6Endomorphisms
IB Linear Algebra
6.1 Invariants
Definition.
If
V
is a (finite-dimensional) vector space over
F
. An endomorphism
of
V
is a linear map
α
:
V → V
. We write
End
(
V
) for the
F
-vector space of all
such linear maps, and I for the identity map V → V .
When we think about matrices representing an endomorphism of
V
, we’ll
use the same basis for the domain and the range. We are going to study some
properties of these endomorphisms that are not dependent on the basis we pick,
known as invariants.
Lemma.
Suppose (
e
1
, ··· , e
n
) and (
f
1
, ··· , f
n
) are bases for
V
and
α ∈ End
(
V
).
If
A
represents
α
with respect to (
e
1
, ··· , e
n
) and
B
represents
α
with respect
to (f
1
, ··· , f
n
), then
B = P
−1
AP,
where P is given by
f
i
=
n
X
j=1
P
ji
e
j
.
Proof.
This is merely a special case of an earlier more general result for arbitrary
maps and spaces.
Definition
(Similar matrices)
.
We say matrices
A
and
B
are similar or conjugate
if there is some P invertible such that B = P
−1
AP .
Recall that
GL
n
(
F
), the group of invertible
n ×n
matrices.
GL
n
(
F
) acts on
Mat
n
(F) by conjugation:
(P, A) 7→ P AP
−1
.
We are conjugating it this way so that the associativity axiom holds (otherwise
we get a right action instead of a left action). Then
A
and
B
are similar iff they
are in the same orbit. Since orbits always partition the set, this is an equivalence
relation.
Our main goal is to classify the orbits, i.e. find a “nice” representative for
each orbit.
Our initial strategy is to identify basis-independent invariants for endomor-
phisms. For example, we will show that the rank, trace, determinant and
characteristic polynomial are all such invariants.
Recall that the trace of a matrix
A ∈ Mat
n
(
F
) is the sum of the diagonal
elements:
Definition (Trace). The trace of a matrix of A ∈ Mat
n
(F) is defined by
tr A =
n
X
i=1
A
ii
.
We want to show that the trace is an invariant. In fact, we will show a
stronger statement (as well as the corresponding statement for determinants):
Lemma.
(i) If A ∈ Mat
m,n
(F) and B ∈ Mat
n,m
(F), then
tr AB = tr BA.
(ii) If A, B ∈ Mat
n
(F) are similar, then tr A = tr B.
(iii) If A, B ∈ Mat
n
(F) are similar, then det A = det B.
Proof.
(i) We have
tr AB =
m
X
i=1
(AB)
ii
=
m
X
i=1
n
X
j=1
A
ij
B
ji
=
n
X
j=1
m
X
i=1
B
ji
A
ij
= tr BA.
(ii) Suppose B = P
−1
AP . Then we have
tr B = tr(P
−1
(AP )) = tr((AP )P
−1
) = tr A.
(iii) We have
det(P
−1
AP ) = det P
−1
det A det P = (det P )
−1
det A det P = det A.
This allows us to define the trace and determinant of an endomorphism.
Definition
(Trace and determinant of endomorphism)
.
Let
α ∈ End
(
V
), and
A
be a matrix representing
α
under any basis. Then the trace of
α
is
tr α
=
tr A
,
and the determinant is det α = det A.
The lemma tells us that the determinant and trace are well-defined. We
can also define the determinant without reference to a basis, by defining more
general volume forms and define the determinant as a scaling factor.
The trace is slightly more tricky to define without basis, but in IB Analysis
II example sheet 4, you will find that it is the directional derivative of the
determinant at the origin.
To talk about the characteristic polynomial, we need to know what eigenvalues
are.
Definition
(Eigenvalue and eigenvector)
.
Let
α ∈ End
(
V
). Then
λ ∈ F
is an
eigenvalue (or E-value) if there is some v ∈ V \ {0} such that αv = λv.
v is an eigenvector if α(v) = λv for some λ ∈ F.
When
λ ∈ F
, the
λ
-eigenspace, written
E
α
(
λ
) or
E
(
λ
) is the subspace of
V
containing all the λ-eigenvectors, i.e.
