Math 3260
Examples on Chapter 5 – Vector spaces
(Anton & Rorres)
Section 5.2 Exercise 15. Find an equation for the plane spanned by the vectors u=(-1,1,1)
and v=(3,4,4).
First a comment. What I find interesting about this exercise is the connection between
linear algebra and analytic geometry.
To write the vector space spanned by the given vectors we need to consider the subspace
of the form {au  bv : a, b  R} . So this is the plane containing the two vectors described
as a subspace of R 3 . But as we know from section 3.5, a plane in R 3 is given by a linear
equation in x,y and z. How do we get that equation?
Write w  au  bv, w  ( x,y,z). Then x  a  3b, y  a  4b, z  a  4b  y . So in order
to get the linear equation representing the plane we need to obtain an equation in x, y and
z independent of a and b. That equation is simply y  z , or y  z  0 .
Try now to work exercise 16.
Exercise 17. Remember that if we consider a homogeneous system of any order, Ax  0 ,
the solutions form a vector space. It is actually enough to check that they form a subspace
in the space of the unknowns. Indeed, if x1 , x2 are solutions, then
A( x1  x2 )  Ax1  Ax2  0 , and for any scalar k, and solution x, A(kx)  kAx  0 .
The same does not hold true for a nonhomogeneous system. For example if we consider a
system Ax  b , and x a solution then A(2 x)  2 Ax  2b  b if b  0 .
Section 5.3. Exercise 11. What this problem says is that the property of linear
independence is inherited by subsets of linear independent vectors. We saw it in class on
a particular example, in exercise 10. We will give now the general proof.
Take a subset of the given set. Without loss of generality we may assume that the vectors
in the subset are v1 ,..., v k with 1  k  r . If the subset were not linearly independent, there
must exist scalars c1 ,..., ck with at least one nonzero, and such that c1 v1  ...  ck v k  0 .
But then we may complete that linear combination to a linear combination of all the
vectors in the set S, by choosing the remaining coefficients to be zero. Since at least one
of the coefficients of the first k vectors is not zero, we obtain a linear combination of the
vectors v1 , v 2 ,..., v r that equals zero, which means that S cannot be a linearly independent
set of vectors. Since this contradicts the hypothesis, it means that our assumption must be
false, and therefore the subset v1 ,..., v k is also linearly independent.
Exercise 14. Every set with more than 3 vectors from P2 is linearly dependent.
We will prove that a set with 4 vectors from P2 is linearly dependent. This will be enough
to prove the above statement, since in view of exercise 13, if a set of 4 vectors is linearly
dependent, any set having more than 4 vectors will also be linearly dependent.
Let us consider 4 polynomials in P2 , p1 , p 2 , p3 , p 4 , pi ( x)  ai  bi x  ci x 2 , and consider
a linear combination of these polynomials that equals the polynomial zero, that is
k1 p1  k 2 p 2  k 3 p3  k 4 p4  0 . As we have see in other cases, this means that the
polynomial representing the above linear combination has all the coefficients zero. This
leads to a homogeneous system for k1 , k 2 , k 3 , k 4 with three equation in four unknowns.
As we know from the theory of homogeneous linear systems, this system has infinitely
many solutions, and therefore it must have at least one solution different from the trivial
solution. But this says that we may obtain the linear combination above with nonzero
coefficients, and therefore the set of four polynomials is linearly dependent.
Exercise 16. Now this problem does not require a very general proof as the above ones,
but we will try to find the linear combination of the vectors that will equal zero. We have
u  v  v w  w u  0.