(5) Since
dim
R
2
=
2
the
third
answer must be wrong. For
the
first
we
do
have
((1,
1),
(1,
1))
=
0,
but
11(1,
1)11
=
11(1, 1)11
=
J2
=I
l.
Compare
to
the
last of
the
four definitions on page
139.
(6)
Apart
from physicists, who have already worked this
out
as
an
exercise
in
Chapter
4, you will not find it so easy to see
the
correctness of
the
third
answer.
It
follows from
(x,y)
=
(1/4)(llx
+
yW
llx

yW).
The
second answer, although
not
correct,
is
not
so far off
the
mark. For such
maps
we
always have
f(
x)
=
At.p(
x)
for all
x
E
V
and
some
A
E
Rand
a suitable orthogonal
map
t.p.
(7) Orthogonal
maps
are always injective (see
the
fact on page 143). There
fore,
we
must
have Ker
Pu
=
U.L
=
0,
if
Pu
is
to
be orthogonal. By
the
corollary on page 143
we
then
have
U
=
V,
and
Pv
=
Idv
is certainly
orthogonal.
(8) See
the
fact
on
page
144.
(9)
The
second answer is
not
as wrong as
the
first, since
(Il,
+)
would indeed
be
a group,
but
the
formulation
is
not
acceptable. N is
not
unfit
to
have
some
group
structure
N
X
N
7
N,
but
since"
+"
here denotes
the
ordinary
addition
of numbers, axiom (3)
is
not
satisfied.
(10) A
2k
X
2k
matrix
is
not
a
k
X
k
matrix, so
the
first answer
must
be
wrong.
The
statement
(_l?k
=
1
in
the
third
is only a decoy, quite irrelevant
to
the
question.
SO(2k)
=I
O(2k)
because, for example,
(
1
1
1)
E
O(2k)
"
SO(2k).
196
CHAPTER
12:
ANSWERS
TO
THE
TESTS
CHAPTER
9
TEST
1
2
3
4
5
6
7
8
9
10
x
x
x
x
x
x
x
x
x
x
x
(1)
If
f(x)
=
AX
is to mean anything,
x
and
f(x)
must belong to
the
same
vector space. Look
at
the definition on page 151.
(2)
fex)
=
AX:::}
fex)
=
(A)(X)
and
f(x)
=
(A)X,
so
x
and
x
are
eigenvectors for 
A
.
(3)
E>.
=
Ker(f

AId)
contains zero as well as the eigenvectors for
A.
eigenvalue A
=
1.
(5) See the corollary on page 153.
(6)
f(x)
=
AX
:::}
x
=
fl(AX)
=
Afl(x)
:::}
fl(X)
=
lx.
Note
that
A
is
in
fact nonzero, otherwise
f
would fail to
be
injective
and
could not
be
an
automorphism.
(7) All three conditions imply diagonalizability (see page 153,
the
corollary on
page 153, and, of course, the definition of diagonalizability
on
page 152),
but
the first two are not
equivalent
to diagonalizability, as
the
example
(
1 0
0)
f
=
0
1 0
:
lR
3
+
lR
3
000
shows.
(8)
We
have only restricted
IF
to be
lR
or
C on page 159, so as
to
avoid
certain technicalities with polynomials over arbitrary fields.
(10) An easy calculation shows
that
v
is
an
eigenvector of
f
for
the
eigenvalue
A,
if
and
only if
<p
1
(v)
is
an
eigenvector of
9
for the eigenvalue
A.
Answers to
Chapter
10
Test
197
CHAPTER
10
TEST
1
2
3
4
5
6
7
8
9
10
x
X
X
X
X
X
X
X
X
X
X
X
(1)
Endomorhpisms having
the
first property are called orthogonal, while
the
third
property imposes no conditions on
f
(symmetry
of
the
inner
product).
(2)
The
Ai
are real numbers, not elements of
V,
therefore
the
first state
ment
is meaningless. For the correctness of
the
two others, see Fact 1
on
page 162.
(3)
The
same argument as used in Fact 2 on page 162 shows
that
the
first two answers are
both
correct: from
u
E
U
and
w
E
U.L
it
follows
that
(f(w),u)
=
(w,f(u))
=
0 for a selfadjoint
f,
and
(f(w),u)
=
(fIf(w),fl(u))
=
(w,fl(u))
=
0 for
an
orthogonal
f.
(Note
that
in
the
latter
case
f
:
V
+
V
and
flU:
U
+
U
are even
isomorphisms. )
(4) For
the
first
matrix
al4
=F
a41
,
for
the
third
al2
=F
a21
.