Dr. Neal, Fall 2008 MATH 307 Subspaces

Let

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V be a vector space. A subset

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W is a subspace of

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V provided (i)

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W is non-empty (ii)

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W is closed under scalar multiplication, and (iii)

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W is closed under addition. In other words,

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W is just a smaller vector space within the larger space V . But adding elements from

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W keeps them in

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W as does multiplying by a scalar. Example 1. Let

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V = R2 (the

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x y plane). Let

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W = {(

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x ,

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y) :

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y = m x }. Then

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W is simply a straight line through the origin. Is

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W a subspace? (i) The point (0, 0) is in

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W ; hence,

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W is non-empty. (ii) If (

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x ,

€

y)

€

∈W , then for any scalar

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c ∈ ℜ, we have c (x, y) = (c x, c y) . But is this point still on the line? Yes because c y = c (m x) = m (c x) ; hence, c (x, y) is still in

€

W . (iii) Now suppose (x1,y1) and (x2, y2 ) are in

€

W . Then y1 = m x1 and y2 = m x2 . Also, (x1,y1) + (x2, y2) = x1 + x2,y1 + y2( ) and y1 + y2 = mx1 +m x2 = m (x1 + x2) ; hence, (x1,y1) + (x2, y2) is still in

€

W . Because Properties (i) , (ii), and (iii) are satisfied,

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W is a subspace of R2 . What if the line did not go through the origin? Consider

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U = {(

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x ,

€

y) :

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y = m x + b} for some

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b ≠ 0. Then for (

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x ,

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y)

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∈U and

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c ≠ 1, c y = c (m x + b) = m(c x) + cb ≠ m x + b ; hence,

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U is not closed under scalar multiplication. It also can be shown that

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U is not closed under addition; but only one property must be disproved to show that

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U is not a subspace. Theorem 3.3. Let

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W be a subspace of a vector space V . Then

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0

€

∈W . Proof. Because

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W must be non-empty, there exists some element

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w

€

∈W . Because

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W is closed under scalar multiplication, we then have

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0 = 0

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w

€

∈W . Corollary. If

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0

€

∉

€

W , then

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W cannot be a subspace. Example 2. Let

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V be the vector space of all 3

€

×3 matrices. Let

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W = {

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A

€

∈ V :

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A−1 exists}. Is

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W a suspace of

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V ? In this vector space, the

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0 element is the 3

€

×3 zero matrix, which is non-invertible. Thus,

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0

€

∉

€

W , and

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W is not a subspace of

€

V .

Dr. Neal, Fall 2008

Proving that a Subset is a Subspace Let

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W be a subset of a vector space

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V . Then

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W must have some defining property that distinguishes its elements from the other elements in

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V . To show that

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W is actually a subspace, we must verify three properties: (i)

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W must be non-empty; (ii)

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W must be closed under scalar multiplication; and (iii)

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W must be closed under addition. (i) The zero vector

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0 must always be in any subspace. So to show

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W is non-empty, it is usually easiest to show that

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0 is in

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W . That is, you must argue that the specific zero vector of the vector space

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V under consideration has the defining property of the set

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W . Then conclude that

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W is non-empty. (ii) To show

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W is closed under scalar multiplication, (a) Let

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w ∈W . (b) Then state what it means for

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w to be in

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W ; i.e., explain the defining property. (c) Let

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c be any scalar (i.e., real number). (d) Argue that

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c w has the defining property of

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W . (e) Conclude that

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W is closed under scalar multiplication. (iii) To show

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W is closed under addition, (a) Let

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w1 and

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w2 be in

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W . (b) Then state what it means for

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w1 and

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w2 to be in

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W ; i.e., explain that they have the defining property. (c) Argue that

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w1 + w2 has the defining property of

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W . (e) Conclude that

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W is closed under addition. (iv) Conclude that

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W is a subspace of

€

V . 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 doing, use correct mathematical logic, do not skip steps, and give conclusions. Note: We also can combine requirements (ii) and (iii) and show instead that for any scalar

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c and vectors

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w1 and

€

w2 in

€

W , that

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c w1 + w2

€

∈W . Example 2. Let

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V = C[a, b] be the set of continuous functions over

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[a, b], where

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a < b. Let

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a < d < b and let

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D = { f ∈ V ′ f (d) exists}. Then

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D is a subspace of

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V . Proof. Let

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f0 be the zero function (i.e.,

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f0(x) = 0 for all

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x ∈ [a, b]). Then

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′ f 0(x) = 0 for all

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x ∈ (a, b); thus,

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′ f 0(d) exists (and equals 0). So

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f0 ∈ D and

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D is non-empty. Let

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f ∈ D. Then

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′ f (d) exists. For any scalar

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c ,

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cf is still a continuous function and

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(cf ′ ) (d) = c( ′ f (d)) still exists. Thus,

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cf ∈ D and

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D is closed under scalar multiplication. Let

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f , g∈ D so that

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′ f (d) and

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′ g (d) exist. Then

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f + g is still a continuous function and

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( f + g ′ ) (d) = ′ f (d) + ′ g (d) still exists. Thus,

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f + g∈ D and

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D is closed under addition.

