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Math 51 — Spring 2009 — Exam II Solutions
Problem 1.
(10 pts.)
Let
L
be a linear transformation from
R
n
to
R
k
. Each of the
statements about
L
below is either
always true
(“T”), or
always false
(“F”), or
sometimes
true and sometimes false, depending on the situation
(“MAYBE”). For each part, decide
which, and justify your answer completely.
Each part is 5 points: 2 for a correct answer, 3 more for correct justiﬁcation.
Note:
if the answer is “MAYBE,” complete justiﬁcation must include two examples — one
where the statement holds true, and another where it is false. If the answer is “TRUE” or
“FALSE,” a proper justiﬁcation cannot be done by giving a single example — only by giving
a proof.
a) If vectors
v
1
,
v
2
,...,
v
m
are linearly independent, then the vectors
L
(
v
1
)
,
L
(
v
2
)
,...,
L
(
v
m
) are linearly independent as well.
T
F
MAYBE
As noted above, we must give two examples: one where this statement holds, and one where
it does not. There are many possibilities for each.
•
Suppose
n
=
k
and
L
=
Id
n
, the identity transformation; then
L
(
v
1
) =
v
1
,
L
(
v
2
) =
v
2
,
...
,
L
(
v
m
) =
v
m
for any choice of vectors. So if we assume
v
1
,
v
2
,...,
v
m
are linearly
independent, then certainly
L
(
v
1
)
,
L
(
v
2
)
,...,
L
(
v
m
) are independent as well.
•
If we take
L
to be the zero transformation (
L
(
v
) =
0
for all
v
), then
L
(
v
1
) =
···
=
L
(
v
m
) =
0
, and so in this case
L
(
v
1
)
,...,
L
(
v
m
) are linearly dependent.
b) If vectors
v
1
,
v
2
,...,
v
m
are linearly dependent, then the vectors
L
(
v
1
)
,
L
(
v
2
)
,...,
L
(
v
m
) are linearly dependent as well.
T
F
MAYBE
Since
v
1
,...,
v
m
are linearly dependent, we may ﬁnd scalars
c
1
,...,c
m
, not all equal to 0,
such that
c
1
v
1
+
···
+
c
m
v
m
=
0
. But then
0
=
L
(
0
) =
c
1
L
(
v
1
) +
···
+
c
m
L
(
v
m
), since
L
is
a linear transformation, and so
L
(
v
1
)
,...,
L
(
v
m
) are linearly dependent (because not all
c
i
are equal to 0).
1

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*Sign up* Problem 2.
(12 pts.)
a) Complete the following sentence to make a true statement: A linear transformation
L
:
R
n
→
R
m
is invertible if and only if
(4 points) Choose your favorite one; they’re all equivalent (but be careful to note the dis-
tinction between the function
L
and its matrix — these are diﬀerent objects):
•
...
L
is both one-to-one
and
onto.
•
...
n
=
m
and
L
is onto.
•
...
n
=
m
and
L
is one-to-one.
•
... the matrix of
L
with respect to the standard basis (that is, the matrix
A
such that
L
(
x
) =
A
x
for all
x
in
R
n
) has reduced row echelon form equal to
I
n
.
•
... the matrix of
L
is square and has rank
n
.
•
... the matrix of
L
is square and has linearly independent columns.

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