Linear Algebra 115
Solutions to Thirteenth Homework
Problem
9 (Section 6
.
1) (a) Assume that
< x, z >
= 0, for all
z
∈
β
, a basis
for a ﬁnite dimensional
V
. Also, assume that
β
=
{
v
1
, v
2
, . . . , v
n
}
. Then we can
write
x
=
a
1
v
1
+
a
2
v
2
+
. . .
+
a
n
v
n
.
Now, we get
< x, x >
=
< x, a
1
v
1
+
a
2
v
2
+
. . .
+
a
n
v
n
>
=
a
1
< x, v
1
>
+
a
2
< x, v
2
>
+
. . .
+
a
n
< x, v
n
>
=
0
,
by our assumption and the fact that
v
1
, . . . , v
n
∈
β
.
But
< x, x >
= 0 iﬀ
x
= 0.
(b) Assume that
< x, z >
=
< y, z >
, for all
z
∈
β
. Then,
< x, z >

< y, z >
= 0, or that
< x

y, z >
= 0, for all
z
∈
β.
Part (a) gives that
x

y
has to be the zero vector, or that
x
=
y
.
±
Problem
11 (Section 6
.
1) We make both computations together:
k
x
±
y
k
2
=
< x
±
y, x
±
y >
=
< x, x >
+
< y, y >
±
< y, x >
±
< x, y >
=
k
x
k
2
+
k
y
k
2
±
< y, x >
±
< x, y >
Adding these two up, we have that
k
x
+
y
k
2
+
k
x

y
k
2
= 2
k
x
k
2
+2
k
y
k
2
.
±
Problem