Geometria Lingotto.
LeLing6: Vector subspaces.
Contents:
¯
• Subspaces.
• Linear systems and linear combinations.
• Sum and intersection of subspaces.
• Spaces of rows and columns of a matrix.
Recommended exercises: GeoLing 8,9.
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1
Linear combinations and subspaces.
Let V be a vector space and v1 , v2 ∈ V two vectors. Two linear combinations c1 v1 + c2 v2
and d1 v1 + d2 v2 can be added:
(c1 v1 + c2 v2 ) + (d1 v1 + d2 v2 ) = (c1 + d1 )v1 + (c2 + d2 )v2 .
The sum of two linear combinations of v1 , v2 is still a linear combination of v1 , v2 .
Thus calling L(v1 , v2 ) the set of all linear combinations of v1 , v2 has the effect that
L(v1 , v2 ) is closed under linear combinations, otherwise said v, w ∈ L(v1 , v2 ) implies
av + bw ∈ L(v1 , v2 ) for any coefficient a, b.
Now we define subspaces.
Definition 1.1. Let V be a vector space and W ⊂ V a non-empty subset. Then W is
called (vector) subspace of V if it is closed under linear combinations, i.e. v, w ∈ W
implies av + bw ∈ W for any a, b.
Clearly
1
L(v1 , v2 ) is a subspace of V. More generally,
Definition 1.2. Take vectors v1 , v2 , · · · , vn ∈ V and let L(v1 , v2 , · · · , vn ) be the set of
their linear combinations, L(v1 , v2 , · · · , vn ) := {c1 v1 + c2 v2 + · · · + cn vn } ⊂ V. Then
L(v1 , v2 , · · · , vn ) is a subspace of V, called the subspace generated or spanned by
v1 , v2 , · · · , vn .
1
the definition is designed precisely for this reason.
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Geometry
1.1 Homogeneous systems and subspaces L(s1 , s2 , · · · , sk )
Geometria Lingotto.


x
Example 1.3. Let C3 := { y } be the vector space of column vectors with three enz  
 
 
1
0
x





0
0 . Then
y  ∈ L(e1 , e3 ) iff y = 0. Therefore
tries. Let e1 =
and e3 =
0
1
z
L(e1 , e3 ) denotes the set of column vectors whose second component “y ” vanishes.
 
x
Example 1.4. Let C3 := { y } the space of column vectors with three components.
z
 
 
 
1
0
0
Call e1 =  0  , e2 =  1  and e3 =  0  . Then L(e1 , e2 , e3 ) = C3 because any
0
0
1
column is a linear combination of e1 , e2 , e3 :


x
 y  = xe1 + ye2 + ze3 .
z
Theorem 1.5. Let V be a K-vector space and W ⊂ V a subset. Then W is a subspace
of V iff:
• 0 ∈ W,
• For any r ∈ K and v ∈ W, r.v ∈ W.
• For any v, w ∈ W, v + w ∈ W.
Note that 0 belongs to every subspace of W. In fact, since the subspace is not empty,
there exists an element v ∈ W. The linear combination 0.v belongs, by definition, to
W, so 0.v ∈ W. It is easy though to show that 0.v = 0 using the axioms of vector
spaces.
1.1
Homogeneous systems and subspaces L(s1 , s2 , · · · , sk )
Considering carefully example (1.3) we notice the solutions of the system {y = 0 are the
columns of the subspace L(e1 , e3 ).
But the solutions of a homogeneous system can be expressed as linear combinations
of a certain number of columns, as the following example shows:
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Geometry
1.2 Sum and intersection of subspaces
Geometria Lingotto.




x + 2y + 3z + 4w
4x + 6y + 7z + 8w
Example 1.6. The system S =
5x
+ 8y + 10z + 12w



10x + 16y + 20z + 24w

=0
2z + 4w

=0
− 25 z − 4w
is solved by 

=0
z
=0
w

2z + 4w
 − 5 z − 4w 
2
 is a linear combination of two columns
The general solution 


z
w




2z + 4w
2
 − 5 z − 4w 
 −5
 2
 = z 2


 1
z
0
w



4



 + w  −4 

 0 
1
 
2
4
 − 5   −4
2  
The set of all solutions is thus the subspace L(
 1 , 0
0
1



).

