A vector space (over the reals) is a set $V$ together with two binary operations which we call addition (+) and scalar multiplication (·). The addition operation is a function which takes two elements of $V$, say $v$ and $w$, and returns a third element of $V$, which we write as $v+w$, while scalar multiplication is a function which takes a real number and an element of $V$,say $s$ and $v$, and returms another element of $V$,which we write as $s\cdot v.$ The operations are called“binary”because they take two arguments. The two operations must satisfy the following properties:
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Problem Ⅰ: Vector Spaces (You are currently browsing this post)
Proof. To establish that $V$ is a vector space over the field $\mathbb{R}$, it suffices to verify that the eight vector space axioms are satisfied. Axiom 1: Commutativity of Addition
\[\forall u, v \in V, \quad u + v = v + u.\]Verification: The standard addition of real numbers is commutative; hence, for any $u, v \in \mathbb{R}$,
\[u + v = v + u.\]Thus, Axiom 1 holds.
Axiom 2: Associativity of Addition
\[\forall u, v, w \in V, \quad (u + v) + w = u + (v + w).\]Verification: The standard addition of real numbers is associative; therefore, for any $u, v, w \in \mathbb{R}$,
\[(u + v) + w = u + (v + w).\]Thus, Axiom 2 is satisfied.
Axiom 3: Existence of Additive Identity
\[\exists \tilde{0} \in V \text{ such that } \forall v \in V, \quad \tilde{0} + v = v.\]Verification: In $\mathbb{R}$, the additive identity is $0$. For any $v \in \mathbb{R}$,
\[0 + v = v.\]Hence, $\tilde{0} = 0$ serves as the additive identity, satisfying Axiom 3.
Axiom 4: Existence of Additive Inverses
\[\forall v \in V, \exists w \in V \text{ such that } v + w = \tilde{0}.\]Verification: For any $v \in \mathbb{R}$, the additive inverse is $-v$. Thus,
\[v + (-v) = 0 = \tilde{0}.\]Therefore, every element $v$ in $V$ has an additive inverse $-v$, satisfying Axiom 4.
Axiom 5: Scalar Multiplication Distributes over Vector Addition
\[\forall s \in \mathbb{R}, \forall u, v \in V, \quad s \cdot (u + v) = s \cdot u + s \cdot v.\]Verification: The distributive property of real numbers ensures that for any $s, u, v \in \mathbb{R}$,
\[s \cdot (u + v) = s \cdot u + s \cdot v.\]Thus, Axiom 5 is satisfied.
Axiom 6: Scalar Addition Distributes over Scalar Multiplication
\[\forall a, b \in \mathbb{R}, \forall v \in V, \quad (a + b) \cdot v = a \cdot v + b \cdot v.\]Verification: The distributive property of real numbers implies that for any $a, b, v \in \mathbb{R}$,
\[(a + b) \cdot v = a \cdot v + b \cdot v.\]Therefore, Axiom 6 holds.
Axiom 7: Compatibility of Scalar Multiplication with Field Multiplication
\[\forall s, c \in \mathbb{R}, \forall v \in V, \quad (s \cdot c) \cdot v = s \cdot (c \cdot v).\]Verification: The associativity of multiplication in $\mathbb{R}$ ensures that for any $s, c, v \in \mathbb{R}$,
\[(s \cdot c) \cdot v = s \cdot (c \cdot v).\]Hence, Axiom 7 is satisfied.
Axiom 8: Scalar Multiplicative Identity
\[\forall v \in V, \quad 1 \cdot v = v.\]Verification: In $\mathbb{R}$, the multiplicative identity is $1$. For any $v \in \mathbb{R}$,
\[1 \cdot v = v.\]Thus, Axiom 8 holds.
Having verified all eight vector space axioms, $V = \mathbb{R}$ with standard addition and scalar multiplication is a vector space over $\mathbb{R}$. $\textbf{Q.E.D.}$
Commentary
The question (a) and the subsequent (b) may seem “obvious,” but proving them from the most basic axioms is neither common nor necessary in the study of engineering and applied sciences. However, I personally believe it contains three layers of meaning:
- Any mathematical theorem is derived through progressive steps from the most fundamental axioms, and the assumptions are similar. There is a strong intuitive, and consequently logical, consideration here. It is not as simple as it seems. (Here’s a famous joke: How do you determine if a 3-year-old child has a talent for mathematics? Ask them why $\textit{1+2 = 2+1}$. If they answer, “Because of the commutative property of addition,” their mathematical talent is average. If they answer, “Because the set of integers forms an Abelian group under addition,” then they are highly gifted.)
