Math 110, Summer 2012: Practice Exam 1 SOLUTIONS

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1 Math, Summer 22: Practice Exam SOLUTIONS Choose 3/5 of the following problems Make sure to justify all steps in your solutions Let V be a K-vector space, for some number field K Let U V be a nonempty subset of V i) Define what it means for U V to be a vector subspace of V Define span K U ii) Prove that span K U is a vector subspace of V iii) Consider the Q-vector space V = Mat 2 (Q) and the subset U = {I 2, A, A 2 }, where A = Find v U such that span Q U = span Q U, where U = U \ {v} Show that U is linearly independent iv) Find a vector w Mat 2 (Q) such that w / span Q U Is the set U {w} linearly independent? Explain your answer v) Extend the set U {w} to a basis B of Mat 2 (Q), taking care to explain how you know the set B you ve obtained is a basis i) U V is a subspace if, for any u, v U, λ, µ K we have λu + µv U We define span K U = { k c j u j V c j K, u j U, k N} j= ii) Let u, v span K U, λµ K Then, we must have where u i, v j U, c i, d j K Then, u = c u + + c m u m, v = d v + + d n v n, λu + µv = λc u + + λc m u m + µd v + + µd n v n = a w + + a k w k, where a l K, w l U Hence, λu + µv U iii) We see that U = {,, } 2 There is a nontrivial linear relation 2 Mat2(Q) = 2 2 Thus, by the Elimination Lemma we know that span Q U = span Q {I 2, A 2 }, = 2A I 2 2 A 2

2 so that we can take v = A U Set U = {I 2, A 2 } Suppose that there is a linear relation λ I 2 + λ 2 A 2 λ + λ = Mat2(Q) = 2 =, 2λ 2 λ + λ 2 so that λ 2 =, λ + λ 2 = Hence, we must have λ = λ 2 = and U is linearly independent iv) If we take w =, then is is easy to see that w / span Q U (it s not possible to find c, c 2 Q such that c I 2 +c 2 A 2 = w) Then, by a result from class (Corollary to Elimination Lemma) we see that U {w} must be linearly independent v) Since U {w} consists of 3 vectors and is linearly independent and Mat 2 (Q) is has dimension 4, we need only find some vector z / span Q U {w} to obtain a basis U {w, z} (since this set is then linearly independent, by a similar argument as in iv)) There are many ways to proceed: we will use the most general Let S denote the standard ordered basis of Mat 2 (Q) Then, [A 2 ] S = 2, [I 2] S =, [w] S = Then, b span Q {[A 2 ] S, [I 2 ] S, [w] S } if and only if the matrix equation 2 x = b, is consistent So we form the augmented matrix b b b 2 2 b 3 b 2 2 b 3 + 2b b 4 b 4 b and there is no pivot in the last column if and only if b 4 = b Thus, b / span Q {[A 2 ] S, [I 2 ] S, [w] S } if and only if b 4 b Thus, since the S-coordinate morphism is an isomorphism we have that e = / span Q U {w} Hence, the set B = {I 2, A 2, w, e }, is linearly independent As this set contains 4 vectors then it must be a basis, by a result from class Alternatively, you can just come up with some matrix B Mat 2 (Q) and show that B / span Q U {w} In any case, you will need to prove that you know the B you choose is not an element of span Q U {w} 2

