eigenvector

Note that all vectors in the direction of v (tv, for any scalar t) are also eigenvectors of M; but they're not really considered "different" for these purposes.

Ax = λx

For some scalar λ, which is called the eigenvalue. That is, x does not change direction when multiplied by A. As an example case, consider the matrix:

  [ 2  1 ]
  [ 0  3 ]

The eigenvalues of this matrix are 2 and 3. For the first eigenvector, corresponding to eigenvalue 2, rearrange the above equation to:

(A - λI)x = 0

Which is just finding the null space of A-λI. That matrix is:

  [ 0  1 ]
  [ 0  1 ]

A basis for the null space of this matrix, and thus the eigenvalue, is:

  [ 1 ]
  [ 0 ]

Do the same for eigenvalue 3 to get:

  [ 1 ]
  [ 1 ]

These form a basis for the column space of A. In general, the eigenvectors will always form this basis, so long as A is non-degenerate.

For further information, see MIT's OpenCourseWare (http://ocw.mit.edu/18/18.06/f02/index.html), which contains a fantastic set of video lectures on the subject.

A nice way to visualise such a transformation is that of a rubber square with an arrow drawn on it (see http://www.physlink.com/Education/AskExperts/ae520.cfm). If you stretch the square along a particular axis, only arrows in certain directions will keep their direction; this is true no matter how hard you stretch the beast. We can then say that those directions (vectors) are eigenvectors for that transformation (stretching), and the eigenvalues (length of the new arrow compared to the old arrow) depend on how hard you stretch (which is inherent in the transformation matrix).