Download Introduction Initializations A Matrix and Its Jordan Form

Classroom Tips and Techniques: Jordan Normal Form of a Matrix
Robert J. Lopez
Emeritus Professor of Mathematics and Maple Fellow
Maplesoft
Introduction
A square matrix is similar either to a diagonal matrix or to a Jordan matrix, that is, a diagonal
matrix with a mix of 1s and 0s on the superdiagonal. The columns of the transition matrix for a
diagonalizable matrix are the eigenvectors of the matrix. The nondiagonalizable square matrix
doesn't have "enough" linearly independent eigenvectors and is similar to its Jordan matrix via a
transition matrix composed of eigenvectors and generalized eigenvectors.
Selecting the "correct" eigenvectors and finding the generalized eigenvectors require
considerable care. The theory underlying the structure supporting the Jordan form is some of
the most abstract mathematics encountered in the first linear algebra course. In fact, many
leading undergraduate texts today simply omit this discussion.
This month's article explores computations by which the matrix of transition to Jordan form can
be constructed, and in so doing, illustrates some of the underlying theory.
Initializations
A Matrix and Its Jordan Form
Table 1 displays the 7×7 matrix , its Jordan normal form , and the transition matrix for the
similarity transform
. The Jordan matrix is a block-diagonal matrix with four distinct
blocks of orders 2×2, 3×3, 1×1, 1×1. The eigenvalues of appear on the main diagonal of ;
the eigenvalue 2 has algebraic multiplicity 7 but geometric multiplicity 4, the same number as
the number of blocks in . Each block accounts for one of the four eigenvectors.
To determine the columns of , it is first necessary to know the sizes of the blocks in . Thus,
something must be learned about the structure of before one can find the transition matrix
that converts
Table 1
to .
The matrix , its transition matrix , and its Jordan normal form
Eigenvectors and Generalized Eigenvectors
An eigenvector of a matrix is a vector that remains invariant under multiplication by . In
particular, it is a vector that satisfies the equation
, for a specific value of the constant
called the eigenvalue. Clearly, if
is an eigenpair for , then so is
, for any scalar
multiple . Thus, an eigenvector is really an "eigendirection," that is, a direction that remains
unchanged under multiplication by .
If
is a transition matrix by which is similar to its Jordan form , then we have
, or
. The eigenvectors and generalized eigenvectors of are the columns of . To see
what does to these columns, it is sufficient to compare to the product , as provided in
Table 2.
=
Table 2
The matrices
and
Careful inspection of the matrices in Table 2 would reveal that if
, then
Columns 1, 3, 6, and 7 of the transition matrix are eigenvectors; columns 2, 3, and 5 are
generalized eigenvectors. Table 3 states this in terms of the matrix .
Table 3
Eigenvectors and generalized eigenvectors of
The eigenspace is spanned by the four eigenvectors
. The generalized eigenvectors
are "chained" to the eigenvectors. There are chains of lengths 2, 3, 1, and 1. At the
base of each chain is an eigenvector. Any remaining members of a chain are generalized
eigenvectors.
To determine the structure of the Jordan form , one must know the number of chains, and the
length of each chain.
Determining the Structure of
Table 4 defines the matrix
identity matrix.
, with the eigenvalue
=
Table 4
The matrix
set equal to 2, and
being the
be the algebraic multiplicity of the eigenvalue 2, and
For the × = 7×7 matrix , let
determine the integer
. Let be the smallest integer for which the rank of
equals . The integer is called the index of . Table 5 shows that
for the matrix
in Table 4.
=
=
=
Table 5
The index of
is 3
The index of can also be found from the minimal polynomial for . The Cayley-Hamilton
theorem states that
, the characteristic polynomial of , annihilates (i.e.,
is the zero matrix). The minimal polynomial is the lowest-degree polynomial
for
which
.
The characteristic and minimal polynomials for
are given in Table 6.
=
=
Table 6
Characteristic and minimal polynomials for
In the minimal polynomial, the degree of the factor containing a particular eigenvalue
the index of the matrix
. The index of
gives
is the length of the longest chain.
Before we can make more progress in deconstructing the structure of , we need the following
definition.
The rank of x, a member in a chain of generalized eigenvectors, is the
smallest integer for which
. Thus,
, but
.
