An eigenspace of an n x n matrix is the subspace spanned by the eigenvectors corresponding to a given eigenvalue. Essentially, an eigenspace is the null space of a matrix A-λI, where λ is an eigenvalue of the matrix A and I is the identity matrix. For an eigenspace V of a matrix A in Rn the dimension of V is less than or equal to n. Since all eigenvectors for a given eigenvalue are linearly independent, the dimension of the eigenspace is equal to the number of linearly independent vectors that satisfy (A-λI) v = 0 which follows from the definition of an eigenvector. Furthermore, the dimension of an eigenspace is less than or equal to the algebraic multiplicity of its respective eigenvalue. It follows then that the sum of the dimensions of all eigenspaces is less than or equal to n (i.e. an n x n matrix has at most n linearly independent eigenvectors and at most n distinct eigenvalues).
For symmetric matrices, eigenspaces for respective eigenvalues are orthogonal.
In simpler terms. Suppose we have an n x n matrix, then the eigenvalues are the solutions to the characteristic equation det(A-λI) = 0, note that the characteristic equation will always be of degree n. For a 3 x 3 matrix A:
1 0 1
2 0 4
0 0 2
The characteristic equation is: (1-λ)(-λ)(2-λ)=0 the values λ=1, λ=2, and λ=0 satisfy this equation with each solution of algebraic multiplicity 1 (since no roots are raised to any power other than 1). Hence there should be 3 distinct, linearly independent eigenvectors one corresponding to each eigenvalue, λ, and each defining its own eigenspace.
The eigenvectors are defined as the null space (or kernel) of the matrix A-λI. For λ = 1, A-λI equals:
0 0 1
2 -1 4
0 0 1
which row reduces to:
0 0 1
2 -1 0
0 0 0
If we consider this matrix to represent a system of linear equations in 3 dimensions all equalling zero, then the solution is any x, y, z such that x = y/2, y = anything, and z = 0. We can represent this solution as a vector subspace which is the eigenspace to the eigenvector 1 of the matrix A. It is the line (1, 2, 0)t where t is any real number. This eigenspace is of dimension 1 since only 1 vector is needed to define it. It is in R3 as the line is defined by 3 coordinates: x, y, and z. By the same methods we could determine the respective eigenspaces for the other two eigenvalues.