Naber G.L. The Geometry of Minkowski Spacetime. An Introduction to the Mathematics of the Special Theory of Relativity
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94 2 Skew-Symmetric Linear Transformations and Electromagnetic Fields
electromagnetic field only if the charges whose motion is to be governed by it
have negligible contribution to the ambient field. It must be possible to regard
the field as “given” and the charged particles as “test charges”. The much
more difficult question of the interactions between the given field and the
fields created by the moving charges will not be considered here (see [Par]).
W
e
argue
now that the form of the Lorentz Law (2.1.1) suggests that the
linear
tr
ansformations F must be of a particular type (“skew-symmetric”).
Indeed, rewriting (2.1.1) at each fixed point of M as
FU =
m
e
dU
dτ
and dotting both sides with U gives
FU · U =
m
e
dU
dτ
·U =
m
e
A · U =0
since a material particle’s world velocity and world acceleration are always
orthogonal (Exercise 1.4.12). Since any unit timelike vector u ∈Mis the
world velocity of some charged particle (construct one!) we find that, for any
such u, Fu ·u = 0. Linearity therefore implies that Fv · v = 0 for all timelike v.
Now, if u and v are timelike and future-directed, then u + v is also timelike
and so 0 = F (u + v) ·(u + v)=(Fu + Fv) ·(u + v)=Fu ·v + Fv ·u = Fu ·v +
u ·Fv .Thus,Fu ·v = −u · Fv.ButM has a basis of future-directed timelike
vectors so it follows that F must satisfy
Fx ·y = −x · Fy (2.1.2)
for all x and y in M. A linear transformation F : M→
Mwhic
h s
atisfies
(2.1.2) for all x and y in M is
said
to be skew-symmetric (with respect to
the Lorentz inner product on M).
At each fixed point in M we therefore elect to model an electromagnetic
field by a skew-symmetric linear transformation F whose job it is to assign
to the world velocity U of a charged particle passing through that point
the change in world momentum
dP
dτ
= eFU that the particle should expect
to experience due to the field. One would picture the electromagnetic field
in toto therefore as a smooth assignment of such a linear transformation
to each point in (some region of) M (although we shall find that nature
imposes a condition—Maxwell’s Equations—on the manner in which such an
assignment can be made). In the next four sections we carry out a general
investigation of skew-symmetric linear transformations on M andthenturn
to some physical applications in the last two sections.
2.2 Elementary Properties 95
2.2 Elementary Properties
Throughout this section F will represent a nonzero, skew-symmetric linear
transformation on M. The most obvious consequence of the definition (2.1.2)
o
f
skew-symmetr
y is that
Fx ·x = x · Fx = 0 (2.2.1)
for all x in M.If{e
a
}
4
a=1
is an arbitrary admissible basis for M and we write
Fe
b
= F
a
b
e
a
= F
1
b
e
1
+ F
2
b
e
2
+ F
3
b
e
3
+ F
4
b
e
4
,then(2.2.1) implies F
a
a
=0
for a =1, 2, 3, 4, i.e., the diagonal entries in the matrix of F are all zero. In
addition, for i, j =1, 2, 3,F
j
i
= −F
i
j
,whereasF
4
i
= F
i
4
.Thus,thematrix
of F relative to any admissible basis has the form
[F
a
b
]=
⎡
⎢
⎢
⎣
0 F
1
2
F
1
3
F
1
4
−F
1
2
0 F
2
3
F
2
4
−F
1
3
−F
2
3
0 F
3
4
F
1
4
F
2
4
F
3
4
0
⎤
⎥
⎥
⎦
. (2.2.2)
Observe that, due to the fact that the inner product on M is indefinite,
the matrix of a skew-symmetric linear transformation on M is not a skew-
symmetric matrix (in the “time” part).
