Math 4310 Solutions to homework 1 Due 9/1/16

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1 Math 0 Solutions to homework Due 9//6. An element [a] Z/nZ is idempotent if [a] 2 [a]. Find all idempotent elements in Z/0Z and in Z/Z. Solution. First note we clearly have [0] 2 [0] so [0] is idempotent in any Z/nZ. Now suppose an element [a] Z/nZ were idempotent and invertible. Then [a] 2 [a] is equivalent to [a] []. Then the only idempotent and invertible element in Z/nZ is [] the multiplicative identity for any n since it is indeed the case that [] 2 []. Then for any n the elements [0] [] are idempotent. Then to find the remaining idempotent elements we will consider the noninvertible elements i.e. the elements [a] for which it is not the case that gcda n. First for Z/0Z we know 7 and 9 are relatively prime to 0 so the corresponding elements of Z/0Z [] [7] [9] cannot be idempotent. Then we note [2] 2 [] [2] [] 2 [6] [6] [] [5] 2 [25] [5] [6] 2 [6] [6] and [8] 2 [6] [] [8]. Then the idempotent elements in Z/0Z are {[0] [] [5] [6]}. Now for Z/Z since is prime we know Z/Z is a field so every non-zero element is invertible so every non-zero element [a] Z/Z has the property that gcda. Then by our discussion above the idempotent elements in Z/Z are {[0] []}. 2. Let M n R denote the n n matrices with coefficients in R for n. This set has two natural operations matrix addition and matrix multiplication. For which n is M n R with these operations a field? Please justify your answer. Solution. We claim the only values of n for which M n R with these operations is a field is n. First note that for any n we clearly have matrix addition associative commutative with identity element the zero matrix every entry is equal to 0 and every matrix having an additive inverse while multiplication is clearly associative with identity element the identity matrix with diagonal entries equal to and the rest equal to 0. Then the only property which could prevent M n R from being a field is whether every nonzero element has a multiplicative inverse. First for n pick an arbitrary nonzero matrix say a so a 0. This is clearly invertible since every nonzero element is invertible in R so there is a R so that a is the inverse of our original matrix. Indeed we have a a for every nonzero element. Since multiplication is also commutative as inherited from R we have M R a field. Now for n > consider the matrix with the entry in row and column equal to and all other entries equal to 0 this construction is valid for any n. We clearly have a non-zero matrix with determinant 0 so it is not invertible. Then for any n > we have M n R is not a field.

2 Math 0 Fall 206 Solution 2 2. Let F Z/5Z be the field with five elements. In this exercise I ll write a for the congruence class [a] 5. Compute the following in F : a 0 b c 2 d 2. Let F Z/5Z be the field with five elements. In this exercise I ll write a for the congruence 2 class [a] 5. Find all vectors v in F 2 such that 2v. 2 We add to both sides of the equation to get 2v. 2 Next we multiply by 2 to get v. This answer is unique. When 2 solving a linear equation over a field F we will have unique solutions. For more general systems of linear equations there will be 0 or many solutions. The precise number of many will depend on the field we are working over. The reason is that the set of all solutions if there are any form a subspace of a vector space over F. 5. Let F be a field V a vector space over F and v x y V vectors. Prove that if v x v y then x y. Solution. Let V be a vector space over F and v x y V vectors. We know by Axiom A that there is a vector v the additive inverse of v. If we know that 2 v x v y then we may add v to both sides of the equation on the left to get v v x v v y. By VS Axiom A2 associativity of this becomes v v x v v y. By VS Axiom A commutativity of this becomes The VS property A of v this is v v x v v y. 0 x 0 y. Finally VS Axiom A tells us that adding 0 to a vector doesn t change it so x y

