Math 312 Homework 1 Solutions


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1 Math 31 Homework 1 Solutions Last modified: July 15, 01 This homework is due on Thursday, July 1th, 01 at 1:10pm Please turn it in during class, or in my mailbox in the main math office (next to 4W1) Please staple your pages before you turn in your homework and make sure the problems are labeled clearly and are, preferably, in sequential order Solutions are in blue 1 Consider the following general system of two equations in the two variables x, y: ax + cy = e bx + dy = f Find a condition on a, b, c, d so that the system above has a solution for all possible e, f Solution: First assume that both a = 0 and b = 0 In this case, we see that the two equations become cy = e and dy = f If c = 0 or d = 0 then this system contains the equation 0 = e or 0 = f, but we can choose e, f arbitrarily so this would be an inconsistent system On the other hand, if c, d 0, then the system becomes y = e/c = f/d but, again, we can vary e, f so that e/c f/d and we obtain an inconsistent system Thus, at least one of a or b is nonzero, and, by symmetry, we can assume without loss of generality that a 0 Now we look at the augmented matrix obtained from the system: ( ) a c e b d f which row reduces to: ( 1 c/a e/a 0 d bc/a f be/a We can choose f so that f be/a 0 so that this system is consistent if and only if d bc/a 0 or equivalently ad bc 0 When this is the case, we see that x = (de cf)/(ad cb) and y = (af be)/(ad cb) is a solution for an arbitrary e, f Let v 1,, v m F n be given and consider F n as a vector space over F Denote by v i = (a 1i,, a ni ), a ij F, and consider the matrix A M n m (F) defined by A ij = a ij Show that 1 )
2 (a) If Span(v 1,, v m ) = F n then the row reduced echelon form of A has n pivots (b) If the row reduced echelon form of A has n pivots then Span(v 1,, v m ) = F n (c) Deduce: if Span(v 1,, v m ) = F n then m n Solution: Say v = (b 1,, b n ) F n is given Observe that v Span(v 1,, v m ) if and only if the system of linear equations Ax = (b 1,, b n ) t =: b has a solution (a) Suppose that the RREF form of A has less than n pivots (note that there can be at most n) Then the RREF form of A must have a row of zeros Denote by B the RREF form of A, which is obtained from A by applying the elementary row operations R 1,, R s Denote by R i the row operations which reverses R i ie R i is the row operation which is the inverse to R i In particular, we can obtain A back from B by applying R s,, R 1 in that order By our assumption, B contains a row of zeros and thus the augmented matrix: C := B corresponds to an inconsistent system Applying the row operations R s,, R 1 to C, we obtain the matrix: D := A which is also inconsistent since it is obtained by elementary row operations from an inconsistent system In particular, (b 1,, b n ) / Span(v 1,, v m ) (b) Conversely, if B has n pivots, then it has a pivot in every row Thus, the row reduction of is precisely D := C := A B and, since B has a pivot in every row, both systems of linear equations are consistent b 1 b n b 1 b n c 1 c n
3 (c) If Span(v 1,, v m ) = F n then B has a pivot in every row However, B is a matrix of size n m By the definition of pivot we see that there are precisely n pivot columns Since B has m columns, we deduce that n m 3 Let F be a field (ie F = Q, R or C) Let A M m n (F) be given Prove that the following sets are subspaces in the corresponding vector space: (a) The set {A B : B M n k (F)}, as a subset of M m k (F) (b) The set {B A : B M k m (F)}, as a subset of M k n (F) (c) The set {B M n k (F) : A B = 0}, as a subset of M n k (F) (d) The set {B M k m (F) : B A = 0}, as