- Even knowing these proof methods, presenting them in a formal manner, particularly using \(\LaTeX\), professional expressions, and more tools, are more challenging tasks.
- When writing papers in the field of computer science, is there still room for expansion in your mathematical derivations?
Proof. Let $X$ be an arbitrary set, and let $F$ denote the set of all functions from $X$ to $\mathbb{R}$, i.e., $F =$ { $f \mid f: X \to \mathbb{R}$ }. Define the operations of addition and scalar multiplication on $F$ as follows:
\[\forall f, g \in F, \quad (f + g)(x) = f(x) + g(x) \quad \text{for all } x \in X,\] \[\forall s \in \mathbb{R}, \forall f \in F, \quad (s \cdot f)(x) = s \cdot f(x) \quad \text{for all } x \in X.\]To establish that $F$ is a vector space over the field $\mathbb{R}$, it is requisite to verify that the eight axioms defining a vector space are satisfied under the aforementioned operations.
Axiom 1: Commutativity of Addition
\[\forall f, g \in F, \quad f + g = g + f.\]Verification: For any $f, g \in F$ and for all $x \in X$,
\[(f + g)(x) = f(x) + g(x) = g(x) + f(x) = (g + f)(x).\]Since this equality holds for all $x \in X$, it follows that $f + g = g + f$. Hence, addition is commutative.
Axiom 2: Associativity of Addition
\[\forall f, g, h \in F, \quad (f + g) + h = f + (g + h).\]Verification: For any $f, g, h \in F$ and for all $x \in X$,
\[((f + g) + h)(x) = (f + g)(x) + h(x) = (f(x) + g(x)) + h(x),\] \[(f + (g + h))(x) = f(x) + (g + h)(x) = f(x) + (g(x) + h(x)).\]By the associativity of real number addition,
\[(f(x) + g(x)) + h(x) = f(x) + (g(x) + h(x)),\]which implies
\[((f + g) + h)(x) = (f + (g + h))(x).\]Since this holds for all $x \in X$, it follows that $(f + g) + h = f + (g + h)$. Hence, addition is associative.
Axiom 3: Existence of Additive Identity
\[\exists \tilde{0} \in F \text{ such that } \forall f \in F, \quad \tilde{0} + f = f.\]Verification: Define $\tilde{0}$ as the zero function, i.e., $\tilde{0}(x) = 0$ for all $x \in X$. For any $f \in F$ and for all $x \in X$,
\[(\tilde{0} + f)(x) = \tilde{0}(x) + f(x) = 0 + f(x) = f(x).\]Thus, $\tilde{0} + f = f$ for all $f \in F$, establishing the existence of an additive identity in $F$.
Axiom 4: Existence of Additive Inverses
\[\forall f \in F, \exists g \in F \text{ such that } f + g = \tilde{0}.\]Verification: For any $f \in F$, define $g = -f$, where $(-f)(x) = -f(x)$ for all $x \in X$. Then, for all $x \in X$,
\[(f + g)(x) = f(x) + (-f(x)) = 0 = \tilde{0}(x).\]Therefore, $f + g = \tilde{0}$, confirming the existence of an additive inverse for every element in $F$.
Axiom 5: Scalar Multiplication Distributes over Vector Addition
\[\forall s \in \mathbb{R}, \forall f, g \in F, \quad s \cdot (f + g) = s \cdot f + s \cdot g.\]Verification: For any $s \in \mathbb{R}$, $f, g \in F$, and for all $x \in X$,
\[(s \cdot (f + g))(x) = s \cdot (f + g)(x) = s \cdot (f(x) + g(x)),\] \[(s \cdot f + s \cdot g)(x) = (s \cdot f)(x) + (s \cdot g)(x) = s \cdot f(x) + s \cdot g(x).\]By the distributive property of real numbers,
\[s \cdot (f(x) + g(x)) = s \cdot f(x) + s \cdot g(x),\]which implies
\[(s \cdot (f + g))(x) = (s \cdot f + s \cdot g)(x).\]Since this holds for all $x \in X$, it follows that $s \cdot (f + g) = s \cdot f + s \cdot g$. Thus, scalar multiplication distributes over vector addition.
