Exercises 1B — Axler, Linear Algebra Done Right, 4th Edition
Exercise 1. Prove that \(-(-v) = v\) for all \(v \in V\), where \(V\) is a vector space over a field \(F\).
Proof.
Let \(w \in V\) be the additive inverse of \((-v)\). Then \(w + (-v) = 0\). Adding \(v\) to both sides, we get \(w + (-v) + v = 0 + v\).
\[ \begin{aligned} w + (-v) + v &= 0 + v \\ \Rightarrow\quad w + (-v + v) &= v && \text{[associativity of addition in } V \text{ \& def. of additive identity]} \\ \Rightarrow\quad w + 0 &= v && \text{[def. of additive inverse]} \\ \Rightarrow\quad w &= v && \text{[def. of additive identity]} \end{aligned} \]
Since \(w\) is the additive inverse of \((-v)\), i.e. \(w = -(-v)\), we conclude \(-(-v) = v\).
\[ \therefore\ -(-v) = v \quad \text{for all } v \in V. \qquad \square \]
Exercise 2. Suppose \(a \in F\), \(v \in V\), and \(av = 0\). Prove that \(a = 0\) or \(v = 0\), where \(V\) is a vector space over a field \(F\).
Proof.
We consider two cases.
Case 1: \(a \neq 0\).
Since \(a \neq 0\), it has a multiplicative inverse \(\tfrac{1}{a} \in F\). Scalar-multiplying both sides of \(av = 0\) by \(\tfrac{1}{a}\):
\[ \begin{aligned} \tfrac{1}{a}(av) &= \tfrac{1}{a} \cdot 0 \\ \Rightarrow\quad \bigl(\tfrac{1}{a} \cdot a\bigr)v &= 0 && \text{[associativity of scalar mult.]} \\ \Rightarrow\quad 1 \cdot v &= 0 && \text{[multiplicative inverse in } F\text{]} \\ \Rightarrow\quad v &= 0. && \text{[scalar mult. identity]} \end{aligned} \]
Case 2: \(v \neq 0\).
Suppose for contradiction that \(a \neq 0\). Then multiply both sides of \(av = 0\) by \(a^{-1}\):
\[ \begin{aligned} a^{-1}(av) &= a^{-1} \cdot 0 \\ \Rightarrow\quad (a^{-1}a)v &= 0 && \text{[associativity of scalar mult.]} \\ \Rightarrow\quad v &= 0, && \text{[multiplicative inverse \& scalar mult. identity]} \end{aligned} \]
which contradicts \(v \neq 0\). Therefore \(a = 0\).
\[ \therefore\ \text{For all } a \in F,\ v \in V,\text{ if } av = 0 \text{ then } a = 0 \text{ or } v = 0. \qquad \square \]
Exercise 3. Suppose \(v, w \in V\). Explain why there exists a unique \(x \in V\) such that \(v + 3x = w\).
Proof.
Existence. Adding the additive inverse of \(v\) to both sides and then scalar-multiplying both sides by the multiplicative inverse of \(3\):
\[ \begin{aligned} \tfrac{1}{3}(v + 3x + (-v)) &= \tfrac{1}{3}(w + (-v)) \\ \Rightarrow\quad \tfrac{1}{3}(3x) &= \tfrac{1}{3}(w + (-v)) && \text{[commutativity \& additive inverse in } V\text{]} \\ \Rightarrow\quad x &= \tfrac{1}{3}(w + (-v)). && \text{[scalar mult. identity]} \end{aligned} \]
So \(x = \tfrac{1}{3}(w + (-v)) \in V\) satisfies the equation, proving existence. \(\square\)
Uniqueness. Assume \(x' \in V\) is another vector satisfying \(v + 3x' = w\). Then:
\[ v + 3x = v + 3x' \qquad \text{(since both equal } w\text{)}. \]
Performing the same operations as above:
\[ \begin{aligned} \tfrac{1}{3}(v + 3x + (-v)) &= \tfrac{1}{3}(v + 3x' + (-v)) \\ \Rightarrow\quad \tfrac{1}{3}(3x) &= \tfrac{1}{3}(3x') \\ \Rightarrow\quad x &= x'. \end{aligned} \]
\[ \therefore\ x \text{ is unique.} \qquad \square \]
Exercise 4. The empty set is not a vector space. Which condition in the definition of a vector space is not satisfied?
