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 -1AP is a diagonal matrix. If V is a finite-dimensionalal vector space, then a linear map T : V → V is called diagonalizable if there exists a 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.
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.
The fundamental fact about diagonalizable maps and matrices is expressed by the following:
- An n-by-n matrix A over the field F is diagonalizable if and only if the sum of the dimensionss of its eigenspaces is equal to n, which is the case if and only if there exists a basis of Fn 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 -1AP 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 dimensionss 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.
The following sufficient (but not necessary) condition is often useful.
- 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.
- 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.
Here is an example of a diagonalizable matrix:
Since the matrix is triangular (specifically upper triangular), the eigenvalues are 5, 0, and -2. Since A is a 3-by-3 matrix with 3 real, distinct eigenvalues, A is diagonalizable over R.
As a rule of thumb, over C almost every matrix is diagonalizable. More precisely: the set of complex n-by-n matrices that are not diagonalizable over C, considered as a subset of Cn×n, is a null set with respect to the Lebesgue measure. One can also say that the diagonalizable matrices form a dense subset with respect 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. As n increases, it becomes (in some sense) less and less likely that a randomly selected real matrix is diagonalizable over R.
An application
Diagonalization can be used
to compute the powers of a matrix A efficiently, provided the matrix is diagonalizable. Suppose we have found that
is a diagonal matrix. Then
and the latter is easy to calculate since it only involves the powers of a diagonal matrix.
For example, consider the following matrix:
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Calculating the various powers of M reveals a surprising pattern:
The above phenomenon can be explained by diagonalizing M. To accomplish this, we need a basis of R2 consisting of eigenvectors
of M. One such eigenvector basis is given by
-
where ei denotes the standard basis of Rn.
The reverse change of basis is given by
Straighforward calculations show that
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Thus, a and b are the eigenvalues corresponding to u and v, respectively.
By linearity of matrix multiplication, we have that
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Switching back to the standard basis, we have
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The preceding relations, expressed in matrix form, are
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thereby explaining the above phenomenon.
