Problem 1
Construct a matrix whose nullspace consists of all combinations of 2
6
6
4
2
2
1
0
3
7
7
5
and
2
6
6
4
3
1
0
1
3
7
7
5
.
1
x.
Xs
Xs
×4
Special solution I
I⑤⑤Isetx=o
pivot
free
S2=[§)
'
Specislsdutiona
set
Xs=o
,
#=L
Sa
-
[kg)Problem 2 [
Construct a matrix whose column space contains 2
4
1
1
5
3
5 and 2
4
0
3
13
5 and whose
nullspace contains 2
4
1
1
2
3
5.
2
A- [I § I]
we must hare A. [If
=
④
Hence
.
÷÷÷÷ :
:÷
C
=
-
3
Thus
.
a- I! ! I;]Problem 3 ]
Let u1 = 2
4
1
0
0
3
5, u2 =
2
4
1
1
0
3
5
,
u
3
=
2
4
1
1
13
5, u4 = 2
4
2
3
4
3
5. Show that u1, u2, u3
are independent but
u1, u
2
, u
3
, u
4
are dependent.
3
Ci)
Us
,
Ua
,
uz
are
independent if
-
f Lik
u,
y
)
,
has three pivots
.
Indeed
,
i: : : ni
: : :3
Cii) A 3×4 matrix always has dependent columns
.
[
Ur Us
Us
Ua
]
is 3×4
,
hence
U2
.
U2 Mesilla are
dependent
.Problem 4 [
For which numbers c, d does the following matrix have rank 2?
A = 2
4
12505
00c22
000d2
3
5
4
Ant: : ⇒ ⇒ sheath
'
:&:
.
Suppose
c
-
to
,
then d must be 0
,
otherwise the kurth
column is a
pivot
as well
,
hence the rank
would be 3
.
Since
,
d
= o
,
the fourth column is a
pivot
,
thus again
we have rank 3
.
So
,
we
need to have C
-
-
o
.
Now
,
the fourth column
is
a
pivot and in order
to not have the fifth column
as
a
pivot
we
must hare 01=2
.Problem 5 []
Find a basis for each of the four fundamental subspaces (column, null, row,
left null) associated with the following matrix:
A = 2
4
01234
01246
00012 3
5
5
free
specialisation
A-
E
: :& I
"
s.
column space can.ci:p
,
,
'
'
II:/
Nuelspace
NCH
-
LI:o)
,
't;)
,
Row
space
CAT
-
-
<
[Ez)
.to?no.1sLeftnueespaceNh-7oE:'qEInE!:ogEgif~
I :÷
.
÷:
*.m*a⇒
.Problem 6 [
Suppose that S is spanned by s1 = 2
6
6
4
1
2
2
3
3
7
7
5
, s2 =
2
6
6
4
1
3
3
2
3
7
7
5
. Find two vectors that
span the orthogonal complement S?. (Hint: this is the same as solving
Ax = 0 for some A)
6
Let A
=
[ kg kg]
.
The left null space is
orthogonal to
the column space
.
Hence
,
we
need to find NEAT
)
free
at
.
I: : : :3 rt: : : ⇒
Special solutions
set!)
sa
-
-
I
.
Hence
5-
=
LE;)
,
[!)
>Problem 7
Suppose P is the subspace of R4 that consists of vectors 2
6
6
4
x
1
x2
x
3
x4
3
7
7
5
that satisfy
x1 + x2 + x3 + x4 = 0. Find a basis for the perpendicular complement
P ?
of P.
7
P is the null space of LI
L
I
1)
=
A
The orthogonal complement of Nat
is the
row
space of A which is C CAT =L
Ez)
>