a subset of M k m (F) Solution: Recall that a set S V is a subspace if the following two conditions hold: S and For all v, v S and a, a F one has av + a v S To show that the sets above are subspaces, we check these two conditions (a) An arbitrary element in this set is of the form A B for some B As 0 = A 0 is contained in the set, it is nonempty Let A B and A B be two arbitrary elements of the set and a, a be two arbitrary scalars Then a(a B) + a (A B ) = A (ab + a B ) is contained in the set Thus, this set is closed under matrix addition and scalar multiplication, as required (b) An arbitrary element in this set is of the form B A for some B As 0 = 0 A is contained in the set, it is nonempty Let B A and B A be two arbitrary elements of the set and a, a be two arbitrary scalars Then a(b A) + a (B A) = (ab + a B ) A is contained in the set Thus, this set is closed under matrix addition and scalar multiplication, as required (c) Observe that A 0 = 0 so that 0 is an element of this set Suppose B, B are contained in the set equivalently, AB = AB = 0 and let a, a be two arbitrary scalars Then A (ab + a B ) = aab + a AB = a0 + a 0 = 0 so that ab + a B is contained in the set as well (d) Observe that 0 A = 0 so that 0 is an element of this set Suppose B, B are contained in the set equivalently, BA = B A = 0 and let a, a be two arbitrary scalars Then (ab + a B ) A = aba + a B A = a0 + a 0 = 0 so that ab + a B is contained in the set as well 4 Exercise 138 from the text, reproduced here for your convenience: Are the following sets subspaces of R 3 under the usual vector addition and scalar multiplication? Explain (a) W 1 = {(a 1, a, a 3 ) : a 1 = 3a, and a 3 = a } 3
4 (b) W = {(a 1, a, a 3 ) : a 1 = a 3 + } (c) W 3 = {(a 1, a, a 3 ) : a 1 7a + a 3 = 0} (d) W 4 = {(a 1, a, a 3 ) : a 1 4a a 3 = 0} (e) W 5 = {(a 1, a, a 3 ) : a 1 + a 3a 3 = 1} (f) W 6 = {(a 1, a, a 3 ) : 5a 1 3a + 6a 3 = 0} Solution: (a) Observe that (0, 0, 0) W 1 so that W 1 is nonempty Suppose (a 1, a, a 3 ), (b 1, b, b 3 ) W 1 are arbitrary Let a, b F be scalars and consider the linear combination: a(a 1, a, a 3 ) + b(b 1, b, b 3 ) = (aa 1 + bb 1, aa + bb, aa 3 + bb 3 ) =: w Observe that aa 1 + bb 1 = a3a + b3b = 3(aa + bb ) and aa 3 + bb 3 = a( a ) + b( b 3 ) = (aa +bb 3 ) so that w W 1 Thus, we see that W 1 is indeed a subspace (b) This is not a subspace as (0, 0, 0) / W (c) This is a subspace, as follows First, we observe that (0, 0, 0) W 3 Suppose v = (a 1, a, a 3 ) and v = (b 1, b, b 3 ) are elements of W 3 and take an arbitrary linear combination av + bv = (aa 1 + bb 1, aa + bb, aa 3 + bb 3 ) We have: (aa 1 + bb 1 ) 7(aa + bb ) + (aa 3 + bb 3 ) = a(a 1 7a + a 3 ) + b(b 1 7b + b 3 ) = 0 Thus, we see that av + bv W 3 as well, proving that W 3 is indeed a subspace (d) In a similar way to part (3), we see that W 4 is a subspace (e) The vector (0, 0, 0) / W 5 so that W 5 is not a subspace (f) This is not a subspace as it is not closed under vector addition For example, v = (0,, 1) W 6 and v = (1, 0, 5/6) W 6 However, v + v = (1,, 1 + 5/6) / W6 since (check this!): 5 (1) 3 ( ) + 6 (1 + 5/6) 0 5 Find examples of the following finite sets of vectors in the corresponding vector space: (a) A collection of mn matrices of dimension m n which span M m n (F) (b) A collection of n (n+1) symmetric n n square matrices which span the subspace Sym n := {A M n n (F) : A t = A} of symmetric matrices (c) A collection of n (n 1) skewsymmetric n n square matrices which span the subspace Skew n := {A M n n (F) : A t = A} of skewsymmetric matrices Solution: 4
