Math 206, Spring 2016
Assignment 10 – Solutions
Due: April 8, 2016
Part A.
(1) Below we give several sets V equipped with addition and scaling operations denoted by ⊕ and . For
each, determine if V endowed with the operations ⊕ and is a vector space. If so, verify the existence
of a neutral element, the existence of additive inverses, associativity of addition, and distributivity of
scalar multiplication over vector addition. If not, give a specific example of vectors failing one of the
axioms.
(a) V = {r ∈ R : r > 0}, with a ⊕ b = ab (the expression on the right is the usual multiplication of real
numbers) and k a = ak (the expression on the right is the usual exponentiation of real numbers).
[You may use familiar properties of real arithmetic if you cite them properly.]
Solution. This is a vector space. We claim that 1 is the neutral element of V . To check this, let
r ∈ V be given (i.e., r is a positive real number). Then we have
1 ⊕ r = 1r = r
using familiar rules from real multiplication. Hence 1 is indeed the neutral element.
Now we show that inverses under ⊕ exist. So let r ∈ V be given; since r > 0, we have 1r is also a
positive real number. We then have
1
1
r ⊕ = r = 1.
r
r
We have already shown that 1 is the neutral element in V , and so this expression shows that 1r is
the inverse of r under ⊕.
Next we verify associativity. So let a, b, c ∈ V be given. We then have
a ⊕ (b ⊕ c) = a ⊕ bc
(definition of ⊕)
= a(bc)
(definition of ⊕)
= (ab)c
(real multiplication is associative)
= (ab) ⊕ c
(definition of ⊕)
= (a ⊕ b) ⊕ c
(definition of ⊕).
Finally we verify that scalar multiplication distributes over vector addition. So let a, b ∈ V be
given, and k ∈ R. We then have
k (a ⊕ b) = k (ab)
(definition of ⊕)
= (ab)k
(definition of )
k k
=a b
k
=a ⊕b
(exponent rules for real arithmetic)
k
(definition of ⊕)
= (k a) ⊕ (k b)
(definition of ).
n×n
(b) V = {M ∈ R
matrix scaling.
: M is invertible}, with M1 ⊕ M2 the usual matrix addition, and k M the usual
Solution. This is not a vector space. Recall that the identity matrix I is invertible; it’s also true
that −I is invertible, since (−I)(−I) = (−1)2 I = I. [Hence −I isn’t just invertible, but is in fact
its own inverse.] The given definition of ⊕ would then give
I ⊕ (−I) = I + (−I) = Z,
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Math 206, Spring 2016
Assignment 10 – Solutions
Due: April 8, 2016
where Z is the n × n matrix for which every entry is zero. But Z 6∈ V (since, for example, it’s rank
is 0 instead of n), and so V is not closed under ⊕.
(c) Suppose that V is a vector space with operations + and ·. Define new operations ⊕ and by
v1 ⊕ v2 = v1 + (−v2 ) (where on the right side, “+” is the original operation on V and −v2 is the
additive inverse of v2 under +), and k v = k · v (i.e., the new scaling operation is the same as
the old scaling operation).
Solution.
2
This is not always a vector space. Let V = R and consider v =
1
1
. Then we
have
1
(v ⊕ v) ⊕ v =
+
1
1
v ⊕ (v ⊕ v) =
⊕
1
−1
1
⊕
=
−1
1
1
−1
+
=
1
−1
−1
−1
+
=
−1
−1
1
0
1
+
=
.
1
0
1
0
0
Hence ⊕ is not associative.
[Note: one can prove a stronger result: if V is any vector space that contains more than just the
neutral element, then the operation ⊕ defined above will also fail to be associative. The proof of
this proceeds much like the above calculation, though v is chosen as any nonzero vector in V . The
two calculations then yield v and v. One then has to argue that −v 6= v, for which the axiom
1 · v = v is useful.
Finally: it’s not correct to say that for all V , the operations ⊕ and fail to make V into a vector
space. The reason is that if we let V = {0}, then the operations actually do make V into a
vector space. This happens essentially because addition and scaling on this vector space are always
stupid.]
(d) Suppose that V1 is a vector space under the operations of + and ·, and that V2 is a vector space
under the operations of and . Define V with operations ⊕ and by
V = V1 × V2 = {(v1 , v2 ) : v1 ∈ V1 , v2 ∈ V2 }
(v1 , v2 ) ⊕ (v10 , v20 ) = (v1 + v10 , v2 v20 )
k (v1 , v2 ) = (k · v1 , k
v2 ).
Solution. This is a vector space. First we verify associativity of addition. Let v, w, u ∈ V be
given. This means that there are v1 , w1 , u1 ∈ V1 and v2 , w2 , u2 ∈ V2 so that v = (v1 , v2 ), w =
(w1 , w2 ) and u = (u1 , u2 ). Observe that
(v ⊕ w) ⊕ u = ((v1 , v2 ) ⊕ (w1 , w2 )) ⊕ (u1 , u2 )
= (v1 + w1 , v2 w2 ) ⊕ (u1 , u2 )
(definition of ⊕)
= ((v1 + w1 ) + u1 , (v2 w2 ) u2 )
(definition of ⊕)
= (v1 + (w1 + u1 ) , v2 (w2 u2 ))
(associativity of +, )
= (v1 , v2 ) ⊕ (w1 + u1 , w2 u2 )
(definition of ⊕)
= (v1 , v2 ) ⊕ ((w1 , w2 ) ⊕ (u1 , u2 ))
(definition of ⊕)
Hence + is associative.
