Let *A* be an *n x n* matrix and consider the set *E* = { **x**ε **R** ^{n }: *A* **x** = λ **x**}. If **x** ε *E*, then so is *t* **x** for any scalar *t*, since

Furthermore, if **x** _{1} and **x** _{2} are in *E*, then

These calculations show that *E* is closed under scalar multiplication and vector addition, so *E* is a subspace of **R** ^{n }. Clearly, the zero vector belongs to *E*; but more notably, *the nonzero elements in E are precisely the eigenvectors of A corresponding to the eigenvalue* λ. When the zero vector is adjoined to the collection of eigenvectors corresponding to a particular eigenvalue, the resulting collection,

forms a vector space called the **eigenspace** of *A* correspondign to the eigenvalue λ. Since it depends on both *A* and the selection of one of its eigenvalues, the notation

will be used to denote this space. Since the equation *A* **x** = λ **x** is equivalent to ( *A* − λ *I*) **x = 0**, the eigenspace *E* _{λ}( *A*) can also be characterized as the nullspace of *A* − λ *I*:

This observation provides an immediate proof that *E* _{λ}( *A*) is a subspace of **R** ^{n }.

Consider the matrix

The determination of the eigenvectors of *A* shows that its eigenspaces are

and

*E* _{−1}( *A*) is the line in **R** ^{2} through the origin and the point (1, 1), and *E* _{−2}( *A*) is the line through the origin and the point (2, 3). Both of these eigenspaces are 1‐dimensional subspaces of **R** ^{2}.

**Example 1**: Determine the eigenspaces of the matrix

First, form the matrix

The determinant will be computed by performing a Laplace expansion along the second row:

The roots of the characteristic equation, are clearly λ = −1 and 3, with 3 being a double root; these are the eigenvalues of *B*. The associated eigenvectors can now be found. Substituting λ = −1 into the matrix *B* − λ *I* in (*) gives

which is the coefficient matrix for the equation ( *B* − λ *I*) **x = 0** with λ = −1, which determines the eigenvectors corresponding to the eigenvalue λ = −1. These eigenvectors are the nonzero solutions of

The identical first and third equations imply that x _{1} + x _{3} = 0— that is, *x* _{3} = − *x* _{1}—and the second equation says *x* _{2} = 0. Therefore, the eigenvectors of *B* associated with the eigenvalue λ = −1 are all vectors of the form ( *x* _{1}, 0, − *x* _{1}) ^{T} = *x* _{1}(1, 0, −1) ^{T} for *x* _{1} ≠ 0. Removing the restriction that the scalar multiple be nonzero includes the zero vector and gives the full eigenspace:

Now, since

the eigenvectors corresponding to the eigenvalue λ = 3 are the nonzero solutions of

These equations imply that *x* _{3} = *x* _{1}, and since there is no restriction on *x* _{2}, this component is arbitrary. Therefore, the eigenvectors of *B* associated with λ = 3 are all nonzero vectors of the form (x _{1},x _{2},x _{1}) ^{T} = x _{1}(1,0,1) ^{T} + x _{2}(0,1,0) ^{T} The inclusion of the zero vector gives the eigenspace:

Note that dim *E* _{−1}( *B*) = 1 and dim *E* _{3}( *B*) = 2.