3 2 i) Let B = (b,, b n ) V be an ordered subset of the K-vector space V Define what it means for B to be an ordered basis of V using the notions of linear independence AND span ii) Consider the maximal linear independence property: let E V be a linearly independent subset of the K-vector space V Then, if E E and E is linearly independent then E = E Prove that if B V is a basis (satisfying the definition you gave in 2a)) then B is maximal linearly independent iii) Consider the Q-vector space Q {,2,3} = {f : {, 2, 3} Q} Let where B = {f, f 2, f 3 } Q {,2,3}, f () =, f (2) =, f (3) =, f 2 () =, f 2 (2) =, f 2 (3) =, f 3 () =, f 3 (2) =, f 3 (3) = Prove that B is a basis of Q {,2,3} iv) Let B be as defined in 2iii) Define the B-coordinate morphism [ ] B : Q {,2,3} Q 3, and determine the B-coordinates of f Q {,2,3}, where f () = 2, f (2) =, f (3) = 2 v) Suppose that C = (c, c 2, c 3 ) Q {,2,3} is an ordered basis such that the change of coordinate matrix P C B = Determine c, c 2, c 3 Q {,2,3}, ie, for each i {, 2, 3}, determine c (i), c 2 (i), c 3 (i) i) B is a basis if it is a linearly independent set and span K B = V Since it is an ordered set it is an ordered basis ii) We know that B is linearly independent, by definition of a basis So we must show that, B B, with B linearly indepndent = B = B Suppose that B B Then, there is some v B such that v / B Also, we can t have v = V as B is linearly independent (any set containing V is linearly dependent) As span K B = V then there are λ,, λ n K such that v = λ b + + λ n b n n = v λ i b i = V i= As each b i B B then we have obtained a nontrivial linear relation (the coefficient of v is ) among vectors in B, contradicting the linear independence of B Hence, our initial assumption that B = B must be false so that B = B iii) Consider the standard ordered basis S = {e, e 2, e 3 } Q {,2,3} We are going to show that B is linearly independent by using the S-coordinate isomorphism: we have [f ] S =, [f 2 ] S =, [f 3 ] S = 3

4 Then, as, then the set {[f ] S, [f 2 ] S, [f 3 ] S } is linearly independent Since the S-coordinate morphism is an isomorphism we have that {f, f 2, f 3 } is also linearly independent Hence, as Q {,2,3} has dimension 3, B must be a basis iv) We use the change of coordinate matrix P B S = P S B = /2 /2 = /2 /2 Then, we have /2 /2 2 3/2 [f ] B = P B S [f ] S = /2 /2 = 3/2 2 2 v) We have Hence, P B C = P C B = = = [[c ] B [c 2 ] B [c 3 ] B ] c = f 2 f 3, c 2 = f + f 3, c 3 = f f 2 + f 3, so that c () =, c (2) =, c (3) =, c 2 () =, c 2 (2) =, c 2 (3) = 2, c 3 () =, c 3 (2) =, c 3 (3) = 3 i) Define the kernel ker f of a linear morphism f : V W and the rank of f, rankf ii) Consider the function x f : Q 3 Q 2 ; x 2 x 3 x x 2 + 2x 3 x + x 3 Explain briefly why f is a linear morphism What is the rank of f? Justify your answer Using only the rank of f prove that f is not injective (do not row-reduce!) iii) Let S (2) Q 2, S (3) Q 3 be the standard ordered bases Find invertible matrices P GL 2 (Q), Q GL 3 (Q) such that Q Ir [f ] S(2) S P =, (3) where r = rankf iv) Prove or disprove: for the P you obtained in 3iii), a column of P is a basis of ker f i) We have ker f = {v V f (v) = W }, and rankf = dim imf, where imf = {w W v V, such that f (v) = w} 4

5 ii) For every x Q 3 we have f (x) = 2 x, so that f must be linear since it is a matrix transformation The rank of f is 2: we have that f is surjective since there is a pivot in every row of the matrix 2 [f ] S(2) S = (3) By the Rank Theorem 3 = dim Q 3 = dim ker f + rankf = dim ker f = Thus, ker fneq{ Q 3} so that f is not injective iii) There is more than one approach to this problem via elementary matrices or following the proof of the classification of morphisms theorem We will follow the latter: first, we find a basis of ker f This is the same as finding the solution set of the matrix equation 2 x = Hence, we have x x ker f = x 2 x + x 3 =, x 2 x 3 = = x x Q x 3 x Thus, the set is a basis of ker f We extend this to a basis of Q3 : for example,,, which can be easily show to be a linearly independent set, hence a basis of Q 3 Now consider the matrix P = Let Then, we must have Q = [f (e ) f (e 2 )] = Q [f ] S(2) S (3) P = v) By construction, we have a column of P is a basis of ker f If you obtained P, Q using elementary matrices (so that you may not have obtained the same P, Q as I have) then you can see that, since Q [f ] S(2) S P =, (3) 5

