Jordan

Jordan Canonical Form.
Kevin Buzzard
December 14, 2012
I didn’t prove uniqueness of Jordan Canonical Form in the lectures; here is how it goes.
Recall the notation: if λ ∈ C and n ∈ Z
≥1
, then J
n
(λ) is an n × n matrix which is upper-
triangular, has λ’s all down the main diagonal, has 1s all down the diagonal just above it, and
then zeros everywhere else.
In mathematical notation: if J
n
(λ) = (a
ij
)
1≤i,j≤n
then
a
ij
=
_
¸
_
¸
_
λ if i = j,
1 if j = i + 1,
0 otherwise.
In pictures:
J
n
(λ) =
_
_
_
_
_
_
_
_
_
λ 1 0 0 . . . 0 0 0
0 λ 1 0 . . . 0 0 0
0 0 λ 1 . . . 0 0 0
.
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.
.
0 0 0 0 . . . 0 λ 1
0 0 0 0 . . . 0 0 λ
_
_
_
_
_
_
_
_
_
Let’s say a matrix is in Jordan Canonical Form if it is of the form J
n
1
(λ
1
)⊕J
n
2
(λ
2
)⊕· · · ⊕J
n
r
(λ
r
)
for some choices of n
i
and λ
i
. Let me stress right now that the λ
i
do not have to be distinct here;
for example the identity matrix is J
1
(1) ⊕J
1
(1) ⊕· · · ⊕J
1
(1).
We proved in lectures:
Theorem (Existence of JCF). If A is any matrix with complex coefficients, then A is similar to
a matrix which is in Jordan Canonical Form.
We say that this matrix is the Jordan Canonical Form of A.
What I did not prove, because I ran out of time, was the uniqueness statement that makes the
theorem even more powerful:
Theorem (uniqueness of JCF). The Jordan Canonical Form of A is unique, up to changing the
order of the factors.
In other words, what I am saying is that if the two matrices
J
n
1
(λ
1
) ⊕J
n
2
(λ
2
) ⊕· · · ⊕J
n
r
(λ
r
)
and
J
m
1
(µ
1
) ⊕J
m
2
(µ
2
) ⊕· · · ⊕J
m
s
(µ
s
)
are similar, then r = s and, possibly after re-arranging the J
m
i
(µ
i
), we have n
i
= m
i
and λ
i
= µ
i
for all i.
Note that conversely, re-arranging the order of the factors in a block direct sum changes a
matrix to a similar matrix: if A is an m× m matrix and B is an n × n matrix, then A ⊕ B is
similar to B ⊕ A because you can conjugate by the change of basis matrix which corresponds to
1
changing the order of the basis (moving the first m entries from the front to the back) (checking
the details of this is a worthwhile exercise).
Note that the two theorems together are equivalent to Theorem 11.3 of the course. Note also
that any proof I did not have time to give in the course will not be examinable, but of course you
will be expected to know and use statements of theorems that I stated in the course!
Proof. Ok so let’s prove the uniqueness theorem right now. Here’s how we’re going to do it. Let’s
say we have a matrix that is actually in Jordan Canonical Form, so
A = J
n
1
(λ
1
) ⊕J
n
2
(λ
2
) ⊕· · · ⊕J
n
r
(λ
r
).
The strategy is to show that I can reconstruct the n
i
and the λ
i
from A using only invariants of A,
that is, things that do not change under similarity.
In fact the char poly of A is easily checked to be (x−λ
1
)
n
1
(x−λ
2
)
n
2
· · · , by Proposition 11.1(1)
and Proposition 11.2(1), so we know that the λ
i
are all eigenvalues of A.
Now let λ = λ
1
be an eigenvalue of A; replacing A by a similar matrix if necessary we can
assume that the J
n
i
(λ
i
) with λ
i
= λ all occur at the start:
A = J
n
1
(λ) ⊕J
n
2
(λ) ⊕· · · ⊕J
n
t
(λ) ⊕J
n
t+1
(λ
t+1
) ⊕· · ·
with all the λ
j
, j > t, not equal to λ.
We can also re-arrange these first t terms so that n
1
≤ n
2
≤ . . . ≤ n
t
. Finally, we can group
terms together if the n
i
are the same: if we use M
d
as a shorthand for M⊕M⊕· · · ⊕M (d times)
then we can assume
A = J
1
(λ)
c
1
⊕J
2
(λ)
c
2
⊕· · · ⊕J
k
(λ)
c
k
⊕J
n
t+1
(λ
t+1
) ⊕· · · .