Now we verify that there exists a neutral element. Let 01 be the neutral element of V1 , and 02
the neutral element of V2 . (These elements exist because V1 and V2 are vector spaces.) We claim
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Math 206, Spring 2016
Assignment 10 – Solutions
Due: April 8, 2016
that (01 , 02 ) ∈ V acts neutrally. To prove this, let v ∈ V be given; as before, this means that
v = (v1 , v2 ) for some v1 ∈ V1 and v2 ∈ V2 . Note that
v ⊕ (01 , 02 ) = (v1 , v2 ) ⊕ (01 , 02 )
(definition of ⊕)
= (v1 + 01 , v2 02 )
= (v1 , v2 )
(definition of neutrality of 01 and 02 ).
Hence (01 , 02 ) acts neutrally, as claimed.
Now we show that additive inverses exist. Let v ∈ V be given; as before, this means v = (v1 , v2 )
for some v1 ∈ V1 and v2 ∈ V2 . Let w1 ∈ V1 be the additive inverse of v1 (which we know exists
since V1 is a vector space), and let w2 ∈ V2 be the additive inverse of v2 (which we know exists
since V2 is a vector space). We claim that (w1 , w2 ) is the additive inverse of v. To verify this,
observe that
v ⊕ (w1 , w2 ) = (v1 , v2 ) ⊕ (w1 , w2 )
(definition of ⊕)
= (v1 + w1 , v2 w2 )
= (01 , 02 )
(definition of additive inverse).
Hence (w1 , w2 ) is an additive inverse, as claimed.
Finally, we show that scaling distributes across vector sums. So let k inR be given, and v, w ∈ V .
As usual, we then have v = (v1 , v2 ) and w = (w1 , w2 ), where v1 , w1 ∈ V1 and v2 , w2 ∈ V2 . We
then have
k (v ⊕ w) = k ((v1 , v2 ) ⊕ (w1 , w2 ))
= k (v1 + w1 , v2 w2 )
(definition of ⊕)
= (k · (v1 + w1 ), k
(definition of )
(v2 w2 ))
= ((k · v1 ) + (k · w1 ), (k
= (k · v1 , k
v2 ) (k
v2 ) ⊕ (k · w1 , k
w2 ))
(since · and
distribute over + and )
w2 )
= (k (v1 , v2 )) ⊕ (k (w1 , w2 ))
This is the desire result.
(definition of ⊕)
(definition of ).
(2) Let V be a vector space with operations ⊕ and , and let W be a vector space with operations and
. We say that a function T : V → W respects addition if for all v1 , v2 ∈ V we have T (v1 ⊕ v2 ) =
T (v1 ) T (v2 ), and we say that T : V → W respects scalars if for all k ∈ R and v ∈ V we have
T (k v) = k T (v). Now define
L(V, W ) = {T : V → W : T respects addition and scalars}.
We define addition and scaling by
• if T, S ∈ L(V, W ), then we define T + S : V → W to be the function satisfying (T + S)(v) =
T (v) S(v) for all v ∈ V
• if k ∈ R and T ∈ L(V, W ), then we define k · T : V → W to be the function satisfying (k · T )(v) =
k T (v) for all v ∈ V .
Carefully prove that for all T, S ∈ L(V, W ) we have T + S ∈ L(V, W ), and that for all k ∈ R we have
k · T ∈ L(V, W ).
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Math 206, Spring 2016
Assignment 10 – Solutions
Due: April 8, 2016
Solution. To show that T + S is linear, first let v1 , v2 ∈ V be given. We need to argue that
(T + S)(v1 ⊕ v2 ) = (T + S)(v1 ) (T + S)(v2 ). We do this as follows
(T + S)(v1 ⊕ v2 ) = T (v1 ⊕ v2 ) S(v1 ⊕ v2 )
(definition of T + S)
= (T (v1 ) T (v2 )) (S(v1 ) S(v2 ))
(linearity of T and S)
= T (v1 ) (T (v2 ) (S(v1 ) S(v2 )))
(associativity of )
= T (v1 ) ((T (v2 ) S(v1 )) S(v2 ))
(associativity of )
= T (v1 ) ((S(v1 ) T (v2 )) S(v2 ))
(commutativity of )
= T (v1 ) (S(v1 ) (T (v2 ) S(v2 )))
(associativity of )
= (T (v1 ) S(v1 )) (T (v2 ) S(v2 ))
(associativity of )
= (T (v1 ) S(v1 )) (T (v2 ) S(v2 ))
(associativity of )
= (T + S)(v1 ) (T + S)(v2 )
(definition of S + T ).
