Download MATH 307 Subspaces

Dr. Neal, Fall 2008
MATH 307
Subspaces
Let V be a vector space. A subset W is a subspace of V provided
€
€
€
€
(i) W is non-empty
(ii) W is closed under scalar multiplication, and
€
€
(iii) W is closed under addition.
In other words, W is just a smaller vector space within the larger space V . But
adding elements from W keeps them in W as does multiplying by a scalar.
€
2
Example 1. Let
€ V = R (the x y€plane). Let W = {( x , y ) : y = m x }. Then W is simply a
straight line through the origin. Is W a subspace?
(i) The point (0, 0) is in W ; hence, W is non-empty.
€ for any scalar
€, €
€c ∈ ℜ
€ c (x, y) = (c x,€c y) . But is this point
x , y ) ∈ W , then
(ii) If ( €
we have
€
still on the line? Yes because c y = c (m x) = m (c x) ; hence, c (x, y) is still in W .
€ (x ,y )€and (x , y ) are in W . Then y = m x and y = m x . Also,
(iii) Now suppose
1 1
2 2
1
1
2
2
€€ €
€y )
and
;
hence,
(x1,y1) + (x 2 , y2 ) = ( x1 + x 2 ,y1 +
y
+
y
=
m
x
+
m
x
=
m
(x
+
x
)
2
1
2
1
2€
1
2
(x1,y1) + (x 2 , y2 ) is still in W .
€
2
Because Properties (i) , (ii), and (iii) are satisfied, W is a subspace of R .
€
What if the line did not go through the origin? Consider U = {( x , y ) : y = m x + b }
€ c ≠ 1, c y = c (m x + b) = m(c x) + c b ≠ m x + b ;
for some b ≠ 0. Then for ( x , y ) ∈ U and
hence, U is not closed under scalar multiplication. It also can be shown that U is not
closed under addition; but only one property must be disproved to show that U is not a
€€
€
€
subspace.
€
€
€
€ €
€
€

€
Theorem 3.3. Let W be a subspace of a vector space V . Then 0 ∈ W .
Proof. Because W must be non-empty, there exists some element w ∈ W . Because W is
closed under scalar multiplication, we then have 0 = €
0w ∈W .
€
€

Corollary.
If 0 ∉ W , then W cannot be a subspace.
€ €
€
€
€ € €
€
Example
2. Let V €be the vector space of all 3 × 3 matrices. Let W = { A ∈ V : A−1 exists}.
€€ €
Is W a suspace of V ? In
 this vector space, the 0 element is the 3 × 3 zero matrix, which
is non-invertible. Thus, 0 ∉ W , and W is not a subspace of V .
€
€
€
€ €
€
€
€
€
€€ €
€
€
Dr. Neal, Fall 2008
Proving that a Subset is a Subspace
Let W be a subset of a vector space V . Then W must have some defining property that
distinguishes its elements from the other elements in V . To show that W is actually a
subspace, we must verify three properties: (i) W must be non-empty; (ii) W must be
closed under scalar multiplication; and (iii) W must be closed under addition.
€
€

€
€
(i) The zero vector 0 must always
be in any subspace. So to show W is non-empty, it is