Answer.
The empty set fails the additive identity axiom: a vector space must contain a zero vector \(0\), but the empty set has no elements at all, so no such element can exist. \(\square\)
Exercise 5. Show that in the definition of a vector space, the additive inverse condition can be replaced with the condition \(0v = 0\) for all \(v \in V\), where the first \(0\) is the zero scalar in \(F\) and the second \(0\) is the zero vector in \(V\).
Proof.
We show that the two conditions are equivalent by proving each implies the other.
Direction 1: Standard axioms (including additive inverse) \(\Rightarrow\) \(0_F v = 0_V\) for all \(v \in V\).
By distributivity (a retained axiom):
\[ \begin{aligned} 0_F v &= (0_F + 0_F)v = 0_F v + 0_F v. \end{aligned} \]
Adding the additive inverse of \(0_F v\) to both sides:
\[ 0_V = 0_F v. \qquad \square \]
(This is also established in Axler 1.30.)
Direction 2: \(0_F v = 0_V\) for all \(v \in V\) \(\Rightarrow\) every vector has an additive inverse.
Since \(F\) is a field, \(-1 \in F\), so \((-1)v \in V\) for any \(v \in V\). Using distributivity and the given condition:
\[ \begin{aligned} v + (-1)v &= 1 \cdot v + (-1)v \\ &= (1 + (-1))v && \text{[distributivity]} \\ &= 0_F v && \text{[additive inverse in } F\text{]} \\ &= 0_V. && \text{[given condition]} \end{aligned} \]
Thus \((-1)v\) is the additive inverse of \(v\) for every \(v \in V\).
\[ \therefore\ \text{The two conditions are equivalent.} \qquad \square \]
Exercise 6. Suppose \(\infty\) and \(-\infty\) are objects not in \(\mathbb{R}\). For \(t \in \mathbb{R}\), define
\[ t \cdot \infty = \begin{cases} -\infty & t < 0, \\ 0 & t = 0, \\ \infty & t > 0, \end{cases} \qquad t \cdot (-\infty) = \begin{cases} \infty & t < 0, \\ 0 & t = 0, \\ -\infty & t > 0, \end{cases} \]
and \(t + \infty = \infty + t = \infty + \infty = \infty\), \(\quad t + (-\infty) = (-\infty) + t = (-\infty) + (-\infty) = -\infty\), \(\quad \infty + (-\infty) = (-\infty) + \infty = 0\).
Is \(\mathbb{R} \cup \{\infty, -\infty\}\) a vector space over \(\mathbb{R}\)?
Solution.
No. With the given definitions, associativity of addition and distributivity of scalar multiplication both fail.
Associativity fails. Take \(1, \infty, -\infty \in \mathbb{R} \cup \{\infty, -\infty\}\). Then:
\[ 1 + \bigl(\infty + (-\infty)\bigr) = 1 + 0 = 1, \]
\[ (1 + \infty) + (-\infty) = \infty + (-\infty) = 0. \]
Since \(1 \neq 0\), addition is not associative.
Distributivity fails. Take \(v = \infty\), \(a = 2\), \(b = -1\). Then:
\[ (a + b)v = (2 + (-1))\infty = 1 \cdot \infty = \infty, \]
\[ av + bv = 2\infty + (-1)\infty = \infty + (-\infty) = 0. \]
Since \(\infty \neq 0\), the distributive law \((a+b)v = av + bv\) does not hold.
\[ \therefore\ \mathbb{R} \cup \{\infty, -\infty\} \text{ is not a vector space over } \mathbb{R}. \qquad \square \]
Exercise 7. Suppose \(S\) is a nonempty set and let \(V^S\) denote the set of all functions from \(S\) to \(V\), where \(V\) is a vector space over a field \(F\). Define addition and scalar multiplication on \(V^S\) pointwise: for \(f, g \in V^S\), \(\lambda \in F\), and \(x \in S\),
\[ (f + g)(x) = f(x) + g(x), \qquad (\lambda f)(x) = \lambda\, f(x). \]
Show that \(V^S\) is a vector space over \(F\).