E
α
(λ) = ker(λι −α).
where ι is the identity function.
Definition
(Characteristic polynomial)
.
The characteristic polynomial of
α
is
defined by
χ
α
(t) = det(tι − α).
You might be used to the definition
χ
α
(
t
) =
det
(
α − tι
) instead. These two
definitions are obviously equivalent up to a factor of
−
1, but this definition
has an advantage that
χ
α
(
t
) is always monic, i.e. the leading coefficient is 1.
However, when doing computations in reality, we often use
det
(
α − tι
) instead,
since it is easier to negate tι than α.
We know that
λ
is an eigenvalue of
α
iff
n
(
α −λι
)
>
0 iff
r
(
α −λι
)
< dim V
iff
χ
α
(
λ
) =
det
(
λι − α
) = 0. So the eigenvalues are precisely the roots of the
characteristic polynomial.
If A ∈ Mat
n
(F), we can define χ
A
(t) = det(tI − A).
Lemma.
If
A
and
B
are similar, then they have the same characteristic polyno-
mial.
Proof.
det(tI − P
−1
AP ) = det(P
−1
(tI − A)P ) = det(tI − A).
Lemma. Let α ∈ End(V ) and λ
1
, ··· , λ
k
distinct eigenvalues of α. Then
E(λ
1
) + ··· + E(λ
k
) =
k
M
i=1
E(λ
i
)
is a direct sum.
Proof. Suppose
k
X
i=1
x
i
=
k
X
i=1
y
i
,
with
x
i
, y
i
∈ E
(
λ
i
). We want to show that they are equal. We are going to find
some clever map that tells us what
x
i
and
y
i
are. Consider
β
j
∈ End
(
V
) defined
by
β
j
=
Y
r6=j
(α − λ
r
ι).
Then
β
j
k
X
i=1
x
i
!
=
k
X
i=1
Y
r6=j
(α − λ
r
ι)(x
i
)
=
k
X
i=1
Y
r6=j
(λ
i
− λ
r
)(x
i
).
Each summand is zero, unless i 6= j. So this is equal to
β
j
k
X
i=1
x
i
!
=
Y
r6=j
(λ
j
− λ
r
)(x
j
).
Similarly, we obtain
β
j
k
X
i=1
y
i
!
=
Y
r6=j
(λ
j
− λ
r
)(y
j
).
Since we know that
P
x
i
=
P
y
i
, we must have
Y
r6=j
(λ
j
− λ
r
)x
j
=
Y
r6=j
(λ
j
− λ
r
)y
j
.
Since we know that
Q
r6=j
(λ
r
− λ
j
) 6= 0, we must have x
i
= y
i
for all i.
So each expression for
P
x
i
is unique.
The proof shows that any set of non-zero eigenvectors with distinct eigenvalues
is linearly independent.
Definition
(Diagonalizable)
.
We say
α ∈ End
(
V
) is diagonalizable if there is
some basis for
V
such that
α
is represented by a diagonal matrix, i.e. all terms
not on the diagonal are zero.
These are in some sense the nice matrices we like to work with.
Theorem.
Let
α ∈ End
(
V
) and
λ
1
, ··· , λ
k
be distinct eigenvalues of
α
. Write
E
i
for E(λ
i
). Then the following are equivalent:
(i) α is diagonalizable.
(ii) V has a basis of eigenvectors for α.
(iii) V =
L
k
i=1
E
i
.
(iv) dim V =
P
k
i=1
dim E
i
.
Proof.
– (i) ⇔ (ii): Suppose (e
1
, ··· , e
n
) is a basis for V . Then
α(e
i
) = A
ji
e
j
,
where
A
represents
α
. Then
A
is diagonal iff each
e
i
is an eigenvector. So
done
–
(ii)
⇔
(iii): It is clear that (ii) is true iff
P
E
i
=
V
, but we know that this
must be a direct sum. So done.
–
(iii)
⇔
(iv): This follows from example sheet 1 Q10, which says that
V
=
L
k
i=1
E
i
iff the bases for
E
i
are disjoint and their union is a basis of
V .