Throughout this post, as always, denotes the ring of
matrices with entries from a field
and
denotes the identity matrix.
Recall that is said to be diagonalizable if it is similar to a diagonal matrix, i.e. if there exist a diagonal matrix
and an invertible matrix
such that
In other words,
is diagonalizable if there exists a basis
of
such that the matrix of the linear map
with respect to
call it
is diagonal. So if
are the diagonal entries of
then
i.e. each
is an eigenvector of
So
is diagonalizable if and only if
has a basis consisting of eigenvectors of
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In practice, it is not easy in general to prove whether or not a given matrix is diagonalizable. There is however the following well-known result which is quite useful.
Theorem. Let
i) If has
distinct eigenvalues in
then
is diagonalizable.
i) is diagonalizable if and only if the minimal polynomial of
has the form
for some distinct elements
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Proof. See, for example, Theorem 3.2 and Theorem 4.11 here, or any decent linear algebra textbook
Note that the first part of the Theorem only gives a sufficient condition for a matrix to be diagonalizable. The condition is not necessary because, for example, the identity matrix is diagonal hence diagonalizable but its eigenvalues, which are all
are not distinct.
Remark 1. Diagonalizability depends on the base field For example consider the matrix
The eigenvalues of are
and so
is diagonalizable, by the first part of the above Theorem. However, if we look at
as an element of
then
is no longer diagonalizable because its eigenvalues are not real.
Corollary (characterizing non-diagonalizable matrices). Let
Then is not diagonalizable if and only if
and
for all
Proof. Let be the characteristic polynomial of
i.e.
Suppose first that
is not diagonalizable. Then
is not diagonal and so it’s not a multiple scalar of the identity matrix. Also, by the first part of the above Theorem, the two eigenvalues of
must be equal, i.e. the two roots of
must be equal. Thus
which gives
Conversely, the condition
means that
has two equal roots, i.e.
for some
Thus the minimal polynomial of
is either
or
If the minimal polynomial of
is
then
contradiction. Thus the minimal polynomial of
is
and hence, by the second part of the above Theorem,
is not diagonalizable.
Remark 2. The above Corollary holds for any algebraically closed field not just
but it does not always hold if
is not algebraically closed. For example, if we choose
then the
matrix given in Remark 1 is not diagonalizable but it does not satisfy the identity
given in the Corollary.
Example. By the above Corollary, the following matrices, as elements of are not diagonalizable
Exercise. Find all that make the matrix
non-diagonalizable.
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