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Linear Algebra
Geometry: Solutions to sheet 8
1.
(a) Let
x
∈
R
n
, then
R
◦
T
(
x
) =
R
(
T
(
x
)),
S
◦
T
(
x
) =
S
(
T
(
x
)) and (
R
+
S
)
◦
T
(
x
) =
(
R
+
S
)(
T
(
x
)) =
R
(
T
(
x
)) +
S
(
T
(
x
)).
(b) Let
x
∈
R
n
, then (
R
◦
S
)
◦
T
(
x
) =
R
◦
S
(
T
(
x
)) =
R
(
S
(
T
(
x
))) and
R
◦
(
S
◦
T
)(
x
) =
R
(
S
◦
T
(
x
)) =
R
(
S
(
T
(
x
))), and hence (
R
◦
S
)
◦
T
=
R
◦
(
S
◦
T
)
.
2.
Let
T
:
R
n
→
R
n
be a linear map, show that
(a) Note that
T
(0) = 0, hence 0
∈
ker
T
. If
x
,
y
∈
ker
T
:=
{
x
∈
R
n
;
T
(
x
) = 0
}
then
T
(
x
+
y
) =
T
(
x
) +
T
(
y
) = 0 + 0 = 0, so
x
+
y
∈
ker
T
, and if
x
∈
ker
T
,
then
T
(
λ
x
) =
λT
(
x
) =
λ
0 = 0, hence
λ
x
∈
ker
T
.
(b) Since 0
∈
V
and
T
(0) = 0 we have 0
∈
T
(
V
), so
T
(
V
)
6
=
∅
. Now assume
y
,
y
0
∈
(
V
), then there are
x
,
x
0
∈
V
with
T
(
x
) =
y
and
T
(
x
0
) =
y
0
. Since
V
is linear,
x
+
x
0
∈
V
and hence with the linearity of
T
we ﬁnd
y
+
y
0
=
T
(
x
) +
T
(
x
0
) =
T
(
x
+
x
0
)
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This note was uploaded on 02/25/2010 for the course MATHEMATIC 11007 taught by Professor Spyros during the Spring '10 term at Bristol Community College.
 Spring '10
 Spyros
 Linear Algebra, Algebra, Geometry

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