Homework week 14
1. Explain why the following set of vectors isNOT basis for the indicated vector spaces
1 2 u 0 3 u 2 7 in V 1 3 2 u 6 1 1 in V b) u
Solutions:
(a) A basis for has two linearly independent vectors therefore this set has one to many elements
(b) A basis for has three independent vectors therefore the set is one element to short
2. Which of the following sets of vectors are basis for ?
a) S 2 1 3 0 b) S 0 0 1 3 Solution:
(a) yes: we can write a generic vector of in the unique form
a b
c 2 1 e 3 0 2c 3e c 0
a) u1
2
1
2
3
2
3
2
3
2
2
and the eqquation this defines is
2c
which has solution if
A
3e
a
3e
b
2 3
0 3
has an inverse
det A
3
which implies that in general we can always find a set of coeficients c e such that the generic
vector a b can be reconstructed in a unique way. This garanties also that
2c
3e
0
3e
0
is satified only by the trivial solution c 0 and e 0 therefore the set S is linearly independent
and so the set of vectors forms a basis for 3
(b) No: We saw a theorem during lectures that states that if the zero is included in the set then
automaticaly the set is linearly dependent.
1
1 0
0 1
0 0
0 0
is a basis for the vector space
0 0
0 0
1 0
0 1
M22
All 2 2 matrices by demonstrating that a) the set is linearly independent and b) that the these
a b
matrices define a generic element of M 22 , lets say the matrix
.Solution:
c d
Taking the generic 2x2 matrix
3. Show that the set S
a b
c d
a
b 00 10 c 01 00 d 00 01 0
00 b0 0c 00 00 0c 0 b
d 1 0
0 0
a
0
a
c
which demonstrates that the basis generates any matrix from M 22 . It only rest to prove that the set of
matrices in S are linearly independent. To do that we write down the equation
0 0
0 0
k1
1 0
0 0
k1 k2
k3 k4
k2
0 1
0 0
k3
0 0
1 0
k4
0 0
0 1
which indicates that the only values of k s that make this equation true are
k1
k2
k3
k4
0
which acording to the definition, implies that the set S is linearly independent. In conclusion the set S is
a basis of M22 .
2