Yes, that is correct. Your matrix is 3 by 3 so the base space is 3 dimensional and you have three independent eigenvectors. Since there are 3 and they are independent (eigevectors corresponding to distinct eigenvalues are always independent), they form a basis for the space. That means that any vector v, can be written as .

Applying A to that, .

That is, Av= 0 if and only if so is a basis for the kernel of A and so is a basis for the image.