scales
An eigenvector is a direction that a matrix does not turn. This guide shows why, how to find eigenvectors by solving (A − λI)v = 0, and how eigenspaces work.
- What eigenvectors are
- Intuition
- Why direction is preserved
- (A − λI)v = 0
- Step by step
- 2×2 worked example
- Eigenspaces
- Multiple independent eigenvectors
- Common mistakes
What eigenvectors are
A non-zero vector \mathbf v is an eigenvector of the square matrix A if
for some number \lambda, its eigenvalue. The zero vector is never an eigenvector, although \lambda=0 is allowed.
Intuition
Picture an arrow from the origin. Apply the matrix. A generic arrow ends up pointing somewhere new, but an eigenvector lands on the same line through the origin. It only becomes longer, shorter, or (for negative \lambda) points the opposite way along that line.
Why direction is preserved
A\mathbf v=\lambda\mathbf v says the output is a scalar multiple of the input, and all scalar multiples of \mathbf v lie on one line. So that whole line is mapped onto itself. Any non-zero multiple c\mathbf v is an eigenvector too, because A(c\mathbf v)=cA\mathbf v=\lambda(c\mathbf v).
The equation (A − λI)v = 0
Rearranging A\mathbf v=\lambda\mathbf v gives
For a known \lambda this is a homogeneous linear system. It has non-zero solutions precisely because \det(A-\lambda I)=0. So the eigenvectors for \lambda are the non-zero vectors in the null space of A-\lambda I.
Step-by-step calculation
- Find the eigenvalues from \det(A-\lambda I)=0.
- For each eigenvalue, form A-\lambda I.
- Solve (A-\lambda I)\mathbf v=\mathbf 0 by row reduction. There is always at least one free variable.
- Write the solution as a multiple of a vector (or several vectors).
- Verify with A\mathbf v=\lambda\mathbf v.
Worked example (2×2)
Find the eigenvectors of A=\begin{bmatrix}4&1\\2&3\end{bmatrix}, whose eigenvalues are 5 and 2.
For \lambda=5: A-5I=\begin{bmatrix}-1&1\\2&-2\end{bmatrix}. The first row gives -x+y=0, so y=x and \mathbf v_1=\begin{bmatrix}1\\1\end{bmatrix}. Check: A\mathbf v_1=(5,5)=5\mathbf v_1. ✓
For \lambda=2: A-2I=\begin{bmatrix}2&1\\2&1\end{bmatrix}. The row 2x+y=0 gives y=-2x, so \mathbf v_2=\begin{bmatrix}1\\-2\end{bmatrix}. Check: A\mathbf v_2=(2,-4)=2\mathbf v_2. ✓
Try it: open the Eigen Vector tool to see these vectors drawn as dashed arrows with A\mathbf v moving along the same line.
Eigenspaces
The set of all eigenvectors for one eigenvalue, together with the zero vector, is the eigenspace E_\lambda=\operatorname{null}(A-\lambda I). It is a subspace. Above, E_5 is the line \{t(1,1)\}. Its dimension is the geometric multiplicity of \lambda.
Multiple independent eigenvectors
An eigenvalue can have more than one independent eigenvector. For A=3I, the matrix A-3I is the zero matrix, so every vector solves the system. The eigenspace is the whole plane, with basis (1,0) and (0,1). Compare \begin{bmatrix}1&1\\0&1\end{bmatrix}: here A-I=\begin{bmatrix}0&1\\0&0\end{bmatrix} forces y=0, leaving only multiples of (1,0). Eigenvectors that belong to different eigenvalues are always linearly independent, which is what makes diagonalization work.
Common mistakes
- Calling \mathbf 0 an eigenvector. Eigenvectors must be non-zero.
- Thinking the eigenvector is unique. Any non-zero multiple works; the eigenspace is what is well defined.
- Pairing a vector with the wrong eigenvalue. Always verify A\mathbf v=\lambda\mathbf v.
- Getting only \mathbf v=\mathbf 0 from the system. That signals an arithmetic error or a wrong \lambda, since A-\lambda I must be singular.
- Forgetting that complex eigenvalues have no real eigenvectors (see transformations).
FAQ
Do I need to normalize eigenvectors?
No. Any non-zero scalar multiple is valid. Normalizing to length 1 is a convenience, and it is required only when you specifically want unit vectors.
Can two different eigenvalues share an eigenvector?
No. If A\mathbf v=\lambda\mathbf v=\mu\mathbf v with \mathbf v\ne\mathbf 0, then \lambda=\mu.