In order to establish contact with the notation usually used in physics we
introduce, in each admissible basis {e
a
}, two 3-vectors
E
= E
1
e
1
+ E
2
e
2
+
E
3
e
3
and
B
= B
1
e
1
+ B
2
e
2
+ B
3
e
3
,whereE
1
= F
1
4
,E
2
= F
2
4
,E
3
=
F
3
4
,B
1
= F
2
3
,B
2
= −F
1
3
and B
3
= F
1
2
.Thus,(2.2.2) can be written
[F
a
b
]=
⎡
⎢
⎢
⎣
0 B
3
−B
2
E
1
−B
3
0 B
1
E
2
B
2
−B
1
0 E
3
E
1
E
2
E
3
0
⎤
⎥
⎥
⎦
. (2.2.3)
If F is thought of as describing an electromagnetic field at some point of
M,then
E
and
B
are regarded as the classical electric and magnetic field
3-vectors at that point as measured in {e
a
}.
We consider two simple examples which, in Section 2.4, we will show to be
fully and uniquely representative in the sense that for any skew-symmetric
F : M→Mthere exists a basis relative to which the matrix of F has one
of these forms, but no basis in which it has the other. First fix an admissible
basis {e
a
} and a positive real number α and define a linear transformation
F
N
on M whose matrix relative to this basis is
⎡
⎢
⎢
⎣
0 000
00α 0
0 −α 0 α
00α 0
⎤
⎥
⎥
⎦
.
96 2 Skew-Symmetric Linear Transformations and Electromagnetic Fields
Then F
N
is clearly skew-symmetric,
E
= αe
3
and
B
= αe
1
,soanobserverin
this frame measures electric and magnetic 3-vectors that are perpendicular
and have the same magnitude.
Next, fix an admissible basis {e
a
} and two non-negative real numbers δ
and and let F
R
: M→Mbe the linear transformation whose matrix
relative to {e
a
} is
⎡
⎢
⎢
⎣
0 δ 00
−δ 000
000
00 0
⎤
⎥
⎥
⎦
.
Again, F
R
is skew-symmetric. Moreover,
E
= e
3
and
B
= δe
3
so an ob-
server in this frame will measure electric and magnetic 3-vectors in the same
direction and of magnitude and δ respectively.
Relative to another admissible basis {ˆe
a
} all of the
ˆ
F
a
b
,
ˆ
E
i
and
ˆ
B
i
are
defined in the same way. Thus, if Λ is the Lorentz transformation associated
with the orthogonal transformation L that carries {e
a
} onto {ˆe
a
}, i.e., the
matrix of L
−1
relative to {ˆe
a
}, then the matrix of F relative to {ˆe
a
} is
Λ[F
a
b
]Λ
−1
, i.e.,
ˆ
F
a
b
=Λ
a
α
Λ
b
β
F
α
β
.
Exercise 2.2.1 With [F
a
b
]asin(2.2.3)andΛ=Λ(β)forsomeβ in (−1, 1),
show that
ˆ
E
1
= E
1
,
ˆ
E
2
= γ(E
2
− βB
3
),
ˆ
E
3
= γ(E
3
+ βB
2
),
ˆ
B
1
= B
1
,
ˆ
B
2
= γ(βE
3
+ B
2
),
ˆ
B
3
= −γ(βE
2
− B
3
).
(2.2.4)
Exercise 2.2.2 Show that, for F
N
, any other admissible observer mea-
sures electric and magnetic 3-vectors that are perpendicular and have the
same magnitude. Hint: This is clear if the Lorentz transformation Λ relating
the two frames is a rotation. Verify the statement for Λ(β) and appeal to
Theorem 1.3.5.
Exercise 2.2.3 Show that, for F
R
, another admissible observer will, in gen-
eral, not measure
E
and
B
in the same direction.
Of particular interest is the special case of (2.2.4) when either
B
or
E
is
zero (so that O observes either a purely electric or a purely magnetic field):
If
B
=
O
,then
ˆ
E
1
= E
1
,
ˆ
E
2
= γE
2
,
ˆ
E
3
= γE
3
,
ˆ
B
1
=0,
ˆ
B
2
= βγE
3
,
ˆ
B
3
= −βγE
2
.
