Math 110 Spring 2008
Solutions to HW9
Section 5.1
20. Let
f
(
t
) = det(
A

tI
) = (

1)
n
t
n
+
···
+
a
1
t
+
a
0
. Then
a
0
=
f
(0) = det(
A

0
I
) = det(
A
).
±
21. a. Following the suggestion in the book, we prove this by induction. We may as well take the base case
to be
n
= 2, in which case the proposition is easily veriﬁed. Now assume it is true for
n

1, and take
a matrix
A
∈
M
n
×
n
(
F
). Computing det(
A

tI
) along the ﬁrst row yields
f
(
t
) = (
A
11

t
) det(
˜
A
11

tI
n

1
) +
p
(
t
)
where
p
(
t
) is the sum of the (1
,j
)cofactors for
j
≥
2. Clearly
p
(
t
) will be a polynomial of degree at
most
n

2, as in any of these cofactors you have a product of at most (
n

2) factors of
t
. (We could
make this a little more rigorous with another proof by induction, but I believe this is plausible enough.)
Now by induction det(
˜
A
11

tI
n

1
) =
Q
n
i
=2
(
A
ii

t
) +
r
(
t
), where
r
(
t
) is a polynomial of degree at
most
n

3. Plugging this back into the expression above yields
f
(
t
) =
n
Y
i
=1
(
A
ii

t
) + (
A
11

t
)
r
(
t
) +
p
(
t
) =
n
Y
i
=1
(
A
ii

t
) +
q
(
t
)
where
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 Spring '08
 GUREVITCH
 Math, Linear Algebra, Algebra

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