5 (a) Consider the set G = {E ij : 1 i m, 1 j n} Observe that: A = i,j A ij E ij for an arbitrary m n matrix A, where A ij denotes the i, j component of A Thus, indeed, Span G = M m n, as required Moreover, we see immediately that #G = m n (b) Consider the set G = {E ij + E ji : 1 i j n} Observe that (E ij ) t = E ji so that (E ij + E ji ) t = E ij + E ji and so G indeed contains symmetric matrices Moreover, we observe that, for an arbitrary symmetric matrix A, one has: A = i A ii (E ii + E ii ) + A ij (E ij + E ji ) i<j Thus, Span G = Sym n, as required Moreover, we see that: #G = n = n(n + 1) (c) Consider the set G = {E ij E ji : 1 i < j n} Observe that (E ij E ji ) t = E ji E ij = (E ij E ji ) so that, indeed, G contains skewsymmetric matrices For any skew symmetric matrix A, one has: A = i<j A ij (E ij E ji ) so that Span G = Skew n Moreover, #G = (n 1) = n(n 1) 6 Let V be a vector space over F and let W 1, W be two subspaces of V Prove or disprove the following statements: (a) The set W 1 W = {w V : w W 1 and w W } is a subspace of V (b) The set W 1 W = {w V : w W 1 or w W } is a subspace of V (c) The set W 1 + W = {w 1 + w : w i W i } is a subspace of V Solution: (a) This is a subspace First note that 0 W 1 W as 0 W 1 and 0 W Suppose w, w W 1 W and a, a are scalars Consider aw + a w Since w, w W 1 and W 1 is a subspace, we see that aw + a w W 1 and, since w, w W with W a subspace we see that aw + a w W as well Thus aw + a w W 1 W 5
6 (b) This is not a subspace Consider, eg, V = R, let W 1 denote the xaxis and W denote the yaxis Then W 1 W is the union of the two coordinate axes, in particular: W 1 W = {(x, y) : x = 0 or y = 0} but then (1, 0) W 1 W, (0, 1) W 1 W while (1, 0)+(0, 1) = (1, 1) / W 1 W (c) This is a subspace First note that 0 = W 1 + W as 0 W 1 W Now suppose w, w W 1 + W are arbitrary Then w = w 1 + w and w = w 1 + w for some w i, w i W i, i = 1, Let a, a be arbitrary scalars Then aw + a w = (aw 1 +a w 1)+(aw +a w ) W 1 +W since aw 1 +a w 1 W 1 and aw +a w W 7 Let V be a vector space over F and let W 1, W be two subspaces of V We say that V is the direct sum of W 1 and W, and write V = W 1 W provided that: W 1 W = {0} is the zero subspace V = W 1 + W ; ie any element of V can be written as a sum w 1 + w where w i W i Prove the following: (a) M n n (R) = Sym n Skew n ; here Sym n denotes the subspace of symmetric matrices and Skew n denotes the subspace of antisymmetric matrices (b) C 0 (R, R) = O E; here C 0 (R, R) denotes the vector space of continuous functions f : R R, O denotes the subspace of odd functions (f such that f( x) = f(x)) and E denotes the subspace of even functions (f such that f( x) = f(x)) Hint: If A M n n (F), consider A + A t and A A t Solution: In both situations, we need to show that an arbitrary vector v V can be written as a sum w 1 + w where w i W i, then show that W 1 W = {0} 1 Let A be an n n matrix Observe that: ( ) t 1 (A + At ) = 1 (A + At ), and ( ) t 1 (A At ) = 1 (A At ) Thus, 1 (A + At ) is symmetric while 1 (A At ) is skewsymmetric Moreover, A = 1 (A + At ) + 1 (A At ) Sym n + Skew n Lastly, we must show that Sym n Skew n = {0} For this, suppose that A is both symmetric and skewsymmetric Namely, A t = A = A so that A = 0 but then A = 0 Thus, we deduce that M n n = Sym n Skew n, by the definition of 6
7 This is similar to the proof above as follows Let f be a continuous function and observe that the function f e (x) = 1 (f(x) + f( x)) is even while f o(x) = 1 (f(x) f( x)) is odd; moreover, f = f e + f o so that, indeed, C 0 (R, R) = O + E On the other hand, if a function f O E, we see that f( x) = f(x) = f(x) so that f(x) = 0 which implies that f(x) = 0 for all xl thus O E = {0} From this we deduce that C 0 (R, R) = O E as required 7
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