Let's assume that you are talking about eigenvalues/eigenvectors of matrices. Note that if and is an eigenvalue of with associated eigenvector (or any normed vector space) then so that is also an eigenvector of associated with and a quick check shows that . Make sense?\
Remark: It was important here to note that by the definition of eigenvector.
Sorry, no, I don't understand. What does the double modulus sign mean and what is x?
Also, if for the purposes of making it easy for me to type, we let X equal the double modulus sign of x, then how do you go from
(1/X)(Ax) to lambda x / X?
Finally, what does x / X mean?
Sorry if I seem a bit slow on this, we didn't have the best of lecturers and I made the mistake of not using my tutorials to their full potential.
The double modulus sign is the usual norm on , namely ...and something is of "length one" if it's norm is one. So, for example if is a point on the unit circle we know it's of the form and so by definition so that has length one. So, anyways, my point was if you find ANY eigenvector associated with eigenvalue then is an eigenvector associated with for any . So, if is an eigenvalue associated with then so is but it's easy to show that so that is an eigenvalue associated with of length one.I was just trying to give you an easy way to find examples.