(2.2.5)
If
E
=
O
we have
ˆ
E
1
=0,
ˆ
E
2
= −βγB
3
,
ˆ
E
3
= βγB
2
,
ˆ
B
1
= B
1
,
ˆ
B
2
= γB
2
,
ˆ
B
3
= γB
3
.
(2.2.6)
2.2 Elementary Properties 97
The essential feature of (2.2.5)and(2.2.6) is that “purely electric” and
“purely magnetic” are not relativistically meaningful notions since they are,
in general, not invariant under Lorentz transformations. How much of an
electromagnetic field is “electric” and how much “magnetic” depends on the
frame of reference from which it is being observed. This is the familiar phe-
nomemon of electromagnetic induction. For example, a charge deemed “at
rest” in one frame will give rise to a purely electric field in that frame, but,
viewed from another frame, will be “moving” and so will induce a nonzero
magnetic field as well.
Since
E
and
B
are spacelike one can, beginning with any admissible ba-
sis {e
1
,e
2
,e
3
,e
4
}, choose a right-handed orthonormal basis {ˆe
1
, ˆe
2
, ˆe
3
} for
Span{e
1
,e
2
,e
3
} such that
E
and
B
both lie in Span{ˆe
1
, ˆe
2
} (so that
ˆ
E
3
=
ˆ
B
3
= 0). Choosing a rotation
R
i
j
i,j=1,2,3
in this 3-dimensional Euclidean
space that accomplishes the change of coordinates ˆx
i
= R
i
j
x
j
,i=1, 2, 3,
the corresponding rotation [R
a
b
]
a,b=1,2,3,4
in L yields a new admissible coor-
dinate system in which the third components of
E
and
B
are zero. The gist
of all this is that one can, with little extra effort, work in a basis relative to
which the matrix of F has the form
⎡
⎢
⎢
⎣
00−B
2
E
1
00 B
1
E
2
B
2
−B
1
00
E
1
E
2
00
⎤
⎥
⎥
⎦
. (2.2.7)
Next we collect a few facts that will be of use in the remainder of the chap-
ter. We define the range and kernel (null space) of a linear transformation
T : M→Mby
rng T = {y ∈M: y = Tx for some x ∈M}
and
ker T = {x ∈M: Tx =0}.
Both rng T and ker T are obviously subspaces of M and, consequently, so
are their orthogonal complements (rng T )
⊥
and (ker T )
⊥
(Exercise 1.1.2).
Proposition 2.2.1 If F : M→Mis any nonzero, skew-symmetric linear
transformation on M,then
(a) ker F =(rngF )
⊥
,
(b) rng F =(kerF )
⊥
,
(c) dim(ker F ) is either 0 or 2 so dim(rng F ) is either 4 or 2, respectively.
Proof: (a) First let x ∈ (rng F )
⊥
.Thenx · Fy =0forally in M.Thus,
Fx · y =0forally in M. But the inner product on M is nondegenerate
so we must have Fx = 0, i.e., x ∈ ker F . Next suppose x ∈ ker F .Then
Fx = 0 implies Fx · y =0forally in M so x · Fy =0forally in M, i.e.,
x ∈ (rng F )
⊥
.
98 2 Skew-Symmetric Linear Transformations and Electromagnetic Fields
Exercise 2.2.4 Show that, for any subspace U of M, (U
⊥
)
⊥
= U and
conclude from (a) that (b) is true. Hint:UsethefactthatU and U
⊥
have
complementary dimensions, i.e., dim U +dimU
⊥
=dimM (see Theorem 16,
Chapter 4 of [La]).
(c)
W
ithout
loss of generality select a basis {e
a
} relative to which the matrix
of F has the form (2.2.7). Then, for any v ∈M, Fv has
co
mponents given by
⎡
⎢
⎢
⎣
00−B
2
E
1
00 B
1
E
2
B
2
−B
1
00
E
1
E
2
00
⎤
⎥
⎥
⎦
⎡
⎢
⎢
⎣
v
1
v
2
v
3
v
4
⎤
⎥
⎥
⎦
=
⎡
⎢
⎢
⎣
−B
2
v
3
+ E
1
v
4
B
1
v
3
+ E
2
v
4
B
2
v
1
− B
1
v
2
E
1
v
1
+ E
2
v
2
⎤
⎥
⎥
⎦
which is zero if and only if
.
