Math 110: Worksheet 1 Solutions

August 30

Thursday Aug. 24

1. Determine whether or not the following sets form vector spaces over the given ﬁelds. 1 a (a) The set V of all matrices of the form where a, b ∈ , over with standard b 1 R R addition and scalar multiplication.

1 a Note that V is not closed under addition: for a, b, c, d ∈ , we have and R b 1 1 c but d 1 1 a 1 c 2 a + c + = ∈/ V. b 1 d 1 b + d 2 We conclude that V is not a vector space with the given operations. 1 a (b) The set V of all matrices of the form where a, b ∈ , over with addition b 1 R R ⊕ and scalar multiplication deﬁned by

1 a 1 c 1 a + c 1 a 1 ka ⊕ = , k = . b 1 d 1 b + d 1 b 1 kb 1

We claim that V is indeed a vector space with the given operations. Note ﬁrst that V is closed under the addtion and scalar multiplication operations: for 1 a 1 c , ∈ V and k ∈ , we have b 1 d 1 R 1 a 1 c 1 a + c ⊕ = ∈ V b 1 d 1 b + d 1 and 1 a 1 ka k = ∈ V. b 1 kb 1

1 VS 1: Observe that 1 a 1 c 1 a + c ⊕ = ( defn. of addition) b 1 d 1 b + d 1 ∵ 1 c + a = ( comm. of addition in ) d + b 1 ∵ R 1 c 1 a = ⊕ ( addition) d 1 b 1 ∵ VS 2: Note that 1 a 1 c 1 e 1 a + c 1 e ⊕ ⊕ = ⊕ ( defn. of addition) b 1 d 1 f 1 b + d 1 f 1 ∵ 1 (a + c) + e = ( addition again) (b + d) + f 1 ∵ 1 a + (c + e) = ( add. assoc. in ) b + (d + f) 1 ∵ R 1 a 1 c + e = ⊕ ( addition) b 1 d + f 1 ∵ 1 a 1 c 1 e = ⊕ ⊕ ( addition) b 1 d 1 f 1 ∵

1 0 VS 3: The matrix ∈ V and acts as the zero vector: 0 1 1 a 1 0 1 a ⊕ = b 1 0 1 b 1

1 a 1 −a VS 4: Given any ∈ V , its additive inverse is as b 1 −b 1 1 a 1 −a 1 0 ⊕ = b 1 −b 1 0 1 VS 5: Observe that 1 a 1 a 1 = . b 1 b 1 VS 6: Let k, l ∈ R, we have 1 a 1 (kl)a (kl) = ( defn. of scalar mult) b 1 (kl)b 1 ∵ 1 k(la) = ( mult. assoc. in ) k(lb) 1 ∵ R 1 la = k ( scalar mult.) lb 1 ∵ 1 a = k l ( scalar mult.) b 1 ∵

2 VS 7: We have 1 a 1 c 1 a + c k ⊕ = k ( addition.) b 1 d 1 b + d 1 ∵ 1 k(a + c) = ( scalar mult.) k(b + d) 1 ∵ 1 ka + kc = ( dist. in ) kb + kd 1 ∵ R 1 ka 1 kc = ⊕ ( addition) kb 1 kd 1 ∵ 1 a 1 c = k ⊕ k ( scalar mult.) b 1 d 1 ∵ VS 8: We have 1 a 1 (k + l)a (k + l) = ( scalar mult.) b 1 (k + l)b 1 ∵ 1 ka + la = ( dist. in ) kb + lb 1 ∵ R 1 ka 1 la = ⊕ ( addition.) kb 1 lb 1 ∵ 1 a 1 a = k ⊕ l ( scalar mult.) b 1 b 1 ∵ (c) The set V of all positive real numbers over R with addition ⊕ and scalar multiplication deﬁned by x ⊕ y = xy, a x = xa.

We show that V is indeed a vector space with the given operations. Note ﬁrst that if x, y ∈ V and a ∈ R, we have x ⊕ y = xy ∈ V, a x = xa ∈ V so V is closed under addition and scalar multiplication. VS 1: We have x ⊕ y = xy (∵ addition) = yx (∵ mult. comm. in R) = y ⊕ x (∵ addition) VS 2: Note that (x ⊕ y) ⊕ z = (xy) ⊕ z (∵ addition) = (xy)z (∵ addition) = x(yz)(∵ mult. assoc. in R) = x ⊕ (yz)(∵ addition) = x ⊕ (y ⊕ z)(∵ addition)

3 VS 3: Observe that 1 ∈ V and

1 ⊕ x = 1x = x.

