section_path: Endomorphisms and square matrices/Eigenvalues and eigenvectors
section_name: Eigenvalues and eigenvectors
paragraph_text: Eigenvalues and eigenvectors
  
  If f is a linear endomorphism of a vector space V over a field F, an eigenvector of f is a nonzero vector v of V such that f (v) = av for some scalar a in F. This scalar a is an eigenvalue of f.
  
  If the dimension of V is finite, and a basis has been chosen, f and v may be represented, respectively, by a square matrix M and a column matrix z; the equation defining eigenvectors and eigenvalues becomes
  
  Using the identity matrix I, whose entries are all zero, except those of the main diagonal, which are equal to one, this may be rewritten
  
  As z is supposed to be nonzero, this means that M – aI is a singular matrix, and thus that its determinant det (M − aI) equals zero. The eigenvalues are thus the roots of the polynomial
  
  If V is of dimension n, this is a monic polynomial of degree n, called the characteristic polynomial of the matrix (or of the endomorphism), and there are, at most, n eigenvalues.
  
  If a basis exists that consists only of eigenvectors, the matrix of f on this basis has a very simple structure: it is a diagonal matrix such that the entries on the main diagonal are eigenvalues, and the other entries are zero. In this case, the endomorphism and the matrix are said to be diagonalizable. More generally, an endomorphism and a matrix are also said diagonalizable, if they become diagonalizable after extending the field of scalars. In this extended sense, if the characteristic polynomial is square-free, then the matrix is diagonalizable.
  
  A symmetric matrix is always diagonalizable. There are non-diagonalizable matrices, the simplest being
  
  (it cannot be diagonalizable since its square is the zero matrix, and the square of a nonzero diagonal matrix is never zero).
  
  When an endomorphism is not diagonalizable, there are bases on which it has a simple form, although not as simple as the diagonal form. The Frobenius normal form does not need of extending the field of scalars and makes the characteristic polynomial immediately readable on the matrix. The Jordan normal form requires to extend the field of scalar for containing all eigenvalues, and differs from the diagonal form only by some entries that are just above the main diagonal and are equal to 1.