Studying carefully the procedure for solving a homogeneous system S unveils that
we are actually searching for column vectors s1 , s2 , s3 , · · · , sk of which the solutions
are linear combinations; solving a homogeneous system means finding the subspace
L(s1 , s2 , · · · , sk ) containing all solutions. The reduction of Gauß-Jordan is a simple
method to determine the columns s1 , s2 , s3 , · · · , sk , given the system S .
1.2
Sum and intersection of subspaces
Let W1 , W2 be subspaces of a vector space V.
Definition 1.7. The sum W1 + W2 is
W1 + W2 := {w1 + w2 : w1 ∈ W1 , w2 ∈ W2 }.
The intersection is
W1
\
W2 := {w : w ∈ W1 and w ∈ W2 }.
Theorem 1.5 ensures that both W1 + W2 and W1
in fact we have:
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T
W2 are subspaces. For W1 + W2
Geometry




1.3 Complementary subspaces
Geometria Lingotto.
• 0 ∈ W1 + W2 , for 0 = 0 + 0.
• If r is a number and w ∈ W1 + W2 , then rw = r(w1 + w2 ), where w = w1 + w2 ,
and using the axiom of distributivity rw = rw1 + rw2 . Hence rw ∈ W1 + W2 ,
because rw1 ∈ W1 and rw2 ∈ W2 .
• If v1 , v2 ∈ W1 +W2 it follows v1 +v2 = (w1 +w2 )+(w10 +w20 ), with v1 = w1 +w2 , v2 =
w10 + w20 . Therefore v1 + v2 = (w1 + w10 ) + (w2 + w20 ) ∈ W1 + W2 , since w1 + w10 ∈ W1
and w2 + w20 ∈ W2 .
If W1 = L(w1 , w2 , · · · , wk ) and W2 = L(v1 , v2 , · · · , vn ), we have
W1 + W2 = L(w1 , w2 , · · · , wk , v1 , v2 , · · · , vn ) .
Now an interesting application of the intersection W1
T
W2 .
Proposition 1.8. Let W1 , W2 beTthe solution sets of the homogeneous systems S1 , S2
respectively. The intersection W1 W2 is the solution set of the combined system, obtained putting the equations of S1 and S2 together.
x + 2y + 3z + 4w = 0
Example 1.9. Call W1 the solution set of S1 =
and W2
4x
+
6y
+
7z
+
8w
=
0
T
5x + 8y + 10z + 12w = 0
that of the system S2 =
. Then W1 W2 is the set of
10x + 16y + 20z + 24w = 0
solutions of the system

x + 2y + 3z + 4w = 0



4x + 6y + 7z + 8w = 0
S=
5x + 8y + 10z + 12w = 0



10x + 16y + 20z + 24w = 0
1.3
Complementary subspaces
Let W1 , W2 be subspaces of the vector space V. One says W1 and W2 are complementary
when:
(a) V = W1 + W2 ,
\
(b) {0} = W1 W2 .
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Geometry
Geometria Lingotto.
In the case, W2 is a complement of W1 and vice versa.
Condition (a) says that any vector is the sum of a vector in W1 and one in W2 ,
i.e. for any v ∈ V there are w1 ∈ W1 and w2 ∈ W2 such that v = w1 + w2 . Condition
(b) implies uniqueness, hence that v can be written as sum of w1 , w2 in a unique way:
if v = w1 + w2 = w10 + w20 Tthen w1 = w10 , w2 = w20 . Namely, w1 + w2 = w10 + w20 implies
w1 − w10 = w2 − w20 ∈ W1 W2 = {0}, hence w1 = w10 and w2 = w20 .
2
Two important subspaces
Given a matrix A = (aij ) we may view it in terms of its rows or its columns.


1 0 −2 −4
4  . If we wish to emphasize the rows, we write
Example 2.1. let A =  0 1 25
0 0 0
0


R1
A =  R2  where R1 is the first row, R1 = (1 0 − 2 − 4), R2 the second one and R3
R3
 
1
the third. Considering columns, instead, we put A = (C1 C2 C3 C4 ), with C1 =  0  ,
0
and so on.
Associated to A ∈ Mn,m are two very important subspaces, indicated as RA and
CA :
RA := L(R1 , R2 , · · · , Rn ) ,
CA := L(C1 , C2 , · · · , Cm )
RA is a subspace of the vector space Rm of row vectors with m entries, so it is called
row space (or space of rows) of A.
Since CA is a subspace of the space Cn of column vectors with n components, and
is called column space (or space of columns) of A.


1 0
Example 2.2. The matrix A =  0 1  has row space RA = L((1 0), (0 1), (0 0)) and
0 0
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Geometry
Geometria Lingotto.

  
1
0



CA = L( 0 , 1 ) as space of columns.
0
0
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Geometry