Proof.
Since \(f(x), g(x) \in V\) and \(V\) is a vector space over \(F\), the operations are well-defined. We verify each axiom pointwise.
Commutativity of addition.
\[ (f + g)(x) = f(x) + g(x) = g(x) + f(x) = (g + f)(x), \]
so \(f + g = g + f\).
Associativity of addition.
\[ \bigl((f+g)+h\bigr)(x) = (f(x)+g(x))+h(x) = f(x)+(g(x)+h(x)) = \bigl(f+(g+h)\bigr)(x), \]
so \((f+g)+h = f+(g+h)\).
Associativity of scalar multiplication.
\[ \bigl((ab)f\bigr)(x) = (ab)\,f(x) = a\bigl(b\,f(x)\bigr) = a\bigl((bf)(x)\bigr) = \bigl(a(bf)\bigr)(x), \]
so \((ab)f = a(bf)\).
Additive identity. Define \(z \in V^S\) by \(z(x) = 0\) for all \(x \in S\). Then:
\[ (f + z)(x) = f(x) + z(x) = f(x) + 0 = f(x), \]
so \(f + z = f\).
Additive inverse. For each \(f \in V^S\), define \(y \in V^S\) by \(y(x) = -f(x)\) for all \(x \in S\). Then:
\[ (f + y)(x) = f(x) + y(x) = f(x) + (-f(x)) = 0 = z(x), \]
so \(f + y = z\).
Multiplicative identity.
\[ (1 \cdot f)(x) = 1 \cdot f(x) = f(x), \]
so \(1f = f\).
Distributivity (scalar over vector addition).
\[ \bigl(a(f+g)\bigr)(x) = a\bigl((f+g)(x)\bigr) = a\bigl(f(x)+g(x)\bigr) = af(x)+ag(x) = (af+ag)(x), \]
so \(a(f+g) = af + ag\).
Distributivity (vector over scalar addition).
\[ \bigl((a+b)f\bigr)(x) = (a+b)\,f(x) = af(x)+bf(x) = (af+bf)(x), \]
so \((a+b)f = af + bf\).
\[ \therefore\ V^S \text{ is a vector space over } F. \qquad \square \]
Exercise 8. Suppose \(V\) is a real vector space. The complexification of \(V\), denoted \(V_c\), equals \(V \times V\), where we write a pair \((u,v) \in V_c\) as \(u + iv\). Define addition and scalar multiplication on \(V_c\) by:
\[ (u_1 + iv_1) + (u_2 + iv_2) = (u_1+u_2) + i(v_1+v_2), \qquad u_1,v_1,u_2,v_2 \in V, \]
\[ (a_R + a_I i)(u + iv) = (a_R u - a_I v) + i(a_R v + a_I u), \qquad a_R, a_I \in \mathbb{R},\ u,v \in V. \]
Prove that \(V_c\) is a vector space over \(\mathbb{C}\).
Proof.
Since \(u, v \in V\) and \(V\) is a real vector space, all operations below in \(V\) are well-defined. We verify each axiom.
