The primary decomposition theorem allows us to use the minimal polynomial of a matrix to write a vector space as a direct sum of invariant subspaces. It is key to understanding and interpreting the information provided by the minimal polynomial.
Let
be the space of
vectors. Let
be a
matrix.
Remember that a subspace
is said to be invariant
under the linear transformation defined by
if and only if
for
any
.
In other words,
is invariant if its elements are transformed by
into other elements of
.
The minimal polynomial of
is the monic annihilating
polynomial (i.e.,
and the leading coefficient of
is
)
that divides every annihilating polynomial of
.
The minimal polynomial is the lowest-degree polynomial in the set of
annihilating polynomials and
it has the property
thatwhere
are the distinct
eigenvalues of
and
for any
,
where
is the
algebraic
multiplicity of
.
We can write the minimal polynomial in terms of
as
where
is the
identity matrix.
In this lecture we are going to prove the Primary Decomposition Theorem, which
states that the vector space
can be written
as
where
denotes the null space
of the matrix
,
that
is,
and
denotes a direct sum.
Thus, by the definition of direct sum, we will be able to uniquely write each
vector
as
where
for
.
Moreover, the vectors
are guaranteed to be linearly
independent whenever they are non-zero.
Furthermore, we will show that the null spaces
are invariant.
The next proposition paves the way to the Primary Decomposition Theorem.
Proposition
Let
be the space of
vectors. Let
be a
matrix. Let
and
be two polynomials such that: 1) they are relatively prime (i.e., their
greatest
common divisor is
);
2) they are not identically equal to zero; 3)
is an annihilating polynomial.
Then,
and
and
are invariant under the linear transformation defined by
.
Since
and
are relatively prime and not identically equal to zero, by
Bezout's
theorem there exists two polynomials
and
such
that
Thus,
Arbitrarily
choose any
and post-multiply both sides of the previous equation by
:
where
we have
defined
The
previous two equations can be pre-multiplied by
and
so as to obtain
which
imply that
Since
was arbitrary, we
have
that
is, any
can be written as the sum of an element of
and an element of
.
In order to show that the sum is direct, we are going to prove that the
representation
is
unique. Suppose there is another
representation
where
and
.
Subtracting equation (2) from equation (3), we
obtain
Post-multiplying
equation (1) by
,
we get
where:
in step
we have used the fact that
and
belong to
;
in step
we have used equation (4); in step
we have used the fact that
and
belong to
.
Thus,
.
In a completely analogous manner, we can prove that
.
As a consequence, the representation as a sum is unique, which implies
that
Moreover,
by the properties of matrix
polynomials, the null spaces
and
are invariant under the linear transformation defined by
.
A simple example follows.
Example
Consider the space
of all
vectors and a
matrix
whose characteristic polynomial
is
In
other words,
has an eigenvalue
with algebraic multiplicity
and another eigenvalue
with algebraic multiplicity
.
The two
polynomials
satisfy
all the assumptions of the previous proposition because, by the
Cayley-Hamilton
theorem, the characteristic polynomial is an annihilating polynomial.
Then,
where
the two spaces
and
are invariant.
Here is the Primary Decomposition Theorem.
Proposition
Let
be the space of
vectors. Let
be a
matrix. Let
be the minimal polynomial of
:
where
are the distinct
eigenvalues of
.
Then,
where
the spaces
,
...,
are invariant under the linear transformation defined by
.
This proposition is proved by applying
recursively the previous proposition. We start with the two polynomials
They
are relatively prime, non-zero, and their product is an annihilating
polynomial since it is equal to the minimal polynomial.
Therefore,
We
now restrict our attention to the space
.
We define the two
polynomials
They
are relatively prime, non-zero, and their product
is, by definition, an annihilating polynomial on the space
.
Therefore,
and
We
then decompose the spaces
,
...,
with the same technique, until we
obtain
Thus, the space
can be written as a direct sum of invariant spaces of the
form
which
contain vectors
such
that
These
vectors are called generalized eigenvectors and the spaces
are called generalized eigenspaces.
In the special case in which
,
then
that
is,
is an eigenvector of
associated to the eigenvalue
and
is the eigenspace of
.
The Primary Decomposition Theorem provides the key to understanding the information provided by the minimal polynomial.
It is natural to make a comparison with the characteristic
polynomialwhich
reveals the eigenvalues
and their algebraic multiplicities
,
but contains no information on the eigenspaces.
The minimal
polynomialalso
reveals the eigenvalues, but not their algebraic multiplicities; however, it
provides an additional piece of information: by looking at the exponents
,
we can understand whether the eigenspaces are "large enough" to contain a
basis of eigenvectors for the space
.