Definition 1
Rank of a generalized eigenvector
In essence, the chain-member x has rank if x is in the null space of , not not in the null space
of
. Assigning a rank to a member of a chain of generalized eigenvectors is simply a way
of recording in which null space that vector falls.
Finding the lengths of chains shorter than the longest requires the numbers
Table 7 provides the values of
, for the matrix
in Table 1. Since
, the rank of
is 7.
=
=
=
Table 7
Determination of
The data in Table 7 indicate that there are four vectors of rank 1 (the four eigenvectors); two
vectors of rank 2, and 1 vector of rank 3. These vectors are arranged in four chains, depicted in
Figure 1.
Figure 1
Overlapping subspaces in
induced by the matrix
The vectors
and
are eigenvectors, and no chain of generalized eigenvectors emanates from
is an eigenvector, and it is the start of a chain of
either of these two vectors. The vector
length 2. Thus, the vector
is the generalized eigenvector at the end of this chain of length 2.
The vector
is an eigenvector, the start of the chain of length 3. The generalized eigenvector
is the second member of this chain, and the generalized eigenvector
is the end of the
longest chain.
is the null space of the matrix . The subspace
is the null space of ,
The eigenspace
but
, that is, the eigenspace
is a subspace of . Likewise,
so that
is a
subspace of . In this example,
, the whole space, so we have
.
Table 3, therefore, can be expressed by the equations in Table 8.
Table 8
Generalized eigenvector chains and the matrix
Finding the Generalized Eigenvectors
The chain-length for the generalized eigenvectors tells us from which
to select the end of the
chain. As can be seen from Table 8, if the generalized eigenvector at the end of the chain is
known, the complete chain can be computed by successively multiplying by . Therefore, we
first obtain bases for the null spaces .
>
>
is spanned by the four eigenvectors in
. The enveloping space
is
The eigenspace
spanned by the six basis vectors in
. We should have anticipated the seven vectors in
since
; hence, the most general vector in
will be
>
>
The Chain of Length 3
The chain of length 3 "ends" with a generalized eigenvector of rank 3. This vector lies in
that
>
, but
must not be the zero vector. Clearly,
, as we see from
so
>
We must insure that
compute
for
to be a generalize eigenvector of rank 3. Consequently, we
>
>
The vector
will be nonzero provided
>
>
Assuming this to be the case, we take
, that is
>
>
as the chain member in
, and
as the member in the eigenspace
. Provided the
condition in (1) prevails, we thus have the length-3 chain
.
An alternate way of looking at the condition
is to observe that must lie in
but not
in . This means the component of X that is in the orthogonal complement of
must not be
zero. So, to project X onto , we form the projection matrix
matrix whose columns are the vectors that span .
>
>
The component of X orthogonal to
>
is then
, where
is a
>
generates seven homogeneous equations, from which
The condition
by giving it the value
>
>
The vector
is nonzero provided
>
>
does not hold. The conditions are the same!
can be eliminated
The Chain of Length 2
The chain of length 2 "ends" at a generalized eigenvector in
that is distinct from the vector
. To find a vector in
that is distinct from , consider as candidates the six basis vectors in
. We test the rank of each matrix whose first column is and whose second column is one
of these basis vectors. If the rank of any of these matrices is 2, then the corresponding basis
vector is a candidate for the generalized eigenvector of rank 2 that ends the chain of length 2.
The relevant calculations are then
>
>
The ranks of the six possible matrices are all 2; any of the six basis vectors in
selected. The anchor of this chain is then the eigenvector
>
>
where the vector selected from
>
is
can be
>
The Transition Matrix
We now have five of the seven columns of the transition matrix. The remaining two columns
are eigenvectors taken from the subspace . There are
>
>
ways to pick two different vectors from a set of four. The explicit combinations are
>
>
A brute-force method for picking two that flesh out a linearly-independent set of seven vectors
for the columns of the transition matrix simply tests the rank of each possible transition matrix
that can be constructed from the five vectors we already have. The resulting transition matrix
is the first such matrix whose rank is 7.
>
>
rather than
To verify that the transition matrix puts into its Jordan form, we show
. The matrix is difficult to invert because it contains many indeterminates; the
following test that shows
is the zero matrix, is therefore more efficient.
