Linear Algebra Final Exam
1:00–3:00, Sunday, June 2
Bradley 102
1
Let
T
:
R
3
→
R
3
be a linear transformation with the property that
T
◦
T
◦
T
= 0 (we’ll refer to
T
◦
T
◦
T
as
T
3
for the rest of this problem).
(a) What exactly does this mean? That is, what is the practical upshot
when it comes to plugging in vectors to
T
3
?
Solution
It means that
T
3
(
v
) =
0
for every vector
v
∈
R
3
.
(b) Suppose that
x
∈
R
3
is such that
T
2
(
x
) =
T
(
T
(
x
))
6
=
0
. If
z
=
cT
2
(
x
),
then what is
T
(
z
)? Let
y
=
c
1
T
(
x
) +
c
2
T
2
(
x
). What is
T
2
(
y
)?
Solution
Well,
T
(
z
) =
T
(
cT
2
(
x
)) =
cT
3
(
x
) =
0
and
T
2
(
y
) =
T
2
(
c
1
T
(
x
) +
c
2
T
2
(
x
)) =
c
1
T
3
(
x
) +
c
2
T
3
(
T
(
x
)) =
0
.
(c) Let’s say
b
1
=
x
b
2
=
T
(
x
)
,
and
b
3
=
T
2
(
x
)
.
Show that
b
1
is not a linear combination of
b
2
and
b
3
.
Solution
From what we’ve seen in (b), any linear combination
y
of
b
2
and
b
3
has the property that
T
2
(
y
) =
0
. Since
T
2
(
b
1
) =
T
2
(
x
)
6
=
0
,
b
1
can’t be a linear combination of
b
2
and
b
3
.
1
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(d) Explain why the set
B
=
{
b
1
,
b
2
,
b
3
}
is a basis for
R
3
.
(
Hint:
Some
of the work you’ve already done might help. )
Solution
We know that
T
(
b
2
) =
T
2
(
x
)
6
=
0
and
T
(
b
3
) =
0
, so
b
2
is not
a multiple of
b
1
.
And we showed in (c) that
b
1
is not a linear
combination of
b
2
and
b
3
, so the whole set must be linearly in
dependent. Since we are working in
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 Fall '08
 Staff
 Linear Algebra, Algebra, linear transformation

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