−B
2
v
3
+ E
1
v
4
=0,
B
1
v
3
+ E
2
v
4
=0
and
.
B
2
v
1
− B
1
v
2
=0,
E
1
v
1
+ E
2
v
2
=0.
Notice that the determinant of the coefficient matrix in the first system is
−
E
·
B
and, in the second,
E
·
B
.If
E
·
B
= 0 both systems have only the
trivial solution so the kernel of F consists of 0 alone and dim(ker F ) = 0. Since
4=dim M =dim(kerF )+dim(rng F ), dim(rng F )=4.If
E
·
B
=0,each
system has nontrivial solutions, say, (v
3
,v
4
)=
v
3
0
,v
4
0
and (v
1
,v
2
)=
v
1
0
,v
2
0
.
Since F = 0, all of the nontrivial solutions to the first system are of the form
b
v
3
0
,v
4
0
,b∈ R, and, for the second, a(v
1
0
,v
2
0
),a∈ R.Thus,thekernelof
F is the set of
⎡
⎢
⎢
⎣
v
1
v
2
v
3
v
4
⎤
⎥
⎥
⎦
= a
⎡
⎢
⎢
⎣
v
1
0
v
2
0
0
0
⎤
⎥
⎥
⎦
+ b
⎡
⎢
⎢
⎣
0
0
v
3
0
v
4
0
⎤
⎥
⎥
⎦
,
so dim(ker F ) = 2 and therefore dim(rng F )=2.
Recall that a real number λ is an eigenvalue of F if there exists a nonzero
x ∈Msuch that Fx = λx and that any such x is an eigenvector of F
corresponding to λ.Theeigenspace of F corresponding to λ is {x ∈M:
Fx = λx}, i.e., the set of eigenvectors for λ together with 0 ∈M, and it is
indeed a subspace of M. A subspace U of M is said to be invariant under F
if F maps U into U, i.e., if F U⊆U. Any eigenspace of F is invariant under
F since Fx = λx implies F (Fx )=F (λx)=λFx.
Proposition 2.2.2 If F : M→Mis any nonzero, skew-symmetric linear
transformation on M,then
(a) Fx = λx im
plies that either λ =0or x is
nu
ll (or both),
(b) F U⊆Uimplies F (U
⊥
) ⊆U
⊥
.
2.3 Invariant Subspaces 99
Proof: (a) Fx = λx implies Fx · x = λ(x · x)soλ(x · x) = 0 and either
λ =0orx · x =0.
(b) Suppose F U⊆Uand let v ∈U
⊥
. Then, for every u ∈U, Fv ·u = − v ·
Fu = 0 because Fu ∈Uand v ∈U
⊥
.Thus,Fv ∈U
⊥
as required.
The eigenvalues of a linear transformation are found by solving its char-
acteristic equation. For a skew-symmetric linear transformation on M and
with the notation established in (2.2.3) this equation is easy to write down
an
d
qu
ite informative.
Theorem 2.2.3 Let F : M→Mbe a skew-symmetric linear transforma-
tion and {e
a
}
4
a=1
an arbitrary admissible basis for M. With the matrix [F
a
b
]
written in the form (2.2.3)andIthe4 × 4 i
d
enti
ty matrix we have
det ([F
a
b
] − λI)=λ
4
+
|
B
|
2
−|
E
|
2
λ
2
− (
E
·
B
)
2
, (2.2.8)
where |
E
|
2
=(E
1
)
2
+(E
2
)
2
+(E
3
)
2
, |
B
|
2
=(B
1
)
2
+(B
2
)
2
+(B
3
)
2
and
E
·
B
= E
1
B
1
+ E
2
B
2
+ E
3
B
3
.
Exercise 2.2.5 Prove Theorem 2.2.3.
