By Theorem 10, D is a diagonalizable operator which we shall call the diagonalizable part of T.

Then there is a diagonalizable operator D on V and a nilpotent operator N in V such that ( A ) Af, ( b ) Af.

The diagonalizable operator D and the nilpotent operator N are uniquely determined by ( A ) and ( B ) and each of them is a polynomial in T.

) Now Af is a diagonalizable operator which is also nilpotent.

but then since the operator is diagonalizable, the minimal polynomial cannot have a repeated root ; ;

Then every linear operator T in V can be written as the sum of a diagonalizable operator D and a nilpotent operator N which commute.

A compact normal operator ( in particular, a normal operator on a finite-dimensional linear space ) is unitarily diagonalizable.

This implies the usual spectral theorem: every normal operator on a finite-dimensional space is diagonalizable by a unitary operator.

The Jordan – Chevalley decomposition expresses an operator as the sum of its semisimple ( i. e., diagonalizable ) part and its nilpotent part.

If the operator is orthogonal ( an orthogonal involution ), it is orthonormally diagonalizable.

Thus T is not diagonalizable.

We have just observed that we can write Af where D is diagonalizable and N is nilpotent, and where D and N not only commute but are polynomials in T.

Now suppose that we also have Af where D' is diagonalizable, N' is nilpotent, and Af.

Since D and D' are both diagonalizable and they commute, they are simultaneously diagonalizable, and Af is diagonalizable.

For diagonalizable matrices, an even better method is to use the eigenvalue decomposition of A.

* U is diagonalizable ; that is, U is unitarily similar to a diagonal matrix, as a consequence of the spectral theorem.

Normality is a convenient test for diagonalizability: a matrix is normal if and only if it is unitarily similar to a diagonal matrix, and therefore any matrix A satisfying the equation A * A = AA * is diagonalizable.

Furthermore the two are simultaneously diagonalizable, that is: both A and B are made diagonal by the same unitary matrix U. Both UAU < sup >*</ sup > and UBU < sup >*</ sup > are diagonal matrices.

The diagonal form for diagonalizable matrices, for instance normal matrices, is a special case of the Jordan normal form.

* An n-by-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 < sup > n </ sup > consisting of eigenvectors 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.

In the case of operators on finite-dimensional vector spaces, for complex square matrices, the relation of being isospectral for two diagonalizable matrices is just similarity.

Now if any of the expressions that the theorem equates to 0 would not reduce to a null expression, in other words if it would be a nonzero polynomial in the coefficients of the matrix, then the set of complex matrices for which this expression happens to give 0 would not be dense in the set of all matrices, which would contradict the fact that the theorem holds for all diagonalizable matrices.

If V is a finite-dimensional vector space, then a linear map T: V → V is called diagonalizable if there exists a basis of V with respect to T which is represented by a diagonal matrix.

A square matrix which is not diagonalizable is called defective.

: which has eigenvalues 1, 2, 2 ( not all distinct ) and is diagonalizable with diagonal form ( similar to A )

Proof: If is diagonalizable, then A is annihilated by some polynomial, which has no multiple root ( since ) and is divided by the minimal polynomial of A.

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.

Given a set S of matrices, each of which is diagonalizable, and any two of which commute, it is always possible to simultaneously diagonalize all of the elements of S. Equivalently, for any set S of mutually commuting semisimple linear transformations of a finite-dimensional vector space V there exists a basis of V consisting of simultaneous eigenvectors of all elements of S. Each of these common eigenvectors v ∈ V, defines a linear functional on the subalgebra U of End ( V ) generated by the set of endomorphisms S ; this functional is defined as the map which associates to each element of U its eigenvalue on the eigenvector v. This " generalized eigenvalue " is a prototype for the notion of a weight.

If is a linear Lie algebra ( a Lie subalgebra of the Lie algebra of endomorphisms of a finite-dimensional vector space V ) over an algebraically closed field, then any Cartan subalgebra of is the centralizer of a maximal toral Lie subalgebra of ; that is, a subalgebra consisting entirely of elements which are diagonalizable as endomorphisms of V which is maximal in the sense that it is not properly included in any other such subalgebra.

Now if A admits a basis of eigenvectors, in other words if A is diagonalizable, then the Cayley – Hamilton theorem must hold for A, since two matrices that give the same values when applied to each element of a basis must be equal.

Not all matrices are diagonalizable, but for matrices with complex coefficients many of them are: the set of diagonalizable complex square matrices of a given size is dense in the set of all such square matrices ( for a matrix to be diagonalizable it suffices for instance that its characteristic polynomial not have multiple roots ).

While this provides a valid proof ( for matrices over the complex numbers ), the argument is not very satisfactory, since the identities represented by the theorem do not in any way depend on the nature of the matrix ( diagonalizable or not ), nor on the kind of entries allowed ( for matrices with real entries the diagonizable ones do not form a dense set, and it seems strange one would have to consider complex matrices to see that the Cayley – Hamilton theorem holds for them ).

Diagonalization is the process of finding a corresponding diagonal matrix for a diagonalizable matrix or linear map.

Conversely, if A is invertible, F is algebraically closed, and A < sup > n </ sup > is diagonalizable for some n that is not an integer multiple of the characteristic of F, then A is diagonalizable.

* Real symmetric matrices are diagonalizable by orthogonal matrices ; i. e., given a real symmetric matrix, is diagonal for some orthogonal matrix.

Except for in characteristic 2, such operators are diagonalizable with 1s and − 1s on the diagonal.

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 ).

Put in another way, a matrix is diagonalizable if each block in its Jordan form has no nilpotent part ; i. e., one-by-one matrix.

* For any abelian group A, one can form the corresponding diagonalizable group D ( A ), defined as a functor by setting D ( A )( T ) to be the set of abelian group homomorphisms from A to invertible global sections of O < sub > T </ sub > for each S-scheme T. If S is affine, D ( A ) can be formed as the spectrum of a group ring.

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 < sup > − 1 </ sup > AP is a diagonal matrix.

* An n-by-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.

* 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.

then it is called a ; this corresponds to having an eigenbasis ( a basis of eigenvectors ), i. e., being a diagonalizable matrix.