over a field is diagonalizable if and only if the sum of the dimensions of its eigenspaces is equal to , which is the case if and only if there exists a basis of consisting of eigenvectors of . If such a basis has been found, one can form the matrix having these basis vectors as columns, and will be a diagonal matrix whose diagonal entries are the eigenvalues of . The matrix is known as a modal matrix for . A linear map is diagonalizable if and only if the sum of the dimensions of its eigenspaces is equal to which is the case if and only if there exists a basis of consisting of eigenvectors of . With respect to such a basis, will be represented by a diagonal matrix. The diagonal entries of this matrix are the eigenvalues of Another characterization: A matrix or linear map is diagonalizable over the field if and only if its minimal polynomial is a product of distinct linear factors over (Put another way, a matrix is diagonalizable if and only if all of its elementary divisors are linear.) The following sufficient (but not necessary) condition is often useful. An matrix is diagonalizable over the field if it has distinct eigenvalues in i.e. if its characteristic polynomial has distinct roots in however, the converse may be false. Consider which has eigenvalues 1, 2, 2 (not all distinct) and is diagonalizable with diagonal form (similar to and change of basis matrix : The converse fails when has an eigenspace of dimension higher than 1. In this example, the eigenspace of associated with the eigenvalue 2 has dimension 2. A linear map with is diagonalizable if it has distinct eigenvalues, i.e. if its characteristic polynomial has distinct roots in . Let be a matrix over If is diagonalizable, then so is any power of it. Conversely, if is invertible, is algebraically closed, and is diagonalizable for some that is not an integer multiple of the characteristic of then is diagonalizable. Proof: If is diagonalizable, then is annihilated by some polynomial which has no multiple root (since and is divided by the minimal polynomial of Over the complex numbers , almost every matrix is diagonalizable. More precisely: the set of complex matrices that are not diagonalizable over considered as a subset of has Lebesgue measure zero. One can also say that the diagonalizable matrices form