Download SOME REMARKS ON BASES Let V denote a vector space. We say

SOME REMARKS ON BASES
Let V denote a vector space. We say that a set of vectors {A1 , A2 , · · · } spans V if
V = Span {A1 , A2 , · · · }.
We may also say that {A1 , A2 , · · · } is a spanning set for V. Notice we are not requiring the set of
vectors to be finite here.
Now, suppose we have a vector space V and we have a spanning set {A1 , A2 , · · · }. What does
the spanning set do for us? The spanning set should be thought of as providing us basic building
blocks for our vector space, in kind of1 the same way the atoms from the periodic table form the
basic building blocks for all other molecules and larger chemical complexes, or how the primary
colors red, blue, and yellow combine to give us all other colors.
Typically, vector spaces are very large complicated things. Even the ”easy” vector spaces, like
R, R2 , R3 , are very large and complicated objects. However, most questions in matrix or linear
algebra decompose in a nice way with respect to linear combinations. This usually means that if
you want to answer some question about a vector space, then it will be enough to be able to answer
the question when restricting to a spanning set. Essentially, the spanning vectors describe all the
different directions in V, and each direction just looks like a copy of R.
x
1 2
, then
and you want to understand the effect of multiplying A by
For example, if A =
y
0 1
one way2 is to use the linearity properties
of matrix multiplication and the fact that {e1 , e2 } is a
1
0
spanning set for R2 , where e1 =
, e2 =
. Then
0
1
x
A
= A(xe1 + ye2 ) = xAe1 + yAe2 .
y
We only have to compute the products of A with the two spanning vectors. We find
1
2
Ae1 =
, Ae2 =
.
0
1
We therefore find that
x
1
2
x + 2y
A
=x
+y
=
.
y
0
1
y
1The
chemistry and painting examples are only metaphorical, and should not be taken too literally.
course, another way is to just to carry out the multiplication process. However, the point of view we develop
here turns out to be a very useful one, despite its apparent inefficiency.
2Of
1
2
SOME REMARKS ON BASES
{A1 , A2 , · · · } being a spanning set for V says that every vector in V can be built up from
{A1 , A2 , · · · }. Working with a set that does not span the space V would be like trying to do
chemistry without carbon, or trying to paint only using red and blue, and the purplish colors we
get by mixing these two. We can do some chemistry without carbon, but we will be in trouble if
we need graphite or carbon dioxide. We can paint some paintings without yellow, but we will be
in trouble if we need yellow or green or orange. We can describe some chunk of our vector space V
with a non-spanning set. However, if we only knew
what Ae1 was in the above example we would
2
be in trouble if we wanted to multiply A by, say,
, because this vector does not sit in the chunk
5
of R2 given by the span of e1 .
Recall that we say V is finite dimensional if we can write
V = Span {A1 , A2 , · · · , Ak }
for some finite collection of vectors {A1 , A2 , · · · , Ak } in V. In other words, V is finite dimensional
provided that V has a finite spanning set. This means that in some sense there are only a finite
number of different directions in our vector space. In the plane we only need two directions, say
North and East. There are plenty of other directions, but they can all be described in terms of
North and East.
Finite dimensional spaces have the advantage that we really only need to understand what happens in a finite number of directions. There doesn’t seem to be much consolation in knowing
that to understand the space of all polynomials we only need to understand all of the monomials
1, x, x2 , · · · . There are just too many directions here! Understanding what happens for the three
standard basis vectors of R3 seems like a much more manageable task.
