Solutions for Math 225 Assignment #2 1

Transcription

1 Solutions for Math 225 Assignment #2 () Determine whether W is a subspae of V and justify your answer: (a) V = R 3, W = {(a,, a) : a R} Proof Yes For a =, (a,, a) = (,, ) W For all (a,, a ), (a 2,, a 2 ) W and R, (a,, a ) + (a 2,, a 2 ) = (a + a 2,, a + a 2 ) W Therefore, W is a subspae of V (b) V = R 3, W = {(a, b, a ) : a, b R}; Proof No, W is not a subspae of V sine (,, ) W but ( )(,, ) = (,, ) W () V = M n n (R), W = {singular n n matries} for n > Proof No, W is not a subspae of V sine A = and B = W but A + B = I W (d) V = R[x, W = {f(x 2 + ) : f(x) R[x} Proof Yes For f(x), f(x 2 + ) W For all f (x 2 + ), f 2 (x 2 + ) W and R, f (x 2 + ) + f 2 (x 2 + ) = (f + f 2 )(x 2 + ) W So W is a subspae of V (2) For eah of the following 3 n matries A, determine whether Col(A) = R 3 : a) b) ) xihen/math2254f/hw2solpdf

2 2 Solution For a m 3 matrix A, Col(A) = R 3 if and only if rank(a) = 3 Find the rank of these matries and we onlude that a) no b) yes ) no (3) Whih of the following statements are true and whih are false? Justify your answer (a) A system Ax = b of linear equations has at least one solution if and only if b Col(A) Proof True Let A = [ v v 2 v n, where v, v 2,, v n are the olumn vetors of A If Ax = b has a solution, let be a solution Then x = 2 b = Ax = [ v v 2 v n 2 = v + 2 v v n So b Span{v, v 2,, v n } = Col(A) If b Col(A), then b = v + 2 v v n = [ v v 2 v n 2 for some, 2,, R Therefore, x = 2 is a solution of Ax = b and hene Ax = b has a solution (b) Let S and S 2 be two subsets of a vetor spae V If Span(S ) Span(S 2 ), then S S 2

3 Proof False Let V = R, S = {} and S 2 = { } Then Span(S ) = Span(S 2 ) = V So Span(S ) Span(S 2 ) but S S 2 () Let S and S 2 be two subsets of a vetor spae V If S S 2, then Span(S ) Span(S 2 ) Proof True For every v Span(S ), v = v + 2 v v n for some, 2,, R and v, v 2,, v n S And sine S S 2, v, v 2,, v n S 2 Therefore, v Span(S 2 ) and hene Span(S ) Span(S 2 ) (d) Nul(A) Nul(BA) for all m n matries A and l m matries B Proof True For x Nul(A), Ax = BAx = and hene x Nul(BA) (4) Prove the following: (a) Every square matrix an be written as the sum of a symmetri matrix and a skew-symmetri matrix (b) Let W be the vetor spae of n n matries and let U and V be the subspaes of W onsisting of symmetri and skew-symmetri matries, respetively Then W = U + V 3 Proof Let A be a square matrix and let B = 2 (A + AT ) and C = 2 (A AT ) Then A = B + C Sine B T = 2 (A + AT ) T = 2 (AT + (A T ) T ) = 2 (A + AT ) = B B is symmetri And sine Sine C T = 2 (A AT ) T = 2 (AT (A T ) T ) = 2 (A AT ) = C C is skew-symmetri So every square matrix is the sum of a symmetri matrix and a skew-symmetri matrix