VS 4: For any x ∈ V , note that x−1 ∈ V so that

x ⊕ x−1 = xx−1 = 1.

VS 5: Note that 1 x = x1 = x.

VS 6: Let a, b ∈ R. We then have

ab (ab) x = x (∵ scalar mult.) b a = (x ) (∵ exponents in R) b = a (x )(∵ scalar mult.) = a (b x)(∵ scalar mult.)

VS 7: Note that

a (x ⊕ y) = a (xy)(∵ addition) a = (xy) (∵ scalar mult.) a a = x y (∵ exponents in R) a a = (x ) ⊕ (y )(∵ scalar mult.) = (a x) ⊕ (a y)(∵ addition)

VS 8: We have

a+b (a + b) x = x (∵ scalar mult.) a b = x x (∵ exponents in R) a b = (x ) ⊕ (x )(∵ addition) = (a x) ⊕ (b x)(∵ scalar mult.)

(d) The set V of solutions of the diﬀerential equation f 00(t) − 4f(t) = t, t ∈ R over R with standard addition and scalar multiplication.

Observe that V is not closed under addition: if f1, f2 ∈ V , then for all t ∈ R

00 00 f1 (t) − 4f1(t) = t, f2 (t) − 4f2(t) = t

so that

00 00 00 (f1 + f2) (t) − 4(f1 + f2)(t) = (f1 (t) − 4f1(t)) + (f2 (t) − 4f2(t)) = 2t 6≡ t.

We conclude that V is not a vector space with the given operations.

4 (e) The set V of 2 × 2 invertible matrices with real entries over R with standard addition and scalar multiplication.

1 0 1 0 Observe that V is not closed under addition: we have , ∈ V but 0 1 0 −1

1 0 1 0 2 0 + = 0 1 0 −1 0 0 is not invertible. We conclude that V is not a vector space with the given operations. 2. By definition, every field F has a multiplicative identity, an element e such that e·x = x for every element x ∈ F . What is the multiplicative identity for R? Prove that the multiplicative identity is unique for any given field.

The multiplicative identity for R is the number 1 as 1 · x = x for all x ∈ R. To show that the identity is unique, let e and e0 be two identities. Consider then the product e · e0. By thinking of e as an identity, we have e · e0 = e0. Likewise, thinking of e0 as an identity leads to e · e0 = e so that e = e0. Thus, the identity is unique.

Tuesday Aug. 29

3. Prove that the set of matrices with zero trace form a subspace of Mn×n(F ). Does the same hold for matrices with zero determinant?

Let T be the set of matrices with zero trace. As Mn×n(F ) is a vector space over F and T is its subset, we merely need to check three properties: • the matrix Z consisting only of zero entries evidently has zero trace so Z ∈ T . • let A, B ∈ T ; it follows then that tr(A) = tr(B) = 0. Note then that

n n n X X X tr(A + B) = (A + B)ii = Aii + Bii = tr(A) + tr(B) = 0. i=1 i=1 i=1

• let A ∈ T and k ∈ F ; we have tr(A) = 0 so that

n n X X tr(kA) = (kA)ii = k Aii = ktr(A) = 0. i=1 i=1

We conclude that T is a subspace of Mn×n(F ). The same cannot be said however about the set of matrices with zero determinant as it is not closed under addition. As an example, let A be the diagonal matrix with A11 = 0 and Aii = 1 for i = 2, . . . , n and let B consist only of zeros except for B11 = 1. Then, det(A) = det(B) = 0 but det(A + B) = 1 6= 0.

5 4. Let B(R) be the set of all bounded functions on R (A function f is bounded if there exists M such that |f(x)| ≤ M for all x. Thus sin(x) is bounded on R but ex is not). Prove that B(R) is a subspace of F(R, R), the set of all functions from R to R.

