Commuting matrices

In linear algebra, two matrices and are said to commute if , or equivalently if their commutator is zero. A set of matrices is said to commute if they commute pairwise, meaning that every pair of matrices in the set commute with each other. Commuting matrices preserve each other's eigenspaces. As a consequence, commuting matrices over an algebraically closed field are simultaneously triangularizable; that is, there are bases over which they are both upper triangular. In other words, if commute, there exists a similarity matrix such that is upper triangular for all . The converse is not necessarily true, as the following counterexample shows: However, if the square of the commutator of two matrices is zero, that is, , then the converse is true. Two diagonalizable matrices and commute () if they are simultaneously diagonalizable (that is, there exists an invertible matrix such that both and are diagonal). The converse is also true; that is, if two diagonalizable matrices commute, they are simultaneously diagonalizable. But if you take any two matrices that commute (and do not assume they are two diagonalizable matrices) they are simultaneously diagonalizable already if one of the matrices has no multiple eigenvalues. If and commute, they have a common eigenvector. If has distinct eigenvalues, and and commute, then 's eigenvectors are 's eigenvectors. If one of the matrices has the property that its minimal polynomial coincides with its characteristic polynomial (that is, it has the maximal degree), which happens in particular whenever the characteristic polynomial has only simple roots, then the other matrix can be written as a polynomial in the first. As a direct consequence of simultaneous triangulizability, the eigenvalues of two commuting complex matrices A, B with their algebraic multiplicities (the multisets of roots of their characteristic polynomials) can be matched up as in such a way that the multiset of eigenvalues of any polynomial in the two matrices is the multiset of the values . This theorem is due to Frobenius.