Answer Key to Final Examination, Version 1, Math. 33A, Spring Quarter, 2003
1.
Suppose matrix
A
is similar to
B
.
Prove the following facts (missing step by step
description of your reasoning costs maximum a half of the allocated points):
(a) There exists an isomorphism from the kernel of
A
onto the kernel of
B
(50 points).
Since
A
is similar to
B
, we have an invertible matrix
S
such that
B
SAS
1
BS
SA
S
1
B
AS
1
. Then if
x
is in the kernel of
A
, then
BS x
SA x
S
0
0 . This
shows
S x
is in the kernel of
B
. Thus we have a linear transformation
T
: Ker
A
Ker
B
given by
T
x
S x
. Since
S
is invertible, Ker
T
Ker
S
0
; so, Ker
T
0
.
If
y
is in the kernel of
B
,
AS
1
y
S
1
B y
S
1
0
0 . Thus
S
1
y
is in the
kernel of
A
. In particular
y
SS
1
y
is in the image of
T
. This shows that Im
T
Ker
B
. The two facts Ker
T
0
and Im
T
Ker
B
combined tell us that
T
is an
isomorphism, and
T
1
is given by the multiplication by
S
1
.
(b) nullity
A
nullity
B
(20 points).
By (a), Ker
A
and Ker
B
has the same dimension, which are the nullity of
A
and
B
,
respectively.
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 Fall '08
 lee
 Math, Linear Algebra, Algebra, Ker B, Ker A, Ker A Ker, Ker T Ker

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