Problem 2.3. Using the fact that V is itself a vector space, prove that, with th

Question

Problem 2.3. Using the fact that V is itself a vector space, prove that, with the vector
addition and scalar multiplication defined above, V/~ is a vector space, i. e., show, in
detail, that the operations on V/~ satisfy the 10 vector space axioms. (Hint: Now that we
know the vector addition and scalar multiplication operations are well-defined, proving that
these axioms hold should reduce to using the fact that V is itself a vector space.)

Definition 7. For a vector space V and subspace Y of V , the vector space V/~ with vector
space operations defined as above is called the quotient space of V by Y and is denoted V=Y .
It is clear from this definition that different choices of subspaces Y can significantly
affect the final quotient space. An interesting fact, though one you will not need to prove,
is that the vector space dimension of the quotient space of a vector space V by a subspace
Y is given by the formula: dim(V/Y ) = dim(V ) - dim(Y ). Of course, one may question
the importance of such a complicated construction. In general, quotient spaces allow us to

Explanation / Answer

(i) ~ is an Equivalence relation on V. x in V, x~x always since (x-x) = 0 which belongs to Y as Y is a subspace of V and so must contain additive identity. If x~y this means (x-y) = m where m Y. thus (y-x) = -m which must also belong to Y as Y us a subspace and must contain the additive inverse of m. Thus x~y implies y~x If x~y and y~z this means (x-y)=m and (y-z)=n where m,n belong to Y. Thus (x-z) = (x-y)+(y-z) = m+n must also belong to Y as Y m,n both belong to Y and Y is a subspace. Hence ~ is reflexive, symmetric and transitive. ~ is an equivalence relation on V. (ii) 1) Associativity [u] + ([v]+[w]) = [(u+(v+w)] = [(u+v)+w)] = ([u]+[v])+[w] (because + is associative in V) 2) Commutativity [u] + [v] = [u+v] = [v+u] = [v] + [u] (bacause + is commutative in V) 3) Identity [0] is the identity element of V/~ as [0]+[v] = [0+v] = [v], [v] + [0] = [v+0] = [v] for all v in V 4) Inverse The inverse of [v] is [-v] as [v] + [-v] = [v+-v] = [0] 5) a*([u]+[v]) = a*[u+v] = [a*(u+v)] = [a*u + a*v] = [a*u]+[a*v] = a*[u] + a*[v] Hence vector multiplication is distributive, when a is from R. 6) (a+ß)[v] = [(a+ß)*v] = [a*v +ß*v] = [a*v] + [ß*v] = a*[v] + ß*[v] Hence scalar multiplication is distributive 7) a*(ß*[v]) = a*([ß*v]) = [a*(ß*v)] = [(a*ß)*(v)] = a*ß*[v] = (a*ß)*[v] Scalar and vector multiplications are compatible 8) 1*[v] = [1*v] = [v] 9) [u] + [v] = [u+v] which belongs to V/~ as u+v belongs to V when u and v belong to V 10) c*[u] = [c*u] which belongs to V/~ as c*u belongs to V when u belongs to V and c belongs to R

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