ma253
, Fall
2010
— Problem Set
1
Solutions
1.
We work with the linear space
R
2
. Fix two numbers
a
and
b
and consider the
function
f
:
R
2
−
→
R
given by
f
x
1
x
2
=
ax
1
+
bx
2
.
Check that
f
is a linear function.
We need to check that
f
distributes over sums and allows us to pull out scalars.
Sums
We start with two vectors
x
1
x
2
and
u
1
u
2
. Their sum is
x
1
+
u
1
x
2
+
u
2
. Now
f
x
1
+
u
1
x
2
+
u
2
=
a
(
x
1
+
u
1
) +
b
(
u
2
+
v
2
) =
ax
1
+
au
1
+
bx
2
+
bu
2
,
while
f
x
1
x
2
+
f
(
u
1
u
2
=
ax
1
+
bx
2
+
au
1
+
bu
2
.
So the two computations give the same answer, which says that
f
distributes over
sums.
Scaling
We start with
x
1
x
2
and a scalar
α
; scaling the vector gives
αx
1
αx
2
. Now we just
compute the two things that are supposed to be equal:
f
αx
1
αx
2
=
aαx
1
+
bαx
2
=
α
(
ax
1
+
bx
2
)
,
while
αf
x
1
x
2
=
α
(
ax
1
+
bx
2
)
.
They are equal, so we’ve shown that we can pull out scalars.
So we’ve shown that
f
satisfies the two requirements to be a linear function, and we’re
done.
2.
Still in
R
2
, look at the special vectors
e
1
=
1
0
and
e
2
=
0
1
. Explain how to get
any vector from these two, using only adding and scaling operations.
Any vector in
R
2
looks like
m
n
for some two numbers
m
and
n
. Now just notice
that
m
n
=
m
1
0
+
n
0
1
=
m
e
1
+
n
e
2
.
Never Give an Iguana Viagra – headline, Reuters, January 25, 2007
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3.
Suppose we know that
f
:
R
2
−
→
R
is a linear function, that
f
(
e
1
) =
a
and that
f
(
e
2
) =
b
. Show that for any vector
v
=
x
1
x
2
we must have
f
(
v
) =
ax
1
+
bx
2
.
We use problem 2 together with the information that
f
is linear.
f
x
1
x
2
=
f
(
x
1
e
1
+
x
2
e
2
)
=
f
(
x
1
e
1
) +
f
(
x
2
e
2
)
because we can distribute over sums
=
x
1
f
(
e
1
) +
x
2
f
(
e
2
)
because we can pull out scalars
=
x
1
a
+
x
2
b
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 Spring '10
 Machaut
 Vector Space, Linear function, shopping list, Rodolfo

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