Math 294
Prelim 2 solution
Spring
2008
1.
The matrix
A
of a linear transformation
T
from
2
3
→
R
R
is
1
1
1
0
0
1
A
=
1a.
Find a basis for
Im A
, what is the dimension of
Im A
T
?
1b.
Find a basis for Ker
A
T
1c
Is the equation
Ax
b
=
G
G
consistent where
(3,3,3)
T
b
=
G
, if not, find the least square solution of
Ax
b
=
G
G
.
Solution
:
(This problem is a modification of a HW in 5.4, pb20)
1a:
the two columns of A are linear independent vectors, so Im A is the span of these two
column vectors.
Since the rankA
T
= rankA, dim
Im A
T
=dim Im A = 2.
1b:
The Ker A
T
is the solution of the equation
1
1
0
1
0
1
0
x
=
G
G
It is also the vector subspace of
3
R
which is orthogonal to Im
A
.
By inspection, the vector
( 1,1,1)
T
−
is orthogonal to the columns of A which spans the ImA.
Since the KerA
T
=
(
)
Im
A
⊥
and
the dim(Im A) = 2, the dim( KerA
T
) = 1, so
(
)
( 1,1,1)
T
T
KerA
span
=
−
.
1c.
A simple calculation using RREF shows that the system
Ax
b
=
G
G
is inconsistent.
The solution
of 1c is a HW problem, pb. 20, section 5.4.
2.
Solution
:
2a:
[
]
1
1
2
3
1
1
2
1
1
1
2
1
0
2
2
M
f
f
f
M
=
=
+
+
⇒
=
−
B
where
1
2
3
1
0
1
1
0
0
,
,
0
0
0
0
0
1
f
f
f
=
=
=
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 Fall '07
 Math, Linear Algebra, Algebra, basis, Ker, F1, linear independence. f1

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