As explained above and in the next section, if
,
then the space
can be written as a direct sum of eigenspaces, which implies that the
eigenvectors of
are sufficient to form a basis for
.
On the contrary, if an exponent
is greater than
,
then the eigenspace associated to
is not "large enough" and we are not able to find a basis of eigenvectors for
(the eigenvalue
is defective). The remedy is to resort to "larger" generalized eigenspaces.
What do we mean by larger? In the lecture on
matrix powers, we have learned
that the larger the power of a matrix is, the larger its null space is.
Therefore, if an
eigenspace
does
not contain enough linearly independent vectors (to be used in the
construction of a basis for
),
then we look at larger powers
(
)
of the matrix
,
so as to obtain larger null spaces
called
generalized eigenspaces.
Thus, the minimal polynomial reveals by how much we need to enlarge the
(generalized) eigenspaces in order to be able to form a basis for the space
.
Example
Let
be a
matrix. Suppose that the characteristic polynomial of
is
and
the minimal polynomial of
is
The
eigenvalue
has algebraic multiplicity equal to
(inferred from the characteristic polynomial), it is not defective (inferred
from the exponent equal to
in the minimal polynomial) and, as a consequence, its geometric multiplicity
is equal to
.
The eigenvalue
has algebraic multiplicity equal to
(inferred from the characteristic polynomial), it is defective (inferred from
the exponent equal to
in the minimal polynomial) and, as a consequence, its geometric multiplicity
must be less than
(hence
).
Remember that a matrix is diagonalizable if and only if
it is similar to a diagonal matrix;
its eigenvalues are not defective (their algebraic and geometric multiplicities coincide);
there exists a
basis of
eigenvectors for
.
Thanks to the Primary Decomposition Theorem, we can provide another criterion for the diagonalizability of a matrix.
Proposition
Let
be a
matrix. Let
be the minimal polynomial of
:
where
are the distinct
eigenvalues of
.
Then,
is diagonalizable if and only if
Let us prove the "if" part, starting from
the assumption that
for every
.
Let
be the space of
vectors.
Then,
In
other words,
is the direct sum of the eigenspaces of
.
Pick any vector
.
Then, we can
write
where
belongs to the eigenspace
for each
.
We can choose a basis
for each eigenspace
and form the
union
which
is a set of linearly
independent vectors and a basis for
because
is a direct sum. The vector
can be re-written in terms of the basis
by writing each
in terms of the basis
.
Since this is true for an arbitrary vector
,
we have that
is a basis of eigenvectors of
for the space
.
Hence,
is diagonalizable. Let us now prove the "only if" part, starting from the
assumption that
is diagonalizable. As proved in the
lecture on the minimal
polynomial,
for every
(otherwise
is not an annihilating polynomial). We are going to prove that, when
is diagonalizable,
becomes an annihilating polynomial by setting
(hence it is the minimal polynomial because any other annihilating polynomial
has higher degree). Since
is similar to a diagonal matrix
(having the eigenvalues
on its main diagonal), we need to check whether
is an annihilating polynomial of
(two similar matrices have the same annihilating polynomials).
But
because
all the diagonal elements of
are products involving factors of the form
,
at least one of which is such that
.
This proves that
is the minimal polynomial of
and has the desired property
(
).
Below you can find some exercises with explained solutions.
Let
be a
matrix. Suppose that the minimal polynomial of
is
Is
diagonalizable?
The matrix
is not diagonalizable because the exponent of the linear factor
is greater than
.
Let
be a
matrix. Suppose that the characteristic polynomial of
is
and
the minimal polynomial of
is
What
is the geometric multiplicity of the eigenvalue
?
From the characteristic polynomial, we
infer that the algebraic multiplicity of
is
.
From the minimal polynomial, we infer that it is not defective. Hence, its
geometric multiplicity equals
.
Let
be the space of
vectors. Let
be a
matrix. Let
be the characteristic polynomial of
:
where
are the distinct eigenvalues of
and
are their respective algebraic multiplicities. Can you modify the Primary
Decomposition Theorem so as to write
as a sum of null spaces by using the characteristic polynomial instead of the
minimal polynomial?
The recursive decomposition used in the
proof of the Primary Decomposition Theorem does not depend on any specific
property possessed by the minimal polynomial and not by the characteristic
one. For example, we can start the decomposition with
They
are relatively prime, non-zero, and their product is an annihilating
polynomial since it is equal to the characteristic polynomial.
Therefore,
We
can then decompose
and keep decomposing the null spaces until we
obtain
Please cite as:
Taboga, Marco (2021). "Primary decomposition theorem", Lectures on matrix algebra. https://www.statlect.com/matrix-algebra/primary-decomposition-theorem.
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