Consequently, the eigenvalues of F are the real solutions to
λ
4
+
|
B
|
2
−|
E
|
2
λ
2
− (
E
·
B
)
2
=0. (2.2.9)
Since the roots of the characteristic polynomial are independent of the choice
of basis and since the leading coefficient on the left-hand side of (2.2.9)is
o
ne
it f
ollows that, while
E
and
B
will, in general, be different in different
admissible bases, the algebraic combinations |
B
|
2
−|
E
|
2
and
E
·
B
are Lorentz
invariants, i.e., the same in all admissible frames. In particular, if both are
zero (i.e., if
E
and
B
are perpendicular and have the same magnitude) in
one frame, the same will be true in any other frame. We shall say that F is
null if |
B
|
2
−|
E
|
2
=
E
·
B
= 0 in any (and therefore every) admissible basis;
otherwise, F is regular. As defined earlier in this section, F
N
is null and F
R
is regular.
Exercise 2.2.6 Show that F is invertible iff
E
·
B
=0.
2.3 Invariant Subspaces
Our objective in this section is to obtain an intrinsic characterization of
“null” and “regular” skew-symmetric linear transformations on M that will
be used in the next section to derive their “canonical forms”. Specifically,
we will show that every skew-symmetric F : M→Mhas a 2-dimensional
100 2 Skew-Symmetric Linear Transformations and Electromagnetic Fields
invariant subspace and that F is regular if and only if U∩U
⊥
= {0} for
some such subspace U (so F is null if and only if U∩U
⊥
= {0} for every
2-dimensional invariant subspace U).
We begin with a few observations on invariant subspaces in general. Let V
be a real vector space of dimension at least 2 and T : V → V a nonzero linear
transformation. Observe that if the characteristic equation det(T − λI)=0
has a real root λ,thenT has an eigenvector, i.e., there is a nonzero v ∈ V
such that Tv = λv. Consequently, Span{v} is a non-trivial invariant subspace
for T . If there are no real roots the situation is less simple.
Lemma 2.3.1 Let V be a real vector space of dimension greater than or
equal to 2 and T : V → V a nonzero linear transformation. If det(T −λI)=0
has a complex solution λ = α + βi, β =0, then there exist nonzero vectors x
and y in V such that
Tx = αx − βy and Ty = αy + βx. (2.3.1)
In particular, Span{x, y} is a nontrivial invariant subspace for T .
Proof: Select a basis for V and let [a
ij
]
i,j=1,...,n
be the matrix of T in this
basis. Since det(T − λI)=0whenλ = α + βi,thesystem
⎧
⎪
⎨
⎪
⎩
a
11
z
1
+ ···+ a
1n
z
n
=(α + βi)z
1
.
.
.
.
.
.
.
.
.
a
n1
z
1
+ ···+ a
nn
z
n
=(α + βi)z
n
as a nontrivial complex solution z
1
= x
1
+ iy
1
,...,z
n
= x
n
+ iy
n
.Thus,
⎧
⎪
⎨
⎪
⎩
a
11
(x
1
+ iy
1
)+···+ a
1n
(x
n
+ iy
n
)=(α + βi)(x
1
+ iy
1
)
.
.
.
.
.
.
.
.
.
a
n1
(x
1
+ iy
1
)+···+ a
nn
(x
n
+ iy
n
)=(α + βi)(x
n
+ iy
n
) .
Separating into real and imaginary parts gives
⎧
⎪
⎨
⎪
⎩
a
11
x
1
+ ···+ a
1n
x
n
= αx
1
− βy
1
.
.
.
.
.
.
.
.
.
a
n1
x
1
+ ···+ a
nn
x
n
= αx
n
− βy
n
(2.3.2)
and
⎧
⎪
⎨
⎪
⎩
a
11
y
1
+ ···+ a
1n
y
n
= αy
1
+ βx
1
.
.
.
.
.
.
.
.
.
a
n1
y
1
+ ···+ a
nn
y
n
= αy
n
+ βx
n
.
