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.

  • $\begingroup$ Yes, if you have $A=PDP^{-1}$, then the columns of $P$ are eigenvectors. Then, by the invertible matrix theorem, the columns form a basis. $\endgroup$ – Michael Burr May 3 '17 at 18:41
  • $\begingroup$ math.stackexchange.com/q/1902923 $\endgroup$ – Jonas Meyer May 3 '17 at 18:43

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


Pick a basis with respect to which the matrix is diagonal. Said basis is an eigenvector basis. It is however true that not all collections of eigenvectors form a basis.


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