Is it possible for an order N square matrix to be diagonalizable but have less than 3 distinct eigenvalues? I've heard that it's possible, but can't think of an example.
Identity doesn't work, since it yields only the zero-vector for its e-value of 1, which is no good.
Edit:
It also doesn't imply that the matrix is invertible, right? Since you can have a singular matrix that is diagonalizable, right?
Edit: Here is a 3x3 matrix that is singular and has 2 distinct e-values, yet still has 3 linearly independent e-vectors, and hence is diagonalizable:
so I guess I have answered my own question.
It's not diagonalizable since its eigenspace is 1 dimensional ( span{(1,0)} ). I.e. only 1 linearly independent e-vector can be found.
I think maybe the solution should have said something like "iff the matrix of eigenvectors (the diagonalizing matrix) is full rank". That'd be equivalent to saying "its eigenspace is the same dimension as the matrix", and correct, right?
Whether or not a matrix is diagonalizable depends, not on its eigenvalues, but on its eigenvectors.
An n by n matrix is diagonalizable if and only if it has n independent eigenvectors. Another way of putting that is that a linear transformation, A, from vector space, V, to itself, can be written as a diagonal matrix if and only if there exist a basis for V consisting of eigenvectors of A.
Of course, if an n by n matrix has n distinct eigenvalues, then it has n independent eigenvectors but the other way is not necessarily true.