To complete our proof of linearity of T + S, let k ∈ R and v ∈ V be given. Then we have
(T + S)(k v) = T (k v) S(k v)
= (k
T (v)) (k
=k
(T (v) S(v))
=k
((T + S)(v))
(definition of T + S)
S(v))
(linearity of T and S)
(since
distributes over )
(definition of T + S).
Hence T + S respects scalars. This completes the proof of linearity of T + S.
Now let ` ∈ R be given and T ∈ L(V, W ), and we argue that ` · T is linear. First, we show it respects
addition. Let v1 , v2 ∈ V be given. We need to argue that (` · T )(v1 ⊕ v2 ) = (` · T )(v1 ) (` · T )(v2 ).
We do this as follows
(` · T )(v1 ⊕ v2 ) = `
T (v1 ⊕ v2 )
=`
(definition of ` · T )
(T (v1 ) T (v2 ))
= (`
T (v1 )) (`
T (v2 ))
= ((` · T )(v1 )) ((` · T )(v2 ))
(linearity of T )
(
distributes across )
(definition of ` · T ).
Hence ` · T respects addition.
To complete our proof of linearity of ` · T , let k ∈ R and v ∈ V be given. Then we have
(` · T )(k v) = `
=`
T (k v)
(k
T (v))
(definition of ` · T )
(linearity of T )
= (`k)
T (v)
(scaling is associative)
= (k`)
T (v)
(real multiplication is commutative)
=k
(`
T (v))
(scaling is associative)
=k
((` · T )(v))
(definition of ` · T ).
Hence ` · T respects scalars. This completes the proof of linearity of ` · T .
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Math 206, Spring 2016
Assignment 10 – Solutions
Due: April 8, 2016
Part B.
(1) (a) Use only vector space axioms to prove the following theorem (which we call the cancellative property): If V is a vector space with operations + and ·, then if v, w1 , w2 ∈ V satisfy
v + w1 = v + w2 ,
then w1 = w2 .
Solution. Since V is a vector space, we know there exists an additive inverse of v; denote the
additive inverse by u. By definition, this means that v + u = 0, where 0 is the neutral element of
V . Observe then that
w1 = 0 + w1
(definition of neutral element)
= (v + u) + w1
(definition of additive inverse)
= (u + v) + w1
(commutativity of addition)
= u + (v + w1 )
(associativity of addition)
= u + (v + w2 )
(hypothesis)
= (u + v) + w2
(associativity of addition)
= (v + u) + w2
(commutativity of addition)
= 0 + w2
(definition of additive inverse)
= w2
(definition of neutral element).
(b) Prove that if v ∈ V is given, then it has a unique additive inverse. [Note: the vector space axioms
say that v is guaranteed to have some additive inverse, but doesn’t stipulate that the inverse in
unique; that’s what you’ll be proving here.]
Solution.
we have
Suppose that u and w are two elements of V with v + u = 0 and v + w = 0. Then
v + u = v + w.
The previous result then gives that u = w.
(c) Carefully use vector space axioms to prove that if v ∈ V is given, then −v = (−1) · v.
Solution. By the previous result, we can show that −v = (−1)·v by proving that v +((−1)·v) =
0. Let’s do it:
v + ((−1) · v) = (1 · v) + ((−1) · v)
= (1 + (−1)) · v
=0·v
=0
(vector space axiom)
(distributivity across scalar sums)
(definition of −1)
(result from class).
[Note that the results which weren’t explicitly written in the language of axioms — our appears
to part (b) and a result from class — were both proved using only axioms of a vector space. It is
in this sense that they are legal to use for this problem, since we could simply cut-and-paste the
previous arguments we have if we wanted to write everything fully in the language of axioms.] http://palmer.wellesley.edu/~aschultz/w16/math206
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Math 206, Spring 2016
Assignment 10 – Solutions
Due: April 8, 2016
(2) Use the subspace theorem to prove that the set of even functions E defined by
E = {f ∈ F (R, R) : f (−x) = f (x) for all x ∈ R}
is a subspace of F (R, R).
Solution. First, let Z be the neutral element of F (R, R). As we saw in class, for all x ∈ R we have
Z(x) = 0. In particular, for any x ∈ R we have that
Z(x) = 0 = Z(−x).
Hence Z ∈ E.
Now let f, g ∈ E be given. To show that f + g ∈ E, let x ∈ R be given. Then we have
(f + g)(−x) = f (−x) + g(−x)
(definition of f + g)
(since f, g ∈ E)
= f (x) + g(x)
= (f + g)(x)
(definition of f + g).
Hence f + g ∈ E.
Finally, let k ∈ R be given, and f ∈ E. We aim to show that the function k · f ∈ E, so let x ∈ R be
given. Then we have
(k · f )(−x) = k(f (−x))
(definition of k · f )
(since f ∈ E)
= k(f (x))
= (k · f )(x)
(definition of k · f ).
Hence k · f ∈ E.
By the subspace test, we see that E is a subspace of F (R, R).
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