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18.102 Introduction to Functional Analysis
Spring 2009
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95
LECTURE
NOTES
FOR
18.102,
SPRING
2009
Lecture
16.
Tuesday,
April
7:
partially
reconstructed
From
last
time
Proposition
23.
The
invertible
elements
form
an
open
subset
GL(
H
)
⊂ B
(
H
)
.
Proof.
Recall
that
we
showed
using
the
convergence
of
the
Neumann
series
that
if
B
∈ B
(
H
)
and
�
B
�
<
1
then
Id
−
B
is
invertible,
meaning
it
has
a
twosided
inverse
in
B
(
H
)
(which
we
know,
from
the
open
mapping
Theorem
to
be
equivalent
to
it
being
a
bijection).
So,
suppose
G
∈
GL(
H
)
,
meaning
it
has
a
twosided
(and
unique)
inverse
G
−
1
∈
B
(
H
) :
(16.1)
G
−
1
G
=
GG
−
1
=
Id
.
Then
we
wish
to
show
that
B
(
G
;
�
)
⊂
GL(
H
)
for
some
� >
0
.
In
fact
we
shall
see
that
we
can
take
�
=
�
G
−
1
�
−
1
.
The
idea
is
that
we
wish
to
show
that
G
+
B
is
a
bijection,
and
hence
invertible.
To
do
so
set
(16.2)
E
=
G
−
1
B
=
G
+
B
=
G
−
1
(Id +
G
−
1
B
)
.
⇒
This
is
injective
if
Id
+
G
−
1
B
is
injective,
and
surjective
if
Id
+
G
−
1
B
is
surjective,
since
G
−
1
is
a
bijection.
From
last
time
we
know
that
(16.3)
�
G
−
1
B
�
<
1 =
⇒
Id +
G
−
1
B
is
invertible.
Since
�
G
−
1
B
� ≤ �
G
−
1
��
B
�
this
follows
if
�
B
�
<
�
G
−
1
�
−
1
as
anticipated.
�
Thus
GL(
H
)
⊂ B
(
H
)
,
the
set
of
invertible
elements,
is
open.
It
is
also
a
group
–
since
the
inverse
of
G
1
G
2
if
G
1
, G
2
∈
GL(
H
)
is
G
−
1
G
−
1
.
2
1
This
group
of
invertible
elements
has
a
smaller
subgroup,
U(
H
)
,
the
unitary
group,
defined
by
(16.4)
U(
H
) =
{
U
∈
GL(
H
);
U
−
1
=
U
∗
}
.
The
unitary
group
consists
of
the
linear
isometric
isomorphisms
of
H
onto
itself
–
thus
(16.5)
(
Uu, Uv
) = (
u, v
)
,
�
Uu
�
=
�
u
� ∀
u, v
∈ H
, U
∈
U(
H
)
.
This
is
an
important
object
and
we
will
use
it
a
little
bit
later
on.
The
unitary
group
on
a
separable
Hilbert
space
may
seem
very
similar
to
the
familiar
unitary
group
of
n
×
n
matrices,
U(
n
)
.
It
is,
of
course
it
is
much
bigger
for
one
thing.
In
fact
there
are
some
other
important
differences
which
I
will
describe
a
little
later
on
(or
get
you
to
do
some
of
it
in
the
problems).
On
important
fact
that
you
should
know,
even
though
I
will
not
prove
it,
is
that
U(
H
)
is
contractible
as
a
metric
space
–
it
has
no
significant
topology.
This
is
to
be
constrasted
with
the
U(
n
)
which
have
a
lot
of
topology,
and
not
at
all
simple
spaces
–
especially
for
large
n.
One
upshot
of
this
is
that
U(
H
)
does
not
look
much
like
the
limit
of
the
U(
n
)
as
n
→ ∞
.
Now,
for
the
rest
of
today
I
will
talk
about
the
opposite
of
the
‘big’
operators
such
as
the
elements
of
GL(
H
)
.
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 Spring '09
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 Linear Algebra, Hilbert space, finite rank, finite rank operators

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