6 the last column of Q [f ] S(2) P is the zero vector in Q 2 Then, if P = [p S (3) p 2 p 3 ], so that i th column of P is p i, then so that Q [f ] S(2) S (3) P = Since Q is invertible we must have [Q [f ] S(2) S (3) p Q [f ] S(2) S (3) p 2 Q [f ] S(2) S (3) p 3 ], = Q [f ] S(2) S (3) p 3 = Q [f (p 3 )] S (2) [f (p 3 )] S (2) = = f (p 3 ) =, and as dim ker f = we see that p 3 must be define a basis of ker f 4 i) Let A Mat n (C) Define what it means for A to be diagonalisable ii) Suppose that P AP = D, with D a diagonal matrix and P GL n (C) Prove that the columns of P are eigenvectors of A For 4iii)-vi) we assume that A Mat 2 (C), A 2 = I 2 and that A is NOT a diagonal matrix iii) Show that the only possible eigenvalues of A are λ = or λ = iv) Let u C 2 be nonzero Show that A(Au + u) = Au + u v) Prove that there always exists some nonzero w C 2 such that Aw w Deduce that λ = must occur as an eigenvalue of A Prove that λ = must also occur as an eigenvalue of A vi) Deduce that A is diagonalisable i) A is diagonalisable if it is similar to a diagonal matrix, ie, if there is P GL n (C) such that P AP = D, with D a diagonal matrix ii) As P AP = D then we see that AP = PD Suppose that D = c c n Now, if P = [p p n ], so the i th column of P is p i, then AP = [Ap Ap n ], and Hence, we have PD = [c p c n p n ] = [Ap Ap n ] Ap i = c i p i, so that the columns of P are eigenvectors of A iii) Suppose that λ is an eigenvalue of A Then, let v be an eigenvector of A with associated eigenvalue λ (so that v C 2) Then, v = I 2 v = A 2 v = A(Av) = A(λv) = λav = λ 2 v Hence, as v C 2, then we must have λ 2 = so that λ can be either or 6

7 iv) We have A(Au + u) = A 2 u + Au = Au + u v) Suppose that for every nonzero w C 2 we have Aw = w Then, we would have Ae = e, Ae 2 = e 2, implying that A = I 2 However, we have assumed that A is not a diagonal matrix Hence, there must exist some nonzero w C 2 such that Aw w Now, let u = Aw + w C 2 Then, by iv) we have Au = u, so that λ = is an eigenvalue of A A similar argument shows that λ = must be an eigenvalue also (here we can find nonzero z C 2 such that Az z Then, A(Az z) = z Az, so that u = Az z is an eigenvector with associated eigenvalue λ = ) vi) We have just shown that A admits two distinct eigenvalues Hence, since A is 2 2 we have that A is diagonalisable (by a result from class) 5 i) Define what it means for a linear endomorphism f End C (V ) to be nilpotent Define the exponent of f, η(f ) ii) Consider the endomorphism f : C 3 C 3 ; x x 2 x 3 x + x 2 x 3 x + x 2 x 3 Show that f is nilpotent and determine the exponent of f, η(f ) iii) Define the height of a vector v C 3 (with respect to f ), ht(v) Find a vector v C 3 such that ht(v) = 2 iv) Find a determine a basis B of C 3 such that [f ] B = v) What is the partition of 3 corresponding to the similarity class of f? i) f is nilpotent if there exists some r N such that f r = f f = EndC (V ) End C (V ) is the zero morphism The exponent of f is the smallest integer r such that f r = while f r ii) You can check that f 2 = End C (C 2 ) and, since f is nonzero, the exponent of f is η(f ) = 2 iii) We define ht(v) to be the smallest integer r such that f r (v) = C 3, while f r (v) C 3 The vector e C 3 is such that ht(e ) = 2, since f (e ) C 3, while f 2 (e ) = C 3 iv) Using the algorithm from the notes you find that B = (f (e ), e, e 2 + e 3 ) is an ordered basis such that [f ] B = Indeed, we have H = ker f = E = span C,, H 2 = C 3 7

8 Then, H 2 = H G 2 = H span C {e }, and, if S = {f (e )} = {e + e 2 }, H = H span C S G = {} span C S span{e 2 + e 3 } Thus, the table you would obtain is e f (e ) e 2 + e 3 v) The partition associated to the nilpotent/similarity class of f is π(f ) :