What defines the number of linearly independent eigenvectors?

Question

I am struggling to understand SVD, so I need quite full understanding about 'diagonalizable'.

I know that if a (N,N) matrix have N linearly independent eigenvectors, then it is diagonalizable.

But, what determines the number of linearly independent eigenvectors?
(Like for singularity of matrix, there is a determinant.)

Should I check every possibility every time when I try to diagonalize?

Answer 1

A general criterion is that the minimal polynomial of the matrix, which is a divisor of its characteristic polynomial and has the same irreducible factors, splits into linear factors and these factors are distinct. So, denoting by $\chi_A(x)$ the characteristic polynomial of the matrix $A$, you may test whether the polynomial, which has the same simple irreducible factors as $\chi_A(x)$: $$p(x)=\frac{\chi_A(x)}{\chi'_A(x)}$$ splits into linear factors and $p(Ax)=0$, whereas for any $q(x)$, a strict divisor of $p(x)$, we have $q(A)\ne 0$.