(2.3.3)
Let x and y be the vectors in V whose components relative to our basis are
x
1
,...,x
n
and y
1
,...,y
n
respectively. Then Tx = αx−βy and Ty = αy + βx
as required. Notice that neither x nor y is zero since if x =0,(2.3.2)andthe
2.3 Invariant Subspaces 101
fact that β =0implyy =0soz
1
= ···= z
n
= 0, which is a contradiction.
Similarly, y = 0 implies x = 0 so, in fact, neither can be zero.
In order to apply this result to the case of interest to us we require two final
preliminary results.
Lemma 2.3.2 Let A and B be real numbers with B =0. Then the equation
λ
4
+ Aλ
2
− B
2
=0has a complex solution.
Proof: Regard λ
4
+ Aλ
2
− B
2
= 0 as a quadratic in λ
2
to obtain λ
2
=
1
2
(−A ±
√
A
2
+4B
2
). Choosing the minus sign gives a negative λ
2
and there-
fore complex λ.
Lemma 2.3.3 Let F : M→Mbe a nonzero, skew-symmetric linear trans-
formation. If the characteristic equation
λ
4
+(|
B
|
2
−|
E
|
2
)λ
2
− (
E
·
B
)
2
=0
has two distinct nonzero, real solutions, then there exists a 2-dimensional
subspace U of M which is invariant under F and satisfies U∩U
⊥
= {0}.
Proof: Let λ
1
and λ
2
be the two distinct nonzero real eigenvalues. Then
there exist nonzero vectors x and y such that Fx = λ
1
x and Fy = λ
2
y.By
Proposition 2.2.2(a), x and y are null. Observe next that x and y are linearly
independent. Indeed, ax + by = 0 implies
aFx + bFy =0,
a(λ
1
x)+b(λ
2
y)=0,
λ
1
(ax )+λ
2
(by)=0,
λ
1
(ax )+λ
2
(−ax )=0,
(λ
1
− λ
2
)ax =0.
Since λ
1
− λ
2
=0, ax = 0, but x is nonzero so a = 0. Similarly, b =0sox
and y are independent. Thus, U = Span{x, y} is 2-dimensional; it is clearly
invariant under F . Now suppose ax + by ∈U∩U
⊥
. Then, in particular,
(ax + by) · x =0,
a(x · x)+b(x · y)=0,
b(x ·y)=0.
But x and y are null and nonparallel so x · y = 0 and therefore b =0.
Similarly, a =0soU∩U
⊥
= {0}.
Theorem 2.3.4 Let F : M→Mbe a nonzero, skew-symmetric linear
transformation on M. If F is regular, then there exists a 2-dimensional sub-
space U of M which is invariant under F and satisfies U∩U
⊥
= {0}.
102 2 Skew-Symmetric Linear Transformations and Electromagnetic Fields
Proof: Relative to any admissible basis, at least one of
E
·
B
or |
B
|
2
−|
E
|
2
must be nonzero and F ’s characteristic equation is
λ
4
+(|
B
|
2
−|
E
|
2
)λ
2
− (
E
·
B
)
2
=0. (2.3.4)
We consider four cases:
1.
E
·
B
=0and|
B
|
2
−|
E
|
2
< 0.
In this case (2.3.4) becomes λ
2
(λ
2
+(|
B
|
2
−|
E
|
2
)) = 0 and the solutions
are λ =0andλ = ±
|
E
|
2
−|
B
|
2
. The latter are two distinct, nonzero real
solutions so Lemma 2.3.3 yields the result.
2.
E
·
B
=0and|
B
|
2
−|
E
|
2
> 0.
The solutions of (2.3.4)arenowλ =0a
ndλ = ±βi,w
hereβ =
|
B
|
2
−|
E
|
2
.
Lemma 2.3.1 implies that there exist nonzero vectors x and y in M such that
Fx = −βy and Fy = βx. (2.3.5)
We claim that x and y are linearly independent. Indeed, suppose ax + by =0
with, say, b =0.Theny = kx,wherek = −a/b.ThenFx = −βy implies
Fx =(−βk)x.ButF ’s only real eigenvalue is 0 and β =0sok =0and
therefore y = 0, which is a contradiction. Thus, b =0.Sincex =0, ax =0
implies a = 0 and the proof is complete.
