Departament of Economic, Financial, and Actuarial Mathematics
Workshop 3
Mathematics I
Workshop 3. Solutions
Linear Span, Basis, and Dimension
Exercise 1.
a) Yes. The line passing through (0,0) and (3,1
2).
b) No. A single vector is not enough to span a 2-dimensional space. We need at least two
vectors.
c) No. The two vectors are linearly dependent. Thus, their linear span is a line.
d) Yes. The vectors are independent. Then, they span a 2-dimensional space.
e) Yes. The same reason as above.
f) Yes. Because among the three there are two linearly independent vectors.
g) No. The vectors are in R2and then, they cannot span R3.
Exercise 2.
a) Yes. The two vectors are linearly independent. Hence, they span a 2-dimensional space,
that is, a plane.
b) No. The two vectors are linearly dependent, which means that they span a line.
c) No. The matrix built from the vectors has rank 2. This means that they span a plane.
d) No. The rank of the matrix built from the vectors has rank 2, which means that their linear
span is a plane in R3.
Exercise 3. Yes.
Exercise 4. A set of vectors is a spanning set of R3if the vectors are in R3(X) and the rank
of the matrix built from the vectors is equal to the dimension of R3, that is, 3. Let
A=
−1 6 1 k
3−4 4 −k
2−2 3 1
.
Studying the 4 minors of the 3-by-4matrix we obtain that rk(A)=3⇔k6=−7/4. Then,
{~u1, ~u2, ~u3, ~u4}is a spanning set of R3for all values of kdifferent from −7/4.
Exercise 5. True. In order a set of vectors to be a basis of Rnwe need it to be a spanning set of
Rnand to be a linearly independent set. Since any vector of Rnis a linear combination of the
vectors in the set it is a spanning set of Rn. Since no vector of the set is a linear combination
of the rest the set is linearly independent.
Exercise 6. For any value of adifferent from 0and 2.
M. Álvarez Mozos 1 ADE - 2013/2014