Whence,

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D is a subspace of

€

C[a, b]. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Example 3. Let

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V = R3 and let

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W = {(

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x ,

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y ,

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z ) :

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z = 0} . Then (0, 0, 0)

€

∈W ; so

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W is non-empty. For vectors

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(x1,y1, 0) and

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(x2,y2, 0) in

€

W and for any scalar

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c , we have

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c (x1,y1, 0) + (x2,y2, 0) = (c x1 + x2, c y1 + y2, 0)

€

∈W . Hence,

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W is a subspace of R3.

Dr. Neal, Fall 2008 Example 4. Let T : Rn→ Rm be a linear transformation. Then

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kerT is a subspace of Rn .

Recall: For T : Rn→ Rm ,

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kerT = {x ∈ Rn T(x) = 0 m}.

Proof. Let

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0 n be the zero vector in Rn and let

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0 m be the zero vector in Rm . Then for any

linear transformation T : Rn→ Rm , we have

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T( 0 n ) =

0 m . Thus,

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0 n ∈ kerT . Hence,

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kerT is non-empty. Let x ∈kerT . Then

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T(x) = 0 m . If

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c is any scalar, then

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c x is still in Rn and

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T(c x) = cT(x) = c 0 m =

0 m ; hence, c x ∈kerT . Thus,

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kerT is closed under scalar multiplication. Let

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u, v ∈ kerT . Then

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T(u) = 0 m = T(v) . Then

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T(u + v) = T(u) + T(v) = 0 m +

0 m =

0 m ;

hence,

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u + v ∈ kerT and

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kerT is closed under addition. Ergo,

€

kerT is a subspace of Rn .

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Recall that a linear transformation T : Rn→ Rm has an

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m × n matrix representation A . Then in matrix form,

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kerT is simply all solutions to the homogeneous system

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AX = 0 . Thus, we can re-state Example 4 in matrix form as: Example 5. Let

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V be the vector space of all

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n ×1 matrices. Let A be an

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m × n matrix. The set of all

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n ×1 matrices X such that

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AX = 0 forms a subspace of

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V . That is, the solution space to the homogeneous system AX = 0 is a subspace of

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V . - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Example 6. Let

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S = {

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u 1, . . .,

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u m} be a collection of

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m vectors in Rn . The set span(S) is the collection of all linear combinations of the vectors in

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S. Then span(S) is a subspace of Rn . Proof. (i) First,

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0 n =

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0 u 1 + . . .+ 0 u m is a linear combination of the vectors in

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S; thus,

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0 n ∈ span(S) and span(S) is non-empty. (ii) Next, suppose

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u ∈ span(S) . Then

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u can be written as

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u = c1 u 1 + . . .+ cm

u m. For any scalar

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c , we then have

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c u = c(c1 u 1 + . . .+ cm

u m) = (cc1) u 1 + . . .+ (ccm )

u m , which is still a linear combination of the vectors in

€

S . Thus,

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c u ∈ span(S) and span(S) is closed under scalar multiplication. (iii) Now let

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u , v ∈ span(S) . Then

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u and

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v can be written as

€

u = c1 u 1 + . . .+ cm

u m and

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v = d1 u 1 + . . .+ dm

u m . Then

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u + v = (c1 u 1 + . . .+ cm

u m ) + (d1 u 1 + . . .+ dm

u m)= (c1 + d1)

u 1 + . . .+ (cm + dm ) u m

which is still a linear combination of the vectors in

€

S. Thus,

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u + v ∈ span(S) and span(S) is closed under addition. Therefore, span(S) is a subspace of Rn .

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

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0 ≤ m < n , Pm is a subspace of Pn . (b) The subset of those polynomials having only non-negative coefficients {a0 + a1x + . . . + anx

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(–∞, ∞) . Indeed, the function f (x) = x − x0 is in the subset, and (c f + g)(x0) = c f (x0 ) + g(x0) = 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

x→af (x) = b is not a subspace for

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b ≠ 0.

Clearly, this subset is not closed under scalar multiplication for

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c ≠ 1. But is it a subspace for

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b = 0? (iii) Let Mn,n be the vector space of all

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n × n matrices. (a) Let

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W = A : Tr(A) = b{ }. Then

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W is a subspace if and only if

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b = 0. (b)

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Sn,n = A : AT = A{ } (

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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

(A + B)T = AT + BT . Hence, Sn,n is a subspace of Mn,n .

Dr. Neal, Fall 2008 MATH 307 Homework on Subspaces

1. Let

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V be the vector space of all 3 x 3 matrices. Let

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W be the subset of

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V consisting of those matrices that have all 0’s down the main diagonal. Prove that

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W is a subspace of

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V . 2. Let T : Rn→ Rm be a linear transformation. The set RangeT is the set of vectors

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b in Rm for which there exists a vector

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v in Rn such that T (v) =

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b. Prove that RangeT is a subspace of Rm . That is, verify that (i) RangeT is non-empty (that there is an element in Rm that has a pre-image); (ii) that RangeT is closed under scalar multiplication; and (iii) that RangeT is closed under addition. 3. Determine if the subsets

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W are subspaces of the vector space

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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. (a) Vector space

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V = 2 x 2 matrices; Subset

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W = {A

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∈ V : det(

€

A) = 10}. (b) Vector space

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V =

€

n × n matrices; Subset

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W = {A

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∈ V : AT = –A }. (c) Vector space

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V = All Real Numbers; Subset

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W = the rational numbers. (d) Vector space V = C[a, b] (i.e., continuous functions on the interval [a, b] .);

Subset W = { f

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∈ C[a, b] : f (x)a

b

∫ dx = 0}.