4 4 For every A W, A = B + C where B is symmetri and C is skew-symmetri, ie, B U and C V Therefore, W = U + V (5) Find V V 2 and V + V 2 for subspaes V, V 2 of V : (a) V = R 3, V = {(x, y, z) : x + y = 2y + z = } and V 2 = {(x, y, z) : x y = 2y z = }; Proof The intersetion is V V 2 = {(x, y, z) : x + y = 2y + z = x y = 2y z = } = {(,, )} Solving x + y = 2y + z =, we obtain V = Span{(,, 2)} Solving x y = 2y z =, we obtain Therefore, V 2 = Span{(,, 2)} V + V 2 = Span{(,, 2), (,, 2)} (b) V = R[x, V = {f(x) R[x : f() = } and V 2 = {f(x) R[x : f(2) = }; Solution The intersetion is V V 2 = {f(x) R[x : f() = f(2) = } = {(x )(x 2)g(x) : g(x) R[x} Every polynomial f(x) R[x an be written as f(x) = f (x) + f 2 (x) where f (x) = f(x) + f()(x 2) and f 2 (x) = f()(x 2) Sine f () = f() + f()( 2) =, f (x) V Sine f 2 (2) = f()(2 2) =, f 2 (x) V 2 Therefore, V +V 2 = V () V = M n n (R), V = {[a ij n n : a ij = for all i > j} and V 2 = {[a ij n n : a ij = for all i < j} (ie V is the set of all upper triangular matries and V 2 is the set of all lower triangular matries)

5 5 Proof The intersetion is V V 2 = {[a ij n n : a ij = for i j} b b = 2 : b, b 2,, b n R b n That is, V V 2 is the subspae of diagonal matries Every n n matrix [a ij n n an be written as [a ij n n = [b ij n n + [ ij n n where { a ij if i j b ij = if i > j and ij = { if i j a ij if i > j Sine [b ij V and [ ij V 2, V + V 2 = M n n (R) (6) Let V and V 2 be two subspaes of a vetor spae V satisfying V V 2 = {} Show that if v + v 2 = w + w 2 for v, w V and v 2, w 2 V 2, then v = w and v 2 = w 2 Proof Sine v + v 2 = w + w 2, v w = w 2 v 2 Sine v, w W and W is a subspae, v w W Sine v 2, w 2 W 2 and W 2 is a subspae, v 2 w 2 W 2 Therefore, v w = w 2 v 2 W W 2 And sine W W 2 = {}, ie, v = w and v 2 = w 2 v w = w 2 v 2 =, (7) Let A be a square matrix Show that if Col(A) = Col(A 2 ), then Col(A 2 ) = Col(A 3 )

6 6 Proof It is enough to prove that Col(AB) = Col(AC) if Col(B) = Col(C) Applying the above to B = A and C = A 2, we onlude that Col(A 2 ) = Col(A 3 ) if Col(A) = Col(A 2 ) Let B = [ u u 2 u m and C = [ v v 2 v n Then Col(B) = Span{u, u 2,, u m }, Col(C) = Span{v, v 2,, v n }, Col(AB) = Col(A [ u u 2 u m ) = Span{Au, Au 2,, Au m }, Col(AC) = Col(A [ v v 2 v n ) = Span{Av, Av 2,, Av n } Sine Col(B) = Col(C), Span{u, u 2,, u m } = Span{v, v 2,, v n } Therefore, u i Span{v, v 2,, v n } for i =, 2,, m and v j Span{u, u 2,, u m } for j =, 2,, n It follows that Au i Span{Av, Av 2,, Av n } for i =, 2,, m and Av j Span{Au, Au 2,, Au m } for j =, 2,, n Consequently, Span{Au, Au 2,, Au m } Span{Av, Av 2,, Av n } and Span{Av, Av 2,, Av n } Span{Au, Au 2,, Au m } That is, Col(AB) Col(AC) and Col(AC) Col(AB) Col(AB) = Col(AC) (8) Let W be a subspae of a vetor spae V over R and v be a vetor in V Show that W {v} is a subspae of V if and only if v W Proof If v W, W {v} = W is a subspae of V Suppose that W = W {v} is a subspae Sine v W and W is a subspae, 2v W = W {v} Therefore, 2v W or 2v {v} If 2v {v}, then 2v = v and hene v = W If 2v W, v = (2v) W sine W is a subspae 2 So