Eigenvectors & Eigenspaces

Eigenvectors solve (A - λ I)v = 0; the eigenspace is (A - λ I).

The idea

An eigenvector of a square matrix $A$ is a nonzero vector $v$ whose direction $A$ preserves: $Av = \lambda v$ for some scalar $\lambda$, so $A$ rescales $v$ by the factor $\lambda$ without turning it. Most vectors are not eigenvectors — a typical vector $u$ is carried to an image $Au$ that points off $u s line, as the figure shows.

The eigenvectors for a fixed $\lambda$ form a subspace once the zero vector is included. If $Av = \lambda v$, then $A(cv) = c\,Av = \lambda(cv)$, so every scalar multiple of $v$ qualifies; and if two vectors satisfy the equation, so does their sum. The zero vector is excluded from the definition of an eigenvector only because $A0 = \lambda 0$ holds for every $\lambda$ and so identifies none.

This subspace is the eigenspace of $\lambda$: $E_{\lambda} = \{\, v : Av = \lambda v \,\} = \ker(A - \lambda I).$ Its dimension is the geometric multiplicity of $\lambda$ — the number of independent directions that $A$ merely scales by $\lambda$.

The kernel description also says how to compute: once $\lambda$ is known, the eigenvectors are the nonzero solutions of the homogeneous system $(A - \lambda I)v = 0$.

Ways to work on it

• Walkthrough. Find an eigenvector of a 2 × 2 matrix for a given eigenvalue.
• Practice. Verify a candidate vector is an eigenvector and read off its eigenvalue.
• Hardest. Eigenspaces and geometric multiplicity for a repeated eigenvalue.

Not sure where to start? Take the ten-question placement test.