Thus, U = Span{x, y} is a 2-dimensional subspace of M that is invariant
under F . We claim that U∩U
⊥
= {0}. Suppose ax + by ∈U∩U
⊥
.
Then ax + by is null so (ax + by) · (ax + by) = 0, i.e.,
a
2
(x · x)+2ab(x · y)+b
2
(y · y)=0.
But x ·y = x · (−
1
β
Fx)=−
1
β
(x · Fx )=0so
a
2
(x · x)+b
2
(y · y)=0,
a
2
x ·
1
β
Fy
+ b
2
y ·
−
1
β
Fx
=0,
a
2
β
x ·Fy −
b
2
β
y · Fx =0,
a
2
β
x ·Fy +
b
2
β
x · Fy =0,
a
2
+ b
2
β
x · Fy =0.
Now, if a
2
+ b
2
=0,thenx · Fy =0sox · (βx)=0andx · x = 0. Similarly,
y · y =0sox and y are orthogonal null vectors and consequently parallel.
But this is a contradiction since x and y are independent. Thus, a
2
+ b
2
=0
so a = b =0andU∩U
⊥
= {0}.
2.3 Invariant Subspaces 103
3.
E
·
B
=0and|
B
|
2
−|
E
|
2
=0.
In this case (2.3.4) becomes λ
4
=(
E
·
B
)
2
so λ
2
= ±|
E
·
B
|. λ
2
= |
E
·
B
|
gives two distinct, nonzero real solutions so the conclusion follows from
Lemma 2.3.3.
4.
E
·
B
=0and|
B
|
2
−|
E
|
2
=0.
Lemma 2.3.2 implies that (2.3.4) has a complex root α + βi(β =0)
.T
hus,
Lemma 2.3.1 yields nonzero vectors x and y in M with
Fx = αx − βy and Fy = αy + βx.
There are two possibilities:
i. x and y are linearly dependent. Then, since neither is zero, y = kx for
some k ∈ R with k =0.Thus,Fx = αx−βy = αx−kβx =(α −kβ)x and
Fy = αy + βx = αy +
β
k
y =
α +
β
k
y.Sinceα +
β
k
= α − kβ and since
0 is not a solution to (2.3.4) in this case we find that F has
tw
o distinct,
nonzero real eigenvalues and again appeal to Lemma 2.3.3.
ii. x and y are linearly independent. Then U = Span{x, y} is a 2-dimensional
subspace of M that is invariant under F.
Exercise 2.3.1 Complete the proof by showing that U∩U
⊥
= {0}.
To complete our work in this section we must show that a nonzero null
skew-symmetric F : M→Mhas 2-dimensional invariant subspaces and that
all of these intersect their orthogonal complements nontrivially. We address
the question of existence first.
Proposition 2.3.5 Let F : M→Mbe a nonzero, null, skew-symmetric
linear transformation on M. Then both ker Fandrng F =(kerF )
⊥
are 2-dimensional invariant subspaces of M and their intersection is a
1-dimensional subspace of M spanned by a null vector.
Proof: ker F and rng F are obviously invariant under F .SinceF is null,
E
·
B
= 0 so, by Exercise 2.2.6, F is not invertible. Thus, dim(ker F ) =0.
Proposition 2.2.1(c) then implies that dim(ker F ) = 2 and, consequently,
dim(rng F )=2.
Now, since rng F ∩ ker F =rngF ∩ (rng F )
⊥
by Proposition 2.2.1(a), if
this intersection is not {0}, it can contain only null vectors. Being a subspace
of M it must therefore be 1-dimensional. We show that this intersection
is, indeed, nontrivial as follows: For a null F the characteristic polynomial
(2.3.4) reduces to λ
4
= 0. The Cayley-Hamilton Theorem (see [H]) therefore
implies that
F
4
=0 (F null). (2.3.6)