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61
If
AB
=
BA
, then
c
w
+
dy
=
ay
+
cz
; that is,
c
(w
−
z
)
=
y
(
a
−
d
)
.
There are two cases to consider. If
a
=
d
, then de
f
ne
B
=
·
01
11
¸
.
The 2,1 entry of
is
d
=
a
, while the 2,1 entry of
is
a
+
c
.As
c
±=
0,
we have
±=
.If
a
±=
d
,de
f
ne
B
=
·
10
¸
.
Now the 2,1 entry of
is
c
+
d
, while the 2,1 entry of
is
c
+
a
. Since
a
±=
d
, these entries are different, and so
±=
in this case as well.
Therefore,
A
is not in the center of GL
(
2
,
R
)
. A similar argument holds if
b
±=
0, (This result is generalized to
n
×
n
matrices in Corollary 4.86. The
proof using linear transformations is much shorter and much simpler.)
2.85
Let
ζ
=
e
2
π
i
/
n
be a primitive
n
th root of unity, and de
f
ne
A
=
h
ζ
0
0
ζ
−
1
i
and
B
=
£
±
.
(i)
Prove that
A
has order
n
and that
B
has order 2.
Solution.
It is clear that
A
n
=
I
=
B
2
.I
f1
≤
k
<
n
, then
A
k
±=
I
, for the 1,1 entry of
A
k
is
ζ
k
±=
1.
(ii)
Prove that
BAB
=
A
−
1
.
Solution.
One multiplies the matrices.
(iii)
Prove that the matrices of the form
A
i
and
i
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This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.
 Fall '11
 KeithCornell

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