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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 €