3 Math 0 Fall 206 Solution 2 6. Let F be a field V a vector space over F v V a vector and F a scalar. Prove that if v 0 then 0 or v 0. Solution. Suppose that there is a scalar ad a vector v so that v 0. We want to prove that either 0 or v 0. If 0 then we re done. Otherwise we must have 0. In particular by Field Axiom M must have a multiplicative inverse which we will denote. By Field Axioms M and M we have. Let w V be an vector. We aim to show that v w w. Indeed v w w v by VS A w v by VS M w v by the above discussion w v by VS M w v by VS M w 0 by assumption 0 w by VS A w by VS A w by VS M w by F M and M w by VS M. We haven t used any property about w so the above list of equalities holds for every w in V. But VS Axiom A says that 0 is the unique vector that as the property 0 w w for each w W. Thus we must have v 0 completing the proof. 7. Let V Z/Z be the integers modulo. Is it possible to define the structure of a vector space on V over F 2 so that vector addition is addition modulo? Solution. Suppose we can define the structure of a vector space on V Z/Z over F 2 so that vector addition is addition modulo. Since [0] 2 is the zero element in F 2 [0] 2 v [0] for any v V by proposition 2... Since [] 2 is the multiplicative identity in F 2 [] 2 v v for any v V by axiom 8. Then we have [0] [0] 2 [] because multiplication by 0 results in 0 [] 2 [] 2 [] because [] 2 [] 2 [0] 2 in F 2 [] 2 [] [] 2 [] by distributive law VS M [] [] by VS M [2] by addition in Z/Z which is impossible. We conclude that there is no such vector space structure on V.

4 Math 0 Fall 206 Solution 2 8. Is there a vector space that contains exactly elements? We note that 7. We know that F Z/7Z is a field with exactly 7 elements. The vector space F is the set of -tuples with entries in F. For each entry there are 7 possibilities so there are possible such vectors. Thus F contains exactly elements and the answer to the question is Yes. 9. Let R denote the positive real numbers. Let { } V R and define vector addition on V by ε ε R to be ε ε γ γ. Prove that V is a vector space over R. and scalar multiplication by Solution. As far as I can tell this is not a subset of any vector space we know of so far. So we must grit our teeth and check the eight axioms. They basically follow from properties of multiplication and exponentiation. Here goes! We should first note that the operations are well-defined on R 2 since the product of two positive numbers is again positive and real powers of positive numbers are defined and positive. Next we check the axioms. A For every u v V u v v u. We have γ γ But on the right-hand side we may apply commutativity of multiplication R to get γ γ γ. This holds for any γ R and so A holds. A2 For every u v w V u v w v u w. We have γ π γ π γ π ν ν ν. γ where the equality is by associativity of multiplication in R. This guarantees A2. A There is a unique element 0 V satisfying 0 v v for all v V. The element 0 satisfies 0 π ν

5 Math 0 Fall 206 Solution 2 5 because a a for any a R. Thus V has a zero vector. As we discussed in class given the other axioms which we shall prove without relying upon uniqueness of 0 the zero vector is automatically unique. A For each v V there exists an element v V satisfying v v 0. For the vector v the numbers are positive so and positive. Moreover 0 exist and are M For each F and u v V u v u v. A basic property of exponentials is that for and γ positive real numbers and ε a real number γ ε ε γ ε. Thus we have ε γ γ γ ε ε ε ε γ ε ε ε ε γ ε ε ε ε M2 For each F and v V v v v. A basic property of exponentials is that for a positive number and ε and µ real numbers εµ ε µ. Thus we have εµ ε εµ µ ε ε µ ε εµ ε µ µ ε γ µ M For each F and v V v v. A basic property of exponentials is that for a positive number and ε and µ real numbers ε µ µ ε µ ε. Thus we have ε µ µ ε ε µ M For each v V v v. ε µ µ ε µ ε µ ε µ A basic property of exponentials is that for a positive number. Thus This completes the rather tedious proof.

6 Math 0 Fall 206 Solution Suppose that V is a vector space and U U 2... U n are subspaces of V. Is U U n a subspace? Is U U 2 a subspace? Please justify your answers. Solution. We claim that U... U n is a subspace. Indeed for arbitrary v w V and F since the U i s are vector spaces and since v w are in each of them 0 v w v are in each of them as well. Therefore 0 v w v are in their intersection which shows that U... U n is a vector space. In general U U 2 needs not be a vector space. For example consider the vector subspaces U : { } x x R 0 and U 2 : { } 0 x R x 0 of R 2. We note that U 0 U 2 but their sum / U U 2. Thus U U 2 cannot be a subspace of R 2. In fact the only time when U U 2 happens to be a subspace is when U U 2 or U 2 U.