Eigenvectors

Precise definition

For an eigenvalue $\lambda$ of $A$, an eigenvector is a nonzero vector $v$ satisfying $Av=\lambda v$. The eigenspace $E_\lambda=\ker(A-\lambda I)$ includes zero but its nonzero members are the eigenvectors associated with $\lambda$.

Notation and mathematical language

Solve the homogeneous system $(A-\lambda I)v=0$ to find a basis of $E_\lambda$. Its dimension is geometric multiplicity. Scaling an eigenvector by any nonzero scalar gives another eigenvector for the same eigenvalue.

Conceptual picture

An eigenvector's line is invariant: applying $A$ keeps the result on that line. A basis of eigenvectors decouples a transformation into independent scaling modes. For symmetric real matrices, eigenspaces of distinct eigenvalues are orthogonal.

Conditions and key results

The zero vector is excluded because it satisfies $A0=\lambda0$ for every scalar and provides no direction. Eigenvectors belonging to distinct eigenvalues are independent. Within a repeated eigenspace, one must choose an independent basis rather than count every scalar multiple.

A reliable strategy

1. Find valid eigenvalues first; do not solve with an arbitrary $\lambda$ unless deriving the characteristic condition.
2. Row-reduce $A-\lambda I$ and parameterize its null space.
3. Choose a nonzero basis for each eigenspace and record its dimension.
4. Verify directly with $Av=\lambda v$ and use independence checks before forming an eigenbasis.

Fully worked example

Interpretation and application

Eigenvectors identify steady patterns, principal axes, modes, and ranking directions. Their coordinates depend on basis and scale; normalization is a convention unless length has a specific meaning.

Common mistakes

Verification and reasonableness

• Multiply $A$ by each proposed vector and compare componentwise with $\lambda v$.
• Confirm eigenspace basis vectors satisfy the null-space equations.
• Check the number of independent eigenvectors against diagonalizability claims.

Practice

1. Find an eigenvector of $\begin{pmatrix}2&0\\0&5\end{pmatrix}$ for $\lambda=5$.
2. Does an eigenspace contain zero?
3. What is geometric multiplicity?

Further deduction

Eigenspaces for different eigenvalues intersect only at zero. If a nonzero $v$ belonged to both $E_\lambda$ and $E_\mu$, then $Av=\lambda v=\mu v$, so $(\lambda-\mu)v=0$ and $\lambda=\mu$. This simple argument supports the theorem that eigenvectors chosen from distinct eigenspaces are linearly independent.

Normalizing an eigenvector changes its length but not its eigendirection. In applications such as principal components, a sign flip $v\mapsto-v$ is equally valid and does not change the represented axis. Comparing computed eigenvectors should therefore account for nonzero scaling and, for repeated eigenvalues, different valid bases of the same eigenspace.

When an eigenvalue lacks enough ordinary eigenvectors, generalized eigenvectors may form a Jordan chain: $(A-\lambda I)v_1=0$ and $(A-\lambda I)v_2=v_1$. They do not satisfy the ordinary eigenvector equation, but they complete a basis in which the matrix has Jordan form. The resulting matrix powers contain polynomial factors such as $n\lambda^{n-1}$, explaining behaviour that eigenvalues alone do not reveal in defective recurrences or differential systems.

Related topics

• Eigenvalues
• Diagonalization
• Kernel and Range