Eigenspace · aiwiki.page

An eigenspace is a linear subspace on which a linear operator acts as multiplication by a single scalar. It contains all eigenvectors associated with a specified eigenvalue, together with the zero vector. In linear algebra, eigenspaces describe directions and higher-dimensional collections of vectors that undergo the same scalar action, providing the underlying structure for diagonalization and spectral decomposition. (math.brown.edu)

Definition and basic properties

Let $T:V\to V$ be a linear map on a vector space over a field $F$ . For a scalar $\lambda\in F$ , define

E_\lambda(T)=\{v\in V:T(v)=\lambda v\} =\ker(T-\lambda I),

where $I$ denotes the identity operator and $\ker$ denotes the kernel. The scalar $\lambda$ is an eigenvalue precisely when this subspace contains a nonzero vector. The zero vector belongs to every kernel but is not an eigenvector. (math.brown.edu)

The subspace property follows directly from linearity. If $u,v\in E_\lambda(T)$ and $a,b\in F$ , then

T(au+bv)=a\lambda u+b\lambda v =\lambda(au+bv).

Thus every linear combination remains in the same eigenspace. When $\lambda=0$ is an eigenvalue, its eigenspace is simply $\ker T$ : the vectors that the operator sends to zero. (textbooks.math.gatech.edu)

Computing an eigenspace

For an $n\times n$ matrix $A$ , the eigenspace is the null space of $A-\lambda I$ , with $I$ now the identity matrix. Computing it requires solving the homogeneous system of linear equations

(A-\lambda I)x=0.

Gaussian elimination identifies the free variables and expresses the solutions as the span of a basis. Finding one eigenvector is insufficient when the eigenspace has dimension greater than one. (math.mit.edu)

The rank–nullity theorem gives

\dim E_\lambda(A)=n-\operatorname{rank}(A-\lambda I),

connecting its dimension to the rank of the shifted matrix. (homepages.ucl.ac.uk)

For example, direct calculation for

A=\begin{pmatrix}2&0&0\\0&2&0\\0&0&3\end{pmatrix}

gives

E_2(A)=\{(x,y,0):x,y\in F\},\qquad E_3(A)=\{(0,0,z):z\in F\}.

The first is a coordinate plane; the second is a coordinate line. Every nonzero vector in the plane is an eigenvector for $2$ , not merely the two coordinate vectors chosen as a basis.

Multiplicity and diagonalization

The dimension of $E_\lambda(A)$ is the geometric multiplicity of $\lambda$ . Its algebraic multiplicity is its multiplicity as a root of the characteristic polynomial

p_A(t)=\det(tI-A),

where $\det$ is the determinant. For any eigenvalue of a finite-dimensional operator,

1\leq\dim E_\lambda(A) \leq\text{algebraic multiplicity of }\lambda.

Repeated roots therefore need not produce equally many independent eigenvectors. (textbooks.math.gatech.edu)

Eigenvectors belonging to distinct eigenvalues are linearly independent. More generally, the sum of the distinct eigenspaces is a direct sum. Diagonalization is possible over $F$ exactly when these eigenspaces together span the whole space:

V=E_{\lambda_1}(T)\oplus\cdots\oplus E_{\lambda_r}(T).

Equivalently, the characteristic polynomial must split into linear factors over $F$ , and geometric and algebraic multiplicities must agree for every eigenvalue. A repeated eigenvalue does not, by itself, prevent diagonalization. (textbooks.math.gatech.edu)

Dependence on the scalar field

The choice of field matters. A matrix with real entries can have no eigenvalues over the real numbers while possessing eigenvalues and eigenspaces over the complex numbers. For example,

R=\begin{pmatrix}0&-1\\1&0\end{pmatrix}

represents a quarter-turn rotation. It has no nonzero real eigenvectors, but over $\mathbb C$ its eigenvalues are $i$ and $-i$ , with eigenspaces

E_i(R)=\operatorname{span}_{\mathbb C}\{(1,-i)\},\qquad E_{-i}(R)=\operatorname{span}_{\mathbb C}\{(1,i)\}.

For real matrices, nonreal eigenvalues and their eigenvectors occur in complex-conjugate pairs. (textbooks.math.gatech.edu)

Coordinate independence

Although matrix coordinates change with the basis, the eigenspace of an operator is intrinsically defined. If

B=P^{-1}AP

for an invertible matrix $P$ , then

E_\lambda(B)=P^{-1}E_\lambda(A).

Indeed, $Bx=\lambda x$ is equivalent to $A(Px)=\lambda(Px)$ . Thus similar matrices represent the same operator in different coordinate systems, and corresponding eigenspaces have equal dimensions. (textbooks.math.gatech.edu)

Orthogonality and spectral decomposition

For a self-adjoint operator on a finite-dimensional inner product space, distinct eigenspaces are mutually orthogonal. The spectral theorem guarantees an orthonormal basis of eigenvectors. This applies to real symmetric matrices and complex Hermitian matrices; complex symmetry alone is insufficient. (math.brown.edu)

Writing $P_\lambda$ for the orthogonal projection onto $E_\lambda(T)$ , the operator has the decomposition

T=\sum_\lambda\lambda P_\lambda,\qquad I=\sum_\lambda P_\lambda.

Within a multidimensional eigenspace, the orthonormal basis is not unique, but the eigenspace and its orthogonal projection are fixed. (math.brown.edu)

Generalized eigenspaces

Ordinary eigenspaces may not span the space. For an operator on an $n$ -dimensional space, the generalized eigenspace associated with $\lambda$ is

G_\lambda(T)=\ker(T-\lambda I)^n.

It contains vectors annihilated by some positive power of $T-\lambda I$ , rather than necessarily by its first power. When the characteristic polynomial splits, generalized eigenspaces form a direct-sum decomposition of $V$ , and their dimensions equal the respective algebraic multiplicities. (math.brown.edu)

For example, direct calculation for

J=\begin{pmatrix}2&1\\0&2\end{pmatrix}

shows that $E_2(J)=\operatorname{span}\{(1,0)\}$ , whereas $(J-2I)^2=0$ , so $G_2(J)=F^2$ . The ordinary eigenspace is one-dimensional despite the eigenvalue’s algebraic multiplicity being two.