So here c
1
∈ Z
≥0
is the number of i ≤ t with n
i
= 1, c
2
is the number with n
i
= 2 and so on. It’s
OK if some of the c
i
are zero, but we can and will assume that the last one isn’t, so c
k
= 0.
If we can prove that we can figure out what the c
i
are using only invariants of A, then we’re
going to be done, because the c
i
are controlling exactly which Jordan blocks with eigenvalue λ
appear in the Jordan Decomposition of A.
But here’s how to do this. Set B = A−λI. Then
B = J
1
(0)
c
1
⊕J
2
(0)
c
2
⊕· · · ⊕J
k
(0)
c
k
⊕J
n
t+1
(λ
t+1
−λ) ⊕· · · .
Note that all those last terms in the · · · are Jordan blocks with eigenvalue λ
i
− λ = 0, because
we put all the λ blocks at the beginning. In particular all of those Jordan blocks have non-zero
determinant.
Now we ask the following question: what is the nullity of B? By Proposition 11.2(3) this is
just the sum of the nullities of each block. But the nullity of J
n
(λ) is 1 if λ = 0 and zero otherwise,
and hence
nullity(B) = c
1
+ c
2
+· · · + c
k
and none of the terms after the · · · contribute.
What about the nullity of B
2
? Well, the nullity of J
n
(0)
2
is usually equal to 2 (with the kernel
being e
1
and e
2
), except that if n = 1 this isn’t right: J
1
(0) = (0) and its square has nullity 1.
Hence
nullity(B
2
) = c
1
+ 2c
2
+ 2c
3
· · · + 2c
k
.
Now you can probably see the trick:
c
1
= 2 nullity(B) −nullity(B
2
)
= 2 nullity(A−λI) −nullity((A−λI)
2
)
In particular, c
1
can be computed intrinsically from A using only invariants, so it will be the same
for any Jordan normal form of A.
2
This trick of course generalises: for example
nullity(B
3
) = c
1
+ 2c
2
+ 3c
3
+ 3c
4
+· · · + 3c
k
and hence
2c
1
+ c
2
= 3 nullity(B) −nullity(B
3
)
and we know c
1
, hence we can work out c
2
.
Continuing in this way, it is not difficult to check that we can compute all of the c
i
in this
manner; we have to compute the nullity of (A −λI)
d
for 1 ≤ d ≤ k to get all the information we
need; we don’t know what k is in practice, but we do know that c
1
+ 2c
2
+ 3c
3
+· · · + kc
k
is the
number of times λ occurs as a root in the char poly of A, and hence the algebraic multiplicity a(λ)
of λ. Moreover, c
k
= 0, hence a(λ) ≥ kc
k
≥ k and we conclude that we will only have to compute
the nullities for d ≤ a(λ) to be sure we know all the c
i
.
In particular, the argument above shows that we can compute the sizes of all the Jordan blocks
with eigenvalue λ in a way that depends only on A up to similarity. Doing this for all λ shows
that the Jordan Canonical Form is unique!
As a consequence, we know that two matrices are similar if and only if they have the same JCF
(up to re-ordering the factors), and hence any data which uniquely determines the JCF determines
the matrix up to similarity. We have figured out the JCF in the proof above, using only invariants
of A, and if we look more carefully about precisely which invariants we used, we can deduce the
following:
Corollary. Let A and B be two n × n matrices with complex coefficients. If A and B do not
have the same characteristic polynomials, then they are definitely not similar. However if p(x) =
(x −λ
1
)
a
1
(x −λ
2
)
a
2
· · · is the char poly of both A and B, then A and B are similar if and only if
nullity((A−λ
i
I)
j
) = nullity((B −λ
i
I)
j
)
for all λ
i
and all 1 ≤ j ≤ a
i
.
Note that the nullity of (A − λ
i
I) is just the geometric multiplicity of the eigenvalue λ
i
, and
the nullity of (A−λ
i
I)
a
i
is the algebraic multiplicity; so the theorem above says that we need to
check slightly more than algebraic and geometric multiplicities, we need to check a slightly more
subtle set of invariants before we can be sure that two matrices are similar.
OK, that’s it, the lecture course finished 3 hours ago and I’m feeling a little like I have lost
something. Have a good Christmas, or whatever you are celebrating, and a happy new year!
Kevin
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