MAT240 Assignment 10 — Partial Solutions
Arthur Fischer
April 1, 2004
Exercise (#17, p.229).
Let
A, B
∈
M
n
×
n
(
F
)
be such that
AB
=

BA
. Prove that if
n
is odd and
F
is not a field of characteristic two, then
A
or
B
is not invertible.
Solution.
Assume as in the statement of the problem that
F
is not a field of characteristic two, that
n
is odd, and that
A, B
∈
M
n
×
n
(
F
) are such that
AB
=

BA
. By various theorems and previous
exercises, we have
det(
A
)
·
det(
B
) = det(
AB
) = det(

BA
) = (

1)
n
det(
BA
) =

det(
B
)
·
det(
A
)
.
Since
F
is not a field of characteristic two, the only fieldelement which is its own negative is 0, and
therefore det(
A
) det(
B
) = 0.
By properties of fields, this only occurs when either det(
A
) = 0 or
det(
B
) = 0, immediately implying that either
A
or
B
is not invertible.
Exercise (#6, p.238).
Prove that if
M
∈
M
n
×
n
(
F
)
can be written in the form
M
=
A
B
0
C
,
where
A
and
C
are square matrices, then
det(
M
) = det(
A
)
·
det(
C
)
.
Solution.
Before we begin, we will introduce a bit of notation. Let
A
be an (
m
×
n
)matrix, and let
i
≤
m
;
k
≤
n
. By
˜
A
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 Fall '09
 Jones
 Linear Algebra, Algebra, Recursion, Det, Structural induction, Mi,1 )

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