Some spanning sets are better than others. For
0, 0], [0, 1, 0], [0, 0, 1]} is a much
√ instance,2 {[1,
√
3
nicer spanning set for R than {[1, 2, 3], [3, −1627, 157], [π, π , π], [2−19 , 9, 3]}. The second is bad
for two reasons. First, given an arbitrary vector in our space, say [1, 36, −7], we would like to be
able to find coefficients expressing our vector in terms of the spanning set. For the first basis this is
trivial; [1, 36, −7] = 1[1, 0, 0] + 36[0, 1, 0] + (−7)[0, 0, 1]. I wouldn’t want to have to find coefficients
for [1, 36, −7] in terms of the second spanning set. Of course, if we absolutely had to we could find
them, it just would be a pain. There is a rather precise reason why the first basis is better than
the second (It has to do with dot products) but we won’t discuss this until chapter 6. Suffice it to
say for now, it is easy to read off the coefficients in the first case. The second reason that the first
spanning set is better than the second is much easier to pin down: The second set is too big! We
have 4 vectors in a 3dimensional space, so there is redundant information in the second spanning
set. More precisely, there must be some linear dependency in the second set of vectors. (What
the dependency is is not so clear due to the ugliness of the vectors, but never the less, there is a
dependency.) Linear dependencies in a spanning set means that the spanning set is not as efficient
as it could be. This leads to the important definition of a basis for a vector space.
SOME REMARKS ON BASES
3
A basis for V is a linearly independent spanning set {A1 , A2 , · · · } for V. Spanning says that
everything in V can be built from {A1 , A2 , · · · }. Independence says we can only build things up
in a single way. Going back to the chemistry example, we know that water can be built up from
things in the periodic table, namely hydrogen and oxygen. We also know that this is the only way.
We can’t build up water from, say, uranium, cobalt and chlorine. We know that yellow and blue
make green. We also know we can’t get green by combining, say purple and orange.
Now, why should we be interested in finding bases for R2 if we already have the standard basis,
which seems to be the best kind of basis we could use? The reason is that the standard basis is not
always the best basis to use. There is no best basis for any vector space. Sometimes we may wish
to choose a basis for our vector space which is adapted to a particular problem we are studying.
Think back to calculus when we study polar coordinates. (This does not quite correspond to finding
a new basis for R2 because polar coordinates are what are called curvilinear coordinates, but the
metaphor is still helpful.) Certain integrals are nasty, if not impossible in cartesian coordinates,
but become almost trivial when we pass to polar coordinates. On the other hand, certain nearly
trivial integrals become impossible when we change to polar coordinates. The general rule is that
if the integrand has some sort of circular structure to it then it is probably advantageous to change
to polar. In matrix algebra, it is usually very helpful to find a basis that is engineered to help solve
whatever problem we are working on. In hindsight, we see that the whole row reduction algorithm
we do when solving a linear system is really just a construction of a nice basis for the null space of
the coefficient matrix.
One example of finding a basis for Rn different from the standard basis arises in solving systems
of equations. Suppose A is some m × n matrix and we want to study the equation Ax = b for
all possible b ∈ Rm . In studying this problem, it is helpful to find a basis adapted to A. Refer
back to the handout I gave on Thursday for a detailed example, but roughly the idea is to start
by finding bases for the column space of A (which is a subspace of Rm ) and the null space of A
(which is a subspace of Rn . Say the column space basis is {b1 , b2 , · · · , bk } and the null space basis
is {y1 , y2 , · · · , yn−k }. Next, for each basis vector for the column space, we find a vector xj ∈ Rn so
that Axj = bj . Then the set {y1 , y2 , · · · , yn−k , x1 , x2 , · · · , xk } is a basis for Rn and is in some sense
much better than the standard basis, as far as working with systems of equations with coefficient
matrix A is concerned. Given b in the column space of A, we want to find all solutions to Ax = b.
To do this, we simply write b in terms of the column space basis,
b = c 1 b1 + c 2 b2 + · · · + c k bk .
Then a particular solution is
x = c1 x1 + c2 x2 + · · · + ck xk .
(Look at how nice this is! All we did is take the coefficients from the column space basis and moved
them over to a different basis!) The general solution is then obtained by adding things from the null
space to this particular solution. This basis is nice for solving systems of equations with coefficient
matrix A because we chose basis elements which had particularly nice vectors when multiplied by
A.