Commutativity of addition.
\[ \begin{aligned} (u_1+iv_1)+(u_2+iv_2) &= (u_1+u_2)+i(v_1+v_2) \\ &= (u_2+u_1)+i(v_2+v_1) && \text{[commutativity in } V\text{]} \\ &= (u_2+iv_2)+(u_1+iv_1). \end{aligned} \]
Associativity of addition.
\[ \begin{aligned} &\bigl((u_1+iv_1)+(u_2+iv_2)\bigr)+(u_3+iv_3) \\ &= \bigl((u_1+u_2)+i(v_1+v_2)\bigr)+(u_3+iv_3) \\ &= \bigl((u_1+u_2)+u_3\bigr)+i\bigl((v_1+v_2)+v_3\bigr) \\ &= \bigl(u_1+(u_2+u_3)\bigr)+i\bigl(v_1+(v_2+v_3)\bigr) && \text{[associativity in } V\text{]} \\ &= (u_1+iv_1)+\bigl((u_2+u_3)+i(v_2+v_3)\bigr) \\ &= (u_1+iv_1)+\bigl((u_2+iv_2)+(u_3+iv_3)\bigr). \end{aligned} \]
Associativity of scalar multiplication.
Define \(I, J : V_c \to V_c\) by \(I(u,v) = (u,v)\) and \(J(u,v) = (-v,u)\). Note \(J^2 = -I\). The scalar multiplication rule is then \(a \cdot x = (a_R I + a_I J)x\) for \(a = a_R + a_I i \in \mathbb{C}\) and \(x \in V_c\). For \(z = z_R + z_I i\) and \(w = w_R + w_I i\):
\[ \begin{aligned} z \cdot (w \cdot x) &= (z_R I + z_I J)(w_R I + w_I J)\,x \\ &= \bigl(z_R w_R I + z_R w_I J + z_I w_R J + z_I w_I J^2\bigr)x \\ &= \bigl((z_R w_R - z_I w_I)I + (z_R w_I + z_I w_R)J\bigr)x \\ &= (zw) \cdot x, \end{aligned} \]
since \(zw = (z_R w_R - z_I w_I) + (z_R w_I + z_I w_R)i\) by definition of complex multiplication.
Additive identity. Let \(0_c = 0 + i0 \in V_c\) where \(0 \in V\). Then:
\[ (u+iv) + 0_c = (u+iv) + (0+i0) = (u+0)+i(v+0) = u+iv. \]
Additive inverse. For \(u + iv \in V_c\), let \((u'+iv') = (-u)+i(-v)\). Then:
\[ (u+iv)+(u'+iv') = (u+(-u))+i(v+(-v)) = 0+i0 = 0_c. \]
Multiplicative identity. Let \(\lambda = 1 + 0i \in \mathbb{C}\). Then:
\[ \lambda(u+iv) = (1\cdot u - 0\cdot v)+i(1\cdot v+0\cdot u) = u+iv. \]
Distributivity (scalar over vector addition). Let \(a = a_R + a_I i\).
\[ \begin{aligned} a\bigl((u_1+iv_1)+(u_2+iv_2)\bigr) &= a\bigl((u_1+u_2)+i(v_1+v_2)\bigr) \\ &= \bigl(a_R(u_1+u_2)-a_I(v_1+v_2)\bigr)+i\bigl(a_R(v_1+v_2)+a_I(u_1+u_2)\bigr) \\ &= \bigl(a_Ru_1-a_Iv_1\bigr)+i\bigl(a_Rv_1+a_Iu_1\bigr) +\bigl(a_Ru_2-a_Iv_2\bigr)+i\bigl(a_Rv_2+a_Iu_2\bigr) \\ &= a(u_1+iv_1)+a(u_2+iv_2). \end{aligned} \]
Distributivity (vector over scalar addition). Let \(a = a_R+a_Ii\), \(b = b_R+b_Ii\).
\[ \begin{aligned} (a+b)(u+iv) &= \bigl((a_R+b_R)u-(a_I+b_I)v\bigr)+i\bigl((a_R+b_R)v+(a_I+b_I)u\bigr) \\ &= \bigl(a_Ru-a_Iv\bigr)+i\bigl(a_Rv+a_Iu\bigr)+\bigl(b_Ru-b_Iv\bigr)+i\bigl(b_Rv+b_Iu\bigr) \\ &= a(u+iv)+b(u+iv). \end{aligned} \]
\[ \therefore\ V_c \text{ is a vector space over } \mathbb{C}. \qquad \square \]