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In linear algebra, a square matrix ''A'' is called diagonalizable if it is similar to a diagonal matrix, i.e., if there exists an invertible matrix ''P'' such that ''P''−1''AP'' is a diagonal matrix. If ''V'' is a finite-dimensional vector space, then a linear map ''T'' : ''V'' → ''V'' is called diagonalizable if there exists an ordered basis of ''V'' with respect to which ''T'' is represented by a diagonal matrix. Diagonalization is the process of finding a corresponding diagonal matrix for a diagonalizable matrix or linear map.〔Horn & Johnson 1985〕 A square matrix that is not diagonalizable is called ''defective.'' Diagonalizable matrices and maps are of interest because diagonal matrices are especially easy to handle: their eigenvalues and eigenvectors are known and one can raise a diagonal matrix to a power by simply raising the diagonal entries to that same power. Geometrically, a diagonalizable matrix is an inhomogeneous dilation (or ''anisotropic scaling'') — it scales the space, as does a ''homogeneous dilation'', but by a different factor in each direction, determined by the scale factors on each axis (diagonal entries). == Characterization == The fundamental fact about diagonalizable maps and matrices is expressed by the following: * An ''n''×''n'' matrix ''A'' over the field ''F'' is diagonalizable if and only if the sum of the dimensions of its eigenspaces is equal to ''n'', which is the case if and only if there exists a basis of ''F''''n'' consisting of eigenvectors of ''A''. If such a basis has been found, one can form the matrix ''P'' having these basis vectors as columns, and ''P''−1''AP'' will be a diagonal matrix. The diagonal entries of this matrix are the eigenvalues of ''A''. * A linear map ''T'' : ''V'' → ''V'' is diagonalizable if and only if the sum of the dimensions of its eigenspaces is equal to dim(''V''), which is the case if and only if there exists a basis of ''V'' consisting of eigenvectors of ''T''. With respect to such a basis, ''T'' will be represented by a diagonal matrix. The diagonal entries of this matrix are the eigenvalues of ''T''. Another characterization: A matrix or linear map is diagonalizable over the field ''F'' if and only if its minimal polynomial is a product of distinct linear factors over ''F''. (Put in 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 ''n''×''n'' matrix ''A'' is diagonalizable over the field ''F'' if it has ''n'' distinct eigenvalues in ''F'', i.e. if its characteristic polynomial has ''n'' distinct roots in ''F''; however, the converse may be false. Let us consider :: : which has eigenvalues 1, 2, 2 (not all distinct) and is diagonalizable with diagonal form ( similar to ''A'') :: : and change of basis matrix ''P'' :: : The converse fails when ''A'' has an eigenspace of dimension higher than 1. In this example, the eigenspace of ''A'' associated with the eigenvalue 2 has dimension 2. * A linear map ''T'' : ''V'' → ''V'' with ''n'' = dim(''V'') is diagonalizable if it has ''n'' distinct eigenvalues, i.e. if its characteristic polynomial has ''n'' distinct roots in ''F''. Let ''A'' be a matrix over ''F''. If ''A'' is diagonalizable, then so is any power of it. Conversely, if ''A'' is invertible, ''F'' is algebraically closed, and ''An'' is diagonalizable for some ''n'' that is not an integer multiple of the characteristic of ''F'', then ''A'' is diagonalizable. Proof: If ''An'' is diagonalizable, then ''A'' is annihilated by some polynomial , which has no multiple root (since ) and is divided by the minimal polynomial of ''A''. As a rule of thumb, over C almost every matrix is diagonalizable. More precisely: the set of complex ''n''×''n'' matrices that are ''not'' diagonalizable over C, considered as a subset of C''n''×''n'', has Lebesgue measure zero. One can also say that the diagonalizable matrices form a dense subset with respect to the Zariski topology: the complement lies inside the set where the discriminant of the characteristic polynomial vanishes, which is a hypersurface. From that follows also density in the usual (''strong'') topology given by a norm. The same is not true over R. The Jordan–Chevalley decomposition expresses an operator as the sum of its semisimple (i.e., diagonalizable) part and its nilpotent part. Hence, a matrix is diagonalizable if and only if its nilpotent part is zero. Put in another way, a matrix is diagonalizable if each block in its Jordan form has no nilpotent part; i.e., each "block" is a one-by-one matrix. 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「diagonalizable matrix」の詳細全文を読む スポンサード リンク
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