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6
SPRING 2012
3.
Invertible Operators, Derivatives
For each
T
∈
L
(
R
n
), let det
T
be the determinant of the
n
×
n
matrix representing
T
with
respect to the standard basis; this de±nes a map det:
L
(
R
n
)
→
R
. In fact, det is continuous.
To see this, note that the map de±ned by
T
°→
(
a
1
,
1
,...,a
n,
1
,a
1
,
2
,...,a
n,
2
,...,a
1
,n
,...,a
n,n
)
whenever [
T
]=(
a
ij
), is a normed vector space isomorphism of (
L
(
R
n
)
,
±·±
2
) onto the
Euclidean space
R
n
2
. But viewed as a function on
R
n
2
, det is just a
n
th
-degree polynomial
in
n
2
variables, and is therefore continuous.
Defnition 3.1.
For each
n
∈
N
,let
GL
(
n
) be the set of invertible elements of
L
(
R
n
).
Arguments similar to the one made above for det (and Cramer’s formula for inverses) show
that the maps (
S,T
)
°→
S
◦
T
from
GL
(
n
)
×
GL
(
n
)to
GL
(
n
)and
T
°→
T
−
1
from
GL
(
n
)to
itself are continuous. Thus
GL
(
n
)isa
topological group
, called the
general linear group
.
Exercise 3.1.
Prove or disprove:
±
T
−
1
±
=
1
±
T
±
for each
T
∈
GL
(
n
)
.
Moreover, since
GL
(
n
) is the inverse image of the open set
R
\{
0
}
under the continuous
map det, it is an open subset of
L
(
R
n
). In other words, for each invertible
S
∈
L
(
R
n
), there
exists
°>
0 such that all
T
∈
L
(
R
n
) which satisfy
±
S
−
T
±
<°
are also invertible. It turns
out that with
S
= Id, we can take
°
= 1 (think of the case
n
=1)
.
Proposition 3.2.
For
T
∈
L
(
R
n
)
, if
±
Id
−
T
±
<
1
, then
T

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