Can we still diagnolize a matrix if its eigenvectors matrix is singular?

[AdrianZ]

well, if a matrix has n linearly independent eigen-vectors then it's easy, what if a matrix is not diagnolizable in that way? Can we still diagnolize it by other means?
And what if a matrix is not diagonalizable at all? Are there still ways to find its exponential matrix?

 

[spikedmath]

If you don't have enough eigenvectors you can't diagonalize it. There is a theorem that says: An n-by-n matrix A is diagonalizable if and only if the sum of the dimensions of its eigenspaces is equal to n.

However, you can 'almost' diagonalize any matrix you want. One way to do this is to use something called Jordan normal forms (http://en.wikipedia.org/wiki/Jordan_normal_form).

Since you are asking about exponential matrices, you might be interested in the Jordan–Chevalley decomposition which allows you to compute the exponential of a matrix by writing it as the sum of a diagonalizable matrix and nilpotent matrix.