Linear Algebra 115
Solutions to First Homework
Problem
8 (Section 1
.
2) We have:
(
a
+
b
)(
x
+
y
)
=
a
(
x
+
y
) +
b
(
x
+
y
)
,
because of axiom VS8
=
ax
+
ay
+
bx
+
by,
because of axiom VS7
Problem
12 (Section 1
.
2) First we observe that the set of even functions
(call it Even) is a subset of
F
(
S, F
). By Theorem 1
.
3 it is a subspace iff three
conditions are met. We are verifying them one by one:
1.
→
0
∈
Even
. Note here that the zero vector is actually the zero function,
i.e.
the one that gives value zero for all arguments.
Call this function
z
(
x
). It is even, since
z
(
x
) = 0 =
z
(

x
).
2. Say
f, g
∈
Even
. Then
f
(
x
) =
f
(

x
)
, g
(
x
) =
g
(

x
). Adding these up we
get
f
(
x
)+
g
(
x
) =
f
(

x
)+
g
(

x
), or in other words
f
+
g
(
x
) =
f
+
g
(

x
),
by the definition of
f
+
g
. But this is like saying the
f
+
g
∈
Even
.
3. Similarly to the above, multiply
f
(
x
) =
f
(

x
) by
c
, to get
cf
(
x
) =
cf
(

x
). From this we conclude that
cf
∈
Even
.
Now, since
Even
is a subset of
F
(
S, F
), it has to be a vector subspace by itself.
Note: There is another way to go for the problem. Verify all 8 axioms and
the fact that
Even
is closed for addition and multiplication. This way is correct,
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 Spring '10
 FUCKHEAD
 Addition, −x, zero function, nonzero points

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