>
>
A Brute-Force Calculation of the Transition Matrix
There is a certain amount of flexibility in the choices of the generalized eigenvectors in the
chains whose members make up the columns of the transition matrix. As an experiment, let's
start with the transition matrix
>
>
all of whose 7×7 = 49 members are indeterminate. By brute-force, set up and solve the
equations inherent in
. The equations are
>
>
The solution is
>
>
That brings the transition matrix
>
>
containing the indeterminates
to the form
>
>
whose number is
>
>
That represents a considerable amount of choice as one selects generalized eigenvectors. How
much flexibility is left in choosing these 21 parameters? Let's randomly assign the following
values to these 21 unknowns via
>
>
to these 21 unknowns. The resulting transition matrix will then be
>
>
whose rank we check with
>
>
and which we test with
>
>
If the rank of
is not 7, a different list
would need to be generated.
>
From Eigenvector to Generalized Eigenvector
An even more interesting question is whether the process of building chains of generalized
eigenvectors can be reversed. In other words, can we start with an eigenvector E and march up
the chain by solving equations of the form
for a generalized eigenvector u? Well, the
answer is a qualified yes, and the relevant calculations, provided below, show how this might be
done.
For the equation
to have a solution, the vector E has to be in the column space of .
Thus, E must be in the null space of because it is an eigenvector, and it has to be in the
column space of in order to support a chain of generalized eigenvectors. To find a basis for
the subspace of
that is common to the column space of , we first find
>
>
a basis for the column space of , then use
>
>
to find a basis for the intersection of
with the column space of . Thus, the most general
eigenvector that can support a chain of generalized eigenvectors is
>
>
The most general solution of the equation
is
>
>
However, the chain can't be advanced to a vector in
because U is not in the column space of
. The free parameters in U show that it contains components from the null space of . Even
if these components were not there, U still would not lie in the column space of , as we can
verify by showing the component of U orthogonal to the column space of is nonzero. The
matrix of projection is
>
>
and the component of U orthogonal to this space is
>
>
In general, this is not the zero vector, even if the free parameters
vanish. Another way to see that U is not in the column space of
for .
, were all to
is to try to solve the equation
>
Error, (in LinearAlgebra:-LA_Main:-LinearSolve) inconsistent
system
>
The system is inconsistent precisely because U is not in the column space of .
An alternate solution of the equation
obtained in Maple as
is obtained with
, the pseudo-inverse of ,
>
>
If were invertible, we would have
. Instead, we have
, the least-squares
solution of minimal length. In effect, the vector U is projected onto the column space of , and
the inverse image of minimal length is found. Thus, use of the pseudo-inverse allows us to
ignore components of U not in the column space of .
The Maple version of
>
is
>
That
is in
is verified by the vanishing of
, as we see from
>
>
However,
is not the end of a length-3 chain that starts with the vector . The chain still
, as
.
needs to be generated back down from
The length-2 chain can also be generated with the pseudo-inverse by starting from , computing
as the end of the chain in , and backing down to the eigenvector
in .
These calculations lead to a transition matrix constructed by the same devices used earlier.
>
>
As we also showed earlier, verification that
form consists in the vanishing of the matrix
is a transition matrix that puts
.
into its Jordan
>
>
Generating a Matrix with a Given Jordan Form
Finally, we address the question of generating a matrix with a known Jordan form . This is
done by defining the matrix as
, where is the given Jordan form, and is a transition
matrix whose columns will be eigenvectors and generalized eigenvectors.
The Jordan matrix is built in Maple with the JordanBlockMatrix command whose use for
generating the Jordan matrix in Table 1 is illustrated below.
>
>
The size of a sub-block and its associated eigenvalue are given by a list, and a list of such lists
determines the full matrix.
The transition matrix can be any nonsingular comformable matrix. However, if the
determinant of is not
, the matrix
will most likely contain fractions, making the
calculations more tedious. Hence, we select from randomly generated matrices with
determinant
and integer entries
or 2. This is done with code such as the following.
>
>
A suitable matrix is recovered from the index printed in front of the matrix. For example, if
the last matrix is chosen for , it can be extracted via
>
>
All that would remain is the construction of
.
The basis that puts into its Jordan form is not unique. Hence, the transition matrix is not
unique. Table 1 shows the matrix that was used to construct for this exploration. The
transition matrices we generated with the various bases we constructed were not necessarily
the same as this .
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