As F(R, R) is a vector space and B(R) is its subset, we just need to check the following three properties:

• the function z ≡ 0 is clearly bounded (as |z(x)| = 0 < 1 for all x) so z ∈ R. • let f, g ∈ B(R). Then there exist M,N such that |f(x)| ≤ M and |g(x)| ≤ N for all x ∈ R. Note then that, by the triangle inequality |(f + g)(x)| = |f(x) + g(x)| ≤ |f(x)| + |g(x)| ≤ M + N

for all x ∈ R; thus, (f + g) is bounded and hence in B(R). • let f ∈ B(R) and a ∈ R. Observe then that for all x ∈ R |(af)(x)| = |af(x)| = |a||f(x)| ≤ |a|M

so af ∈ B(R). We conclude that B(R) is a subspace of F(R, R).

5. Let W1 and W2 be subspaces of a vector space V . Prove that W1 + W2 = W2 if and only if W1 is a subspace of W2.

Suppose ﬁrst that W1 is a subspace of W2. Let t ∈ W1 + W2; there then exist w1 ∈ W1 and w2 ∈ W2 such that t = w1 + w2. As W1 is a subspace of W2, it follows that w1 ∈ W2 as well and hence t = w1 + w2 ∈ W2 so that W1 + W2 ⊂ W2. As we also have W2 ⊂ W1 + W2, we conclude that W1 + W2 = W2.

Conversely, suppose that W1 + W2 = W2; we want to show that W1 is a subspace of W2. Let w1 ∈ W1 and w2 ∈ W2; then, w1 + w2 ∈ W1 + W2. As W1 + W2 = W2, there exists some t ∈ W2 such that

w1 + w2 = t ⇒ w1 = t + (−w2) ∈ W2.

We conclude that W1 ⊂ W2 and, in particular, that W1 is a subspace of W2.

6. Let v1 = (0, 1) and v2 = (1, 1) and deﬁne W1 = {tv1 : t ∈ R} and W2 = {tv2 : t ∈ R}. Also, let V = R2 over R with standard operations.

(a) Show that W1 and W2 are subspaces of V .

As W1 and W2 are subsets of V which itself is a vector space, we just need to check the following three properties: (we treat both the spaces at the same time)

• 0 ∈ Wi by setting t = 0 in the deﬁnitions. • let x, y ∈ Wi. There then exist t, s ∈ R such that x = tvi and y = svi so that x + y = tvi + svi = (t + s)vi ∈ Wi.

6 • let x ∈ Wi and a ∈ R. Note then that

ax = a(tvi) = (at)vi ∈ Wi.

We conclude that both W1 and W2 are subspaces of V .

(b) Show that V = W1 ⊕ W2.

We need to show that (i) W1 ∩ W2 = {0} and (ii) W1 + W2 = V . For (i), note that if u ∈ W1 ∩ W2, then for some t, s ∈ R, we have u = tv1 and u = sv2 so that

tv1 = sv2 ⇒ (0, t) = (s, s).

It follows that s = 0 ⇒ t = 0 so u must be the zero vector.

For (ii), let x = (a, b) ∈ R. We want show that x = w1 + w2 for some w1 ∈ W1 and w2 ∈ W2. Note that setting w1 = (0, b − a) and w2 = (a, a) accomplishes this as wi ∈ Wi for i = 1, 2 and

w1 + w2 = (0, b − a) + (a, a) = (a, b) = v.

As both (i) and (ii) hold, we conclude that V = W1 ⊕ W2. 7. Let E and O denote respectively the subsets consisting of all the even and odd functions in V := F(R, R). In the homework, you are supposed to show that both E and O are subspaces of V . Assuming that, prove that V = E ⊕ O.

As in the previous problem, we just need to show that (i) E ∩ O = {0} and (ii) E + O = V . For (i), let f ∈ E ∩ O. Then, for any x ∈ R, we have f(−x) = f(x) and f(−x) = −f(x) so that f(x) = −f(x) ⇒ f(x) = 0 ⇒ f ≡ 0. For (ii), let f ∈ V . We need to show that f = g + h where g ∈ E and h ∈ O. Deﬁne for all x ∈ R f(x) + f(−x) f(x) − f(−x) g(x) = , h(x) = . 2 2 Note then that g(x) + h(x) = f(x) for all x. Furthermore, for any x ∈ R, we have f(−x) + f(x) g(−x) = = g(x) 2 f(−x) − f(x) h(−x) = = −h(x). 2 This shows that g ∈ E and h ∈ O and hence establishes (ii). We conclude that V = E ⊕ O.