Math 2270, Fall 2015
Instructor: Thomas Goller
20 October 2015
Quiz #5 Solutions
(1) For each of the following, determine whether H is a subspace. You do not need to justify
your answer. (3 points)
⇢
7
(a) H = Span
in R2 .
8
Solution: Yes
Comment: Spans are subspaces and every subspace is a span!
(b) H is the set of all polynomials in P3 such that p~(2) = 0.
Solution: Yes
Comment: Check the three conditions for a subset to be a subspace.
(c) H is the set of all matrices in M2⇥3 of the form

a 0 b
.
0 a c
Solution: Yes
Comment: Check that the 0-matrix fits the form and that the form is preserved
under vector addition and scalar multiplication.
(2) Let A =

1 5 0
0 0 1
1 0
. List vectors that span Nul A. (2 points)
1 0
Solution: The free variables in the equation A~x = ~0 are x2 , x4 , and x5 , and the general
solution is
2
3
2 3
2 3
2 3
5x2 + x4
5
1
0
6
7
6
7
6
7
6
7
x2
6
7
617
607
607
6 7
6 7
6 7
x4 7
~x = 6
6
7 = x2 6 0 7 + x4 6 1 7 + x5 6 0 7 ,
4
5
405
415
405
x4
x5
0
0
1
2 3 2 3 2 3
5
1
0
6 1 7 6 0 7 607
6 7 6 7 6 7
7 6 7 6 7
so the vectors 6
6 0 7 , 6 17 , 607 span Nul A.
4 0 5 4 1 5 405
0
0
1
Comment: Do a quick check to make sure the vectors you found satisfy A~x = ~0.
Math 2270, Fall 2015
Instructor: Thomas Goller
20 October 2015
82 3 2 3 2 39
2
1 =
< 1
(3) Determine whether 405 , 435 , 4 3 5 is a basis for R3 . (2 points)
:
;
2
4
2
Possible solution: The vectors are not a basis because they are dependent:
2 3
2 3 2 3
1
1
2
4 3 5 = 3 405 + 435 .
2
2
4
Possible solution:
1 2
0 3
2 4
1
1 2
3 = 0 3
2
0 0
1
3 = 0,
0
so the vectors are dependent and fail to span R3 , hence they do not form a basis.
(4) Bonus problem: Determine whether
M2⇥2 . (1 bonus point)
⇢



1 0
0 0
0 1
1 1
,
,
,
0 0
1 0
1 1
0 1
is a basis for
Possible solution: The vectors are not a basis because they are dependent:




1 1
0 0
1 0
0 1
+
=
+
.
0 0
1 1
1 0
0 1
⇢



1 0
0 1
0 0
0 0
Possible solution: Consider the basis B =
,
,
,
0 0
0 0
1 0
0 1
ing B-coordinates for the vectors in the proposed list, we get
2 3
2 3
2 3
0
1
1




617
607
607
1 1
1
0
0
0
0 1
7,
7,
7,
=6
=6
=6
4
5
4
5
4
5
0 0 B
0
1 0 B
1
1 1 B
1
0 1
0
0
1
These coordinate vectors do not form a basis of R4 because
1
1
0
0
1
0
1
0
0
0
1
1
0
1
1
0
=
0
0
1
0
1
1
1
0
0
0
1
1
0
1
1
0
=
0
0
1
0
1
1
0
0
0
0
1
1
. Comput-
B
2 3
0
617
7
=6
405 .
1
0
1
=0
1
1
since two rows are the same. Thus the proposed vectors are not a basis of M2⇥2 .