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Transcript
9 Subspaces and Bases
We have seen a number of situations when it is convenient to use a different coordinate
system from the original coordinate system. In this chapter we explore this further and
consider coordinate systems for planes in three dimensions and generalizations of this.
9.1 Subspaces
Subspaces are generalizations of lines and planes through the origin. A line through the
origin consists of all multiples of a fixed vector while a plane through the origin consists
of all linear combinations of two fixed vectors. In general, a subspace consists of all
linear combinations of a fixed set of vectors. However, the standard defintion of a
subspace is a little different.
Definition 1. A collection S of vectors is a subspace if it has the following two
properties.
1.
2.
If u is in S and t is a number then tu is in S
If u and v are in S then u + v is in S
By repeatedly applying these two properties it is not hard to show that if u1, …, un are in
S and c1, …, cn are numbers then c1u1 +  + cnun is in S, i.e. a linear combination of
vectors in S is again in S.
There are two common ways that one encounters subspaces. The first is the set of all
linear combinations of a fixed set of vectors.
Proposition 1. Let u1, …, un be a fixed set of vectors and let S consist of all linear
combinations of u1, …, un, i.e. i.e. all vectors v such that v = c1u1 +  + cnun for some
numbers c1, …, cn. Then S is a subspace.
Proof. Suppose v is in S and t is a number. Since v is in S one has v = c1u1 +  + cnun
for some numbers c1, …, cn. So tv = t(c1u1 +  + cnun) = (tc1)u1 +  + (tcn)un =
a1u1 +  + anun where aj = tcj for each j. Since the aj are numbers it follows that tv is in
S. Suppose v and w are in S. So v = c1u1 +  + cnun and w = b1u1 +  + bnun for some
numbers c1, …, cn and b1, …, bn. So v + w = (c1u1 +  + cnun) + (b1u1 +  + bnun) =
(c1 + b1)u1 +  + (cn + bn)un = a1u1 +  + anun where aj = cj + bj for each j. Since the aj
are numbers it follows that v + w is in S. So S is a subspace. //
9.1 - 1
 c1 + 2c2 
3c - 4c
Example 1. Let S be the set of vectors v such that v =  - 5c11 + 6c22  for some numbers c1


 - c1 - 3c2 
and c2. Show that S is a subspace.
 1
 2
3 
-4

Given v of the above form we can rewrite v as v = c1 - 5 + c2 6  = c1u1 + c2u2
 
 
-1
-3
1
2
 
 
3 
-4

where u1 = - 5 and u2 =  6 . So S is a subspace by the previous proposition.
 
 
-1
-3
Corollary 2. Let A be a matrix and let S be the set of all vectors v such that v = Au for
some vector u. Then S is a subspace.
Proof. S consists of all linear combinations of the columns of A so S is a subspace by the
previous proposition. //
Corollary 3. Consider a system of linear equations
a11x1 + a12x2 +  + a1nxn = b1
a21x1 + a22x2 +  + a2nxn = b2


am1x1 + am2x2 +  + ammxn = bm
 b1 
 x1 
b2 
x

Let S be the set of vectors b = … such that there is a solution x =  …2  to the above
 
 
 bm 
 xn 
equations. Then S is a subspace.
 aa
Let A = 
…
a
11
Proof.
21
m1
a12  a1n
a22  a2n
am2  amn

 be the coefficient matrix.

Then S is the set of all
vectors b such that there is a vector x such that Ax = b. So S is a subspace by the previous
corollary. //
Definition 2. If A is a matrix, then the column space of A is the set of all vectors v such
that there is a vector u such that Au = v.
In other words, it is the set of all v such that one can find a solution to the equation Au = v.
If we define a function f by f(u) = Au, then the column space of A would be what is often
called the range of f.
9.1 - 2
The second common way subspaces arise is as the set of all solutions to a set of
homogeneous linear equations.
 aa
Let A = 
…
a
11
Proposition 4.
21
m1
a12  a1n
a22  a2n
am2  amn

 be a matrix and let S be the set of all vector

 x1 
x
x =  …2  such that
 
 xn 
(1)
a11x1 + a12x2 +  + a1nxn = b1
a21x1 + a22x2 +  + a2nxn = b2


am1x1 + am2x2 +  + ammxn = bm
Then S is a subspace.
 x1 
x
Proof. Note that x =  …2  is a solution to the system (1) if an only if Av = 0. So the
 
 xn 
proposition is actually a corollary to the following proposition. //
Proposition 5. Let A be a matrix and let S be the set of all vectors x such that Ax = 0.
Then S is a subspace.
Proof. Suppose x is in S and t is a number. Since x is in S one has Ax = 0. So
A(tx) = t(Ax) = t0 = 0. So tx is in S. Suppose x and y are in S. So Ax = 0 and Ay = 0. So
A(x + y) = Ax + Ay = 0 + 0. So x + y is in S. So S is a subspace. //
Definition 3. If A is a matrix then the null space of A is the set of all vectors x such that
Ax = 0.
One question that arises is the following. Given a subspace S that is the set of vectors
 x1 
x
x =  …2  that satisfy a homogeneous system of equations (1), can we find a fixed set of
 
 xn 
9.1 - 3
vectors u1, …, un such that S consists of all linear combinations of u1, …, un? The answer
is yes and we do it simply by solving the set of equations using elimination as in chapter
3. We illustrate this by means of an example.
x
y
Example 2. Let S be the subspace consisting of vectors v =  z  such that
 
w
x + 2y + 3z + 5w = 0
2x + 4y + 8z + 4w = 0
(2)
3x + 6y + 12z + 6w = 0
Find u1, …, un such that S consists of all linear combinations of u1, …, un.
1 2 3 5 | 0
The augmented matrix is M =  2 4 8 4 | 0 . We use elementary row operations to
 3 6 12 6 | 0 
bring it to reduced row echelon form.
 1 2 3 5 | 0  R2 - 2R1  R2  1 2 3 5 | 0  R2/2  R3  1 2 3 5 | 0 
 2 4 8 4 | 0  R3 - 3R1  R3  0 0 2 - 6 | 0  R2/3  R3  0 0 1 - 3 | 0 
 3 6 12 6 | 0 
0 0 3 -9 | 0
0 0 1 -3 | 0


R1 - 3R2  R1  1 2 0 4 | 0 
R3 - R2  R3  0 0 1 - 3 | 0 
0 0 0 0 | 0

So a set of equations equivalent to (2) is
x + 2y
+ 4w = 0
z - 3w = 0
0 = 0
or
x = - 2y - 4w
z =
3w
So
x
 - 2y - 4w 
-2
-4
y
y
1 
0




v = z =
= y 0 + w  3  = yu1 + wu2
 
 3w 
 
 
w
 w 
 0 
 1 
where
u1
-2
1
=  0 
 
 0 
u2
-4
0
=  3 
 
 1 
So S consists of all linear combinations of u1 and u2. So S is a plane lying in four
dimensional space.
9.1 - 4