Let V be a vector space over a field, and let it be finitely dimensional. Let $\phi \in End_V(\mathbb{K})$ and let A be the transformation matrix.
It is well known that if A has eigenvectors which form a basis of V, it is diagonalizable.
Is it also true, that if A is diagonalizable, its eigenvectors form a Basis of V?
I am skeptical, because not every diagonalizable matrix has distinct eigenvalues and I do not know given that is the case if the sum of the eigenspaces then add up to V.
Best Answer
Not in this precise formulation. For example, there may simply be more eigenvectors than we need for a basis. For the identity map, all non-zero vectors are eigenvectors (for eigenvalue $1$), but in order for all these vectors to form a basis, they must be linearly independent (which in this cse implies $K=\Bbb F_2$ and $\dim V=1$).
But what we do have is: An endomorphism is diagonalizable if and only if there exists a basis consisting of eigenvectors of it. In fact, this is immediate from a basis with respect to which the matrix is diagonal