MATH 115 SOLUTION SET 7
ANSWERS TO SELECTED PROBLEMS
§
5.1
7.
The point of this problem is to extend the definition of determinant from
matrices to linear transformations. The result is always that a property satisfied
by matrices is always satisfied by linear transformations as well:
a. If
Q
is the change of basis matrix from
β
to
γ
, then [
T
]
β
=
Q
[
T
]
γ
Q

1
, so that
[
T
]
β
and [
T
]
γ
are similar and therefore have the same determinant.
b.
T
is invertible if and only if the matrix [
T
]
β
is invertible, which is so if and
only if det[
T
]
β
6
= 0. But det[
T
]
β
is the same as det
T
.
c. We have det(
T

1
) = det[
T

1
]
β
= det[
T
]

1
β
= (det
T
)

1
.
d.
We have det(
TU
) = det([
TU
]
β
) = det([
T
]
β
[
U
]
β
) = det([
T
]
β
) det([
U
]
β
) =
det(
TU
).
e. We have det(
T

λI
V
) = det([
T

λI
V
]
β
) = det([
T
]

λI
). The second equality
comes from Thm. 2.8.
/
8. a.
T
is invertible if and only if det
T
6
= 0, which is true if and only if 0 is not
a root of the characteristic polynomial det(
T

λI
V
), which is true if and only if 0
is not an eigenvalue of
T
.
b. We have that
λ
is an eigenvalue of
T
if and only if for some
v
∈
V
, we have
Tv
=
λv
. But since
T
is invertible, every eigenvalue is nonzero and we may rewrite
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '10
 SHALOM
 Linear Algebra, Algebra, Determinant, Transformations, Matrices, #, β, Det, characteristic polynomial det

Click to edit the document details