Math
2940
Solutions, Fall
2011
Section
4
.
4
4
)(a)
Q
=
1
0
0
1
0
0
so
QQ
T
=
1
0
0
1
0
0
1
0
0
0
1
0
=
1
0
0
0
1
0
0
0
0
. A square matrix
Q
will
never
work.
(b) For this problem one of the two vectors
must
be 0. Choose any nonzero vector for the
other.
(c) 1
/
√
3(1
,
1
,
1) is our first vector. The other two must lie in the plane
x
+
y
+
z
= 0 (why?)
So use 1
/
√
2(1
,
0
,

1) for the second vector. If the third is (
a, b, c
), then to be perpen
dicular to the first, we need
a
+
b
+
c
= 0 and to be perpendicular to the second we need
a

c
= 0 so (1
,

2
,
1) suggest itself. We normalize to get 1
/
√
6(1
,

2
,
1).
5
) We take (1
,

1
,
0) as our first, and if (
a, b, c
) is the second, we need
a
+
b
+ 2
c
= 0 and
a
=
b
so (1
,
1
,

1) works. Normalizing we get 1
/
√
2(1
,

1
,
0) and 1
/
√
3(1
,
1
,

1).
10
) (a) Dotting the equation
c
1
q
1
+
c
2
q
2
+
c
3
q
3
= 0 with
q
i
gives
c
i
= 0.
This holds for
i
= 1
,
2
,
3.
(b) Say
Q
is
m
×
n
with
m
≥
n
. Then
Q
T
Q
=
I
n
, the
n
×
n
identity matrix. To solve
Qx
= 0
we multiply both side by
Q
T
on the left to get
Q
T
Qx
=
Ix
=
x
= 0 is the only solution.
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 '05
 HUI
 Linear Algebra, Algebra, Vectors, Vector Space, Dot Product, Euclidean vector, ﬁrst vector

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