€
€
usually easiest to show that 0 is in€ W . That is, you must argue that the specific zero
vector of the vector space V under consideration has the defining property of the set
W . Then conclude
that W is non-empty.
€
€
€
€
(ii) To show W is closed
under
scalar multiplication, (a) Let w ∈ W . (b) Then state
what it means for€w to be in W ; i.e., explain the defining property. (c) Let c be any
scalar (i.e., real€number). (d) Argue that c w has the defining property of W . (e)
Conclude that W is closed under scalar multiplication.
€
€
€
€
(iii) To show W is closed under addition, (a) Let w1 and w 2 be in W€. (b) Then state
€ the defining
€ in W ; i.e., explain that they have
what it means for w1 and w 2 to be
€ (c) Argue that w + w has the defining property of W . (e) Conclude that W
property.
1
2
is closed under addition.
€
€
€
€
€ of V .
€ that W
€ is a subspace
(iv) Conclude
€
€
€
Throughout, you may use different symbols that are more appropriate to the
context of the problem. But throughout, use complete sentences to explain what you are
€ logic, do not skip steps, and give conclusions.
doing, use€correct mathematical
€
€
Note: We also can combine requirements (ii) and (iii) and show instead that for any
scalar c and vectors w1 and w 2 in W , that c w1 + w 2 ∈ W .
€
€
€
€
€
€
€
€
Example 2. Let V = C[a, b] be the set of continuous functions over [a, b], where a < b .
€
€
′(d) exists}
Let a < d < b€and let€D = { f ∈ V f €
. Then D is a subspace of V .
Proof. Let f 0 be the zero function (i.e., f 0 (x) = 0 for all x ∈ [a, b]). Then f€
0′ (x) = 0 for all
€
€
′
D
;
thus,
exists
(and
equals
0).
So
and
is
non-empty.
x ∈ (a, b)
f 0 (d)
f0 ∈ D
€
€
€
Let f ∈ D . Then f ′(d) exists. For any scalar c , cf is still a continuous function and
€)′(d) = c( f ′(d)) still exists. Thus,
€
€ scalar multiplication.
(cf
cf ∈ D and€D is closed under
€
€
€
Let f , g ∈ D so that f ′(d) and g′(d) exist. Then f + g is still a continuous function and
€
( f + g)′(d)€= f ′(d) + g′(d) still exists. Thus, €f + g ∈ D and D is closed under addition.
€
€
Whence, D is a subspace of C[a, b] .
- - - - - - - - -€
- - - - - - - -€
- - - - - - - - - - - - - -€- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - €
€
3
Example 3. Let V = R and let W = {( x , y , z ) : z = 0} . Then (0, 0, 0) ∈ W ; so W is nonempty. For vectors (x1, y€1, 0) and (x 2 , y 2 , 0) in W and for any scalar c , we have
€
3
c (x1, y1, 0) + (x 2 , y 2 , 0) = (c x1 + x 2 , c y1 + y 2 , 0) ∈ W . Hence, W is a subspace of R .
€ € € €
€
€
€
€
€
€
€
€
€
€
Dr. Neal, Fall 2008
n
m
n
Example 4. Let T : R → R be a linear transformation. Then ker T is a subspace of R .

n
m
Recall: For T : R → R , ker T = {x ∈ R n T(x) = 0 m } .
€
€


n
m
Proof. Let 0 n be the zero vector in R and let 0 m be the zero vector in R . Then for any



n
m
linear transformation T : R → R €
, we have T(0 n ) = 0 m . Thus, 0 n ∈ ker T . Hence, ker T
is non-empty.

n
T(x)
=
0
€ Let x ∈ ker T . Then
m .€ If c is any scalar, then c x is still in R and

T(c x) = c T(x) = c 0 m = 0 m ; hence, c€x ∈ ker T . Thus,€ ker T is closed €under scalar
multiplication.




Let u, v ∈ ker T . Then T(u)€= 0 m = T(v) . Then T(u + v) = T(u) + T(v) = 0 m + 0 m = 0 m ;
€
€
n
hence, u + v ∈ ker T and ker T is closed under addition.
Ergo, ker T is a subspace of R .
€
------------------------------------------------------------------€
€
n € m
→ R has an m × n matrix representation A .
T
R
Recall that a linear
transformation
:
€ Then in matrix€form, ker T is simply all solutions to€the homogeneous system AX = 0 .
Thus, we can re-state Example 4 in matrix form as:
×1 matrices. Let A be an m × n matrix.
Example 5. Let V be the vector space of all n €
The set of all€n ×1 matrices X such that AX = 0 forms a subspace of €V . That is, the
solution space to the homogeneous system A X = 0 is a subspace of V .
------------------------------------------------------------------€n
€
€


Example
€ 6. Let S = { u1, . . ., um } be
€ a collection of m vectors in
€ R . The set span(S) is
the collection of all linear combinations of the vectors in S€
. Then span(S) is a subspace
n
of R .
€
€
€
 € 

Proof. (i) First, 0 n = 0 u1 + . . . + 0 um is a linear €combination of the vectors in S ; thus,

