Math 2940, Prelim 2 Solutions
October 27, 2011
You are NOT allowed calculators or the text. SHOW ALL WORK!
1) (a) (10 points) Find a basis for the subspace of all vectors in
R
4
orthogonal to both
(1
,
0
,
2
,
3) and (1
,
0
,
0
,
−
1).
(b) (8 points) Let
W
⊆
R
4
be the subspace spanned by
{
(0
,
2
,
0
,
7)
,
(1
,
1
,
1
,
4)
,
(
−
1
,
1
,
−
1
,
3)
,
(3
,
1
,
3
,
5)
}
.
What is the dimension of
W
? Find a matrix
A
whose left null space is exactly
W
.
Solution:
(a) Row reducing the matrix
1
0
0
−
1
1
0
2
3
we get the row reduced echelon
form
1
0
0
−
1
0
0
1
2
so the null space of our original matrix has basis
{
(0
,
1
,
0
,
0)
,
(1
,
0
,
−
2
,
1)
}
.
The null space is necessarily the orthogonal complement of the row space.
(b) Row reducing the matrix
1
1
1
4
−
1
1
−
1
3
0
2
0
7
3
1
3
5
we get the row reduced echelon form
1
0
1
1
/
2
0
1
0
7
/
2
. A basis for the null space of our matrix is then
{
(
−
1
/
2
,
−
7
/
2
,
0
,
1)
,
(
−
1
,
0
,
1
,
0)
}
.
Writing these as columns gives us ourt matrix
−
1
/
2
−
1
−
7
/
2
0
0
1
1
0
.
2) (8 points) a) Find the matrix
P
that projects vectors
v
∈
R
3
onto the plane
x
+2
y
−
z
= 0.
b) (8 points) Express the vector (
−
2
,
−
2
,
1) as the sum of a vector in the plane from part
(a), and a vector normal to that plane.
Solution:
(a) The plane is a two dimensional subspace of
R
3
.
The vectors (1
,
0
,
1) and
(0
,
1
,
2) are independent and in the plane so they form a basis. Set
A
=
1
0
0
1
1
2
so the
projection matrix is
P
=
A
(
A
T
A
)
−
1
A
T
=
1
0
0
1
1
2
1
0
1
0
1
2
1
0
0
1
1
2
−
1
1
0
1
0
1
2
=
1
6