Search results
Results From The WOW.Com Content Network
The fundamental fact about diagonalizable maps and matrices is expressed by the following: An matrix 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 .
Let A be a square n × n matrix with n linearly independent eigenvectors q i (where i = 1, ..., n).Then A can be factored as = where Q is the square n × n matrix whose i th column is the eigenvector q i of A, and Λ is the diagonal matrix whose diagonal elements are the corresponding eigenvalues, Λ ii = λ i.
Two diagonalizable matrices and commute (=) if they are simultaneously diagonalizable (that is, there exists an invertible matrix such that both and are diagonal). [ 4 ] : p. 64 The converse is also true; that is, if two diagonalizable matrices commute, they are simultaneously diagonalizable. [ 5 ]
An n × n matrix A is diagonalizable if and only if the sum of the dimensions of the eigenspaces is n. Or, equivalently, if and only if A has n linearly independent eigenvectors. Not all matrices are diagonalizable; matrices that are not diagonalizable are called defective matrices. Consider the following matrix:
An n × n matrix A is diagonalizable if there is a matrix V and a diagonal matrix D such that A = VDV −1. This happens if and only if A has n eigenvectors which constitute a basis for C n. In this case, V can be chosen to be the matrix with the n eigenvectors as columns, and thus a square root of A is = ,
If is diagonalizable it may be decomposed as = where is an orthogonal matrix =, and is a diagonal matrix of the eigenvalues of . In the special case that A {\displaystyle A} is real symmetric, then Q {\displaystyle Q} and Λ {\displaystyle \Lambda } are also real.
An idempotent matrix is always diagonalizable. [3] Its eigenvalues are either 0 or 1: if is a non-zero eigenvector of some idempotent matrix and its associated eigenvalue, then = = = = =, which implies {,}.
This is because any function of a non-defective matrix acts directly on each of its eigenvalues, and the conjugate transpose of its spectral decomposition is , where is the diagonal matrix of eigenvalues. Likewise, if two normal matrices commute and are therefore simultaneously diagonalizable, any operation between these matrices also acts on ...