0 n ∈ span(S) and span(S) is non-empty.
€




(ii) Next, suppose u ∈ span(S) . Then u can be written as u = c1 u1 + . . .€
+ c m um . For any
€ have
scalar c ,€we then





c u = c(c1u1 + . . . + c m um ) = (cc1 ) u1 + . . . + (cc m ) um ,
€
€
€

€
which is still a linear combination of the vectors in S . Thus, c u ∈ span(S) and span(S)
is closed under scalar multiplication.
€


 



(iii) Now let u, v ∈ span(S) . Then u and v can be written as u = c1u1 + . . . + c m um and



€
€
v = d1u1 + . . . + dm um . Then
€
€
 




u + v = (c1u1 + . . . + c m um ) + (d1u1 + . . . + dm um )
€
€ 
€
= (c1 + d1 ) u1 + . . . + (c m + dm ) um
 
which is still a linear combination of the vectors in S . Thus, u + v ∈ span(S) and span(S)
n
is closed under addition. Therefore, span(S) is a subspace of R .
€
€
€
Dr. Neal, Fall 2008
Other Examples of Subspaces / Non-Subspaces
(i) Let Pn be the vector space of real polynomials having degree ≤ n .
(a) For 0 ≤ m < n , Pm is a subspace of Pn .
€
(b) The subset of those polynomials having only non-negative coefficients
{a€0 + a1x + . . . + an x n : all ai ≥ 0} is not a subspace because it is not closed under
multiplication by negative scalars.
(ii) Let F(–∞, ∞) be the vector space of all real-valued functions f with domain (–∞, ∞).
(a) The subset of continuous functions C(–∞, ∞) is a subspace (the sum of
continuous functions is still continuous as is the scalar product, and there does exist at
least one continuous function).
(b) Fix one x0 . The subset of functions f such that f (x0 ) = 0 is a subspace of
f (x) = x − x0
is in the subset, and
F(–∞, ∞) . Indeed, the function
(c f + g)(x 0 ) = c f (x0 ) + g(x 0 ) = 0 for functions f and g in the subset; hence, the subset is
non-empty and closed.
(c) The subset of functions f such that lim f (x) = b is not a subspace for b ≠ 0.
x →a
Clearly, this subset is not closed under scalar multiplication for c ≠ 1. But is it a
subspace for b = 0?
€
(iii) Let Mn, n be the vector space of all n × n matrices.
€
€
(a) Let W = { A : Tr(A) = b } . Then W is a subspace if and only if b = 0.
€
T
(b) Sn,n = A : A = A ( n × n symmetric matrices). Then Sn, n contains the n × n
€
€
€
identity,
and Sn, n is closed by means of the theorems (c A)T = c AT and
{
}
Sn, n is a subspace of Mn, n .
(A + B)T = AT + BT . Hence,
€
€
€
Dr. Neal, Fall 2008
MATH 307
Homework on Subspaces
1. Let V be the vector space of all 3 x 3 matrices. Let W be the subset of V consisting of
those matrices that have all 0’s down the main diagonal. Prove that W is a subspace of
V.
€
€
€
€
n
m
T
2. Let T : R → R be a linear transformation. The set Range
€ is the set of vectors b in
m
n
R for which there exists a vector v in R such that T (v) = b . Prove that Range T is a
m
subspace of R . That is, verify that (i) Range T is non-empty (that there is an element
m
€
in R that has a pre-image); (ii) that Range T is closed under scalar multiplication;
and
(iii) that Range T is closed €
under addition.
€
3. Determine if the subsets W are subspaces of the vector space V . If so, prove it. If
not, give an example of a vector or vectors in the subset for which scalar multiplication
or addition fails to be closed.
€ 2 x 2 matrices; Subset W = { A ∈ V : €
(a) Vector space V =
det( A ) = 10}.
T
(b) Vector space V = n × n matrices; Subset W = { A ∈ V : A = – A }.
€
€
€
€
(c) Vector space V = All Real Numbers; Subset W = the rational numbers.
€ V = C[a, b] (i.e., continuous
€ space
€
€ functions on the interval [a, b] .);
(d) Vector
€ W = { f ∈ C[a, b] :
Subset
b
∫ f (x) dx =€ 0}.
a
€