Diagonalizable
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-dimensional 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 dimensions 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 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.
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:
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
Thus, a and b are the eigenvalues corresponding to u and v, respectively.
By linearity of matrix multiplication, we have that
Switching back to the standard basis, we have
The preceding relations, expressed in matrix form, are
thereby explaining the above phenomenon.
Referenced By
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