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### Finding the Jordan Canonical Form

```
Date: 04/29/98 at 00:56:41
From: Lisa Russo
Subject: Jordan Canonical Form

I need help in finding the Jordan Canonical Form of a matrix. The
proof that my book illustrates is tedious and hard to follow.

For example, find the Jordan Canonical Form of:

A = [(3 1 2),(0 3 0),(0 0 3)]

I approached the problem by finding [lambda I-A].  This gives me the
characteristic polynomial, and the eigenvalues.  I am stuck after
that point.
```

```
Date: 04/29/98 at 11:21:32
From: Doctor Rob
Subject: Re: Jordan Canonical Form

First you need to find the elementary divisors of A. They will be
divisors of the minimal polynomial. One way to find them is to start
with the matrix t*I-A and reduce it using polynomial elementary row
and column operations to Smith Normal Form, that is, diagonal, with
each diagonal entry dividing the next:

[(t-3 -1 -2),(0 t-3 0),(0 0 t-3)]
-->  [(-1 t-3 -2),(t-3 0 0),(0 0 t-3)]       (swap columns 1 and 2)
-->  [(1 t-3 -2),(3-t 0 0),(0 0 t-3)]        (multiply column 1 by -1)
-->  [(1 0 -2),(3-t (t-3)^2 0),(0 0 t-3)]    (add 3-t times column 1
to col 2)
-->  [(1 0 0),(3-t (t-3)^2 6-2*t),(0 0 t-3)] (add 2 times col 1 to
col 3)
-->  [(1 0 0),(0 (t-3)^2 6-2*t),(0 0 t-3)]   (add t-3 times row 1 to
row 2)
(Here the first row and column are done.)

-->  [(1 0 0),(0 (t-3)^2 0),(0 0 t-3)] (add 2 times row 3 to row 2)
-->  [(1 0 0),(0 0 t-3),(0 (t-3)^2 0)] (swap rows 2 and 3)
-->  [(1 0 0),(0 t-3 0),(0 0 (t-3)^2)] (swap columns 2 and 3)

The invariant factors of t*I-A are the polynomials on the diagonal:
1, t-3, and (t-3)^2. The elementary divisors of t*I-A are the powers
of irreducible polynomials dividing the invariant factors, so they are
(t-3)^2 and (t-3). Consider each in turn. The hypercompanion matrix of
one elementary divisor is [(3)], and the hypercompanion matrix of the
other elementary divisor is [(3 1),(0 3)], so the Jordan canonical
form is J = [(3 0 0),(0 3 1),(0 0 3)], the block-diagonal matrix
obtained by putting each of these hypercompanion matrices on the
diagonal of a matrix which is zero elsewhere.

There remains to figure out what the matrix R is such that:

J = R^(-1)*A*R.

You can start this process by finding a vector v[1] such that:

C[1] = {v[1], A*v[1], A^2*v[1], ..., A^(r[1]-1)*v[1]}

are all independent and r[1] is maximal with this property. r[1] will
actually be the degree of the minimal polynomial. Then find v[2] such
that v[2] is independent of all the vectors in C[1], and:

C[2] = {v[2], A*v[2], ... , A^(r[2]-1)*v[2]}

are all linearly independent of each other and the vectors in C[1],
and r[2] is maximal with this property. Continue until you have n
vectors in your sets C[1], C[2], ... .  Then the matrix S whose
columns are the vectors in ... , C[2], C[1] will be such that
S^(-1)*A*S is in Rational Canonical Form.

In your case, you can choose v[1] = (0 1 0)', for example, and
A*v[1] = (1 3 0)', A^2*v[1] = (6 9 0)' = 6*A*v[1] - 9*v[1],
so r[1] = 2, (A-3*I)^2*v[1] = 0, and v[1] belongs to the elementary
divisor (t-3)^2.

Then you can choose w = (0 0 1)', and then:

A*w = (2 0 3) = 2*A*v[1] - 6*v[1] + 3*w

so r[2] = 1. Then the vector v[2] = w - 2*v[1] satisfies
A*v[2] = 3*v[2], so v[2] belongs to the invariant factor t-3. The
matrix S then should be:

S = [(0 0 1),(-2 1 3),(1 0 0)]

S^(-1) = [(0 0 1),(-3 1 2),(1 0 0)]

S^(-1)*A*S = [(3 0 0),(0 0 -9),(0 1 6)]

To convert this into Jordan Form, the first basis element can remain
the same, but the last two must change. One transformation that
works is:

T = [(1 0 0),(0 -3 1),(0 1 0)]

R = S*T
= [(0 1 0),(-2 0 1),(1 0 0)]

R^(-1) = [(0 0 1),(1 0 0),(0 1 2)]

R^(-1)*A*R = [(3 0 0),(0 3 1),(0 0 3)] = J

-Doctor Rob,  The Math Forum
Check out our web site! http://mathforum.org/dr.math/
```
Associated Topics:
College Linear Algebra

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