So you’ve probably learned that if u is an eigenvector, then multiplying u by any scalar gives you another eigenvector with the same eigenvalue. That means that the set of all a*u where a is any scalar forms a 1-dimensional space (a line if this is a real vector space). This is an eigenspace of dimension one. The full definition of an eigenspace is as the set of all eigenvectors of a given eigenvalue. Now, if an eigenvalue has multiple independent eigenvectors, then the set of all eigenvectors for that eigenvalue is is still a linear space, but of dimension more than one. So for a real vector space, if an eigenvalue has two sets of independent eigenvectors, its eigenspace will be a 2-dimensional plane.
You want an answer?
So you’ve probably learned that if u is an eigenvector, then multiplying u by any scalar gives you another eigenvector with the same eigenvalue. That means that the set of all a*u where a is any scalar forms a 1-dimensional space (a line if this is a real vector space). This is an eigenspace of dimension one. The full definition of an eigenspace is as the set of all eigenvectors of a given eigenvalue. Now, if an eigenvalue has multiple independent eigenvectors, then the set of all eigenvectors for that eigenvalue is is still a linear space, but of dimension more than one. So for a real vector space, if an eigenvalue has two sets of independent eigenvectors, its eigenspace will be a 2-dimensional plane.
That’s pretty much it.
Neat actually, and it fits into my understanding of linear algebra pretty well