Math 235
Assignment 11 Solutions
1.
Consider
C
3
with its standard inner product. Let
~
z
=
1 +
i
2

i

1 +
i
,
~w
=
1

i

2

3
i

1
.
a) Evaluate
h
~
z, ~w
i
and
h
~w,
2
i~
z
i
.
Solution: We have
h
~
z, ~w
i
= (1 +
i
)(1 +
i
) + (2

i
)(

2 + 3
i
) + (

1 +
i
)(

1) = 2
i

1 + 8
i
+ 1

i
= 9
i
h
~w,
2
i~
z
i
=
2
i
h
~
z, ~w
i
=

2
i
(

9
i
) =

18
b) Find a vector in span
{
~
z, ~w
}
that is orthogonal to
~
z
.
Solution: A vector orthogonal to
~
z
in span
{
~
z, ~w
}
is
~v
= perp
~
z
~w
=
~w

h
~w,~
z
i
k
~
z
k
2
~
z
=
1

i

2

3
i

1
+
9
i
9
1 +
i
2

i

1 +
i
=
0
1

i

2

i
c) Write the formula for the projection of
~u
onto
S
= span
{
~
z, ~w
}
.
Solution: Note that span
{
~
z, ~w
}
= span
{
~
z,~v
}
where
~v
is the vector from part b) and
{
~
z,~v
}
is orthogonal over
C
so we have
proj
S
~u
=
< ~u,~
z >
k
~
z
k
2
~
z
+
< ~u,~v >
k
~v
k
2
~v
2.
Let
V
be an inner product space, with complex inner product
h
,
i
. Prove that if
< ~u,~v >
= 0, then
k
~u
+
~v
k
2
=
k
~u
k
2
+
k
~v
k
2
. Is the converse true?
Solution: If
< ~u,~v >
= 0, then we have
k
~u
+
~v
k
2
=
< ~u
+
~v,~u
+
~v >
=
< ~u,~u
+
~v >
+
< ~v,~u
+
~v >
=
< ~u,~u >
+
< ~u,~v >
+
< ~v,~u >
+
< ~v,~v >
=
k
~u
k
2
+ 0 +
0 +
k
~v
k
2
=
k
~u
k
2
+
k
~v
k
2
The converse is not true. One counter example is: Consider
V
=
C
with its standard inner
product and let
~u
= 1 +
i
and
~v
= 1

i
. Then
k
~u
+
~v
k
2
=
k
2
k
2
= 4 and
k
~u
k
2
+
k
~v
k
2
=
2 + 2 = 4, but
< ~u,~v >
= (1 +
i
)(1 +
i
) = 2
i
6
= 0.
1