Axiom 6: Scalar Addition Distributes over Scalar Multiplication
\[\forall a, b \in \mathbb{R}, \forall f \in F, \quad (a + b) \cdot f = a \cdot f + b \cdot f.\]Verification: For any $a, b \in \mathbb{R}$, $f \in F$, and for all $x \in X$,
\[((a + b) \cdot f)(x) = (a + b) \cdot f(x),\] \[(a \cdot f + b \cdot f)(x) = (a \cdot f)(x) + (b \cdot f)(x) = a \cdot f(x) + b \cdot f(x).\]By the distributive property of real numbers,
\[(a + b) \cdot f(x) = a \cdot f(x) + b \cdot f(x),\]which implies
\[((a + b) \cdot f)(x) = (a \cdot f + b \cdot f)(x).\]Since this equality holds for all $x \in X$, it follows that $(a + b) \cdot f = a \cdot f + b \cdot f$. Therefore, scalar addition distributes over scalar multiplication.
Axiom 7: Compatibility of Scalar Multiplication with Field Multiplication
\[\forall a, b \in \mathbb{R}, \forall f \in F, \quad a \cdot (b \cdot f) = (a \cdot b) \cdot f.\]Verification: For any $a, b \in \mathbb{R}$, $f \in F$, and for all $x \in X$,
\[(a \cdot (b \cdot f))(x) = a \cdot (b \cdot f)(x) = a \cdot (b \cdot f(x))\] \[((a \cdot b) \cdot f)(x) = (a \cdot b) \cdot f(x).\]By the associative property of real number multiplication,
\[a \cdot (b \cdot f(x)) = (a \cdot b) \cdot f(x),\]which implies
\[(a \cdot (b \cdot f))(x) = ((a \cdot b) \cdot f)(x).\]Since this holds for all $x \in X$, it follows that $a \cdot (b \cdot f) = (a \cdot b) \cdot f$. Hence, scalar multiplication is compatible with field multiplication.
Axiom 8: Existence of Scalar Multiplicative Identity
\[\forall f \in F, \quad 1 \cdot f = f.\]Verification: For any $f \in F$ and for all $x \in X$,
\[(1 \cdot f)(x) = 1 \cdot f(x) = f(x).\]Therefore, $1 \cdot f = f$ for all $f \in F$, establishing the existence of a scalar multiplicative identity in $F$.
After verifying all eight axioms, we conclude that $F =$ { $f \mid f: X \to \mathbb{R}$ } with standard function addition and scalar multiplication forms a vector space over $\mathbb{R}$. $\textbf{Q.E.D.}$
Let $V$ be a vector space over the field $\mathbb{R}$, equipped with the operations of vector addition $+$ and scalar multiplication $\cdot$. Denote by $\vec{0}$ the additive identity in $V$, satisfying
\[\forall v \in V, \quad \vec{0} + v = v.\]Theorem 1.c.1. In any vector space $V$ over $\mathbb{R}$, the additive identity is unique.
Proof. Suppose $i \in V$ is an element such that there exists $w \in V$ with
\[i + w = w.\]Consider the additive identity $\vec{0}$ in $V$, which satisfies
\[\forall v \in V, \quad \vec{0} + v = v.\]In particular, for the vector $w \in V$, we have
\[\vec{0} + w = w.\]Given that both $i$ and $\vec{0}$ satisfy the property of being additive identities with respect to the vector $w$, we have
\[i + w = w \quad \text{and} \quad \vec{0} + w = w.\]Therefore, we can equate the two expressions:
\[i + w = \vec{0} + w.\]By the associativity and cancellation properties of vector addition, we can deduce the following:
\[i + w = \vec{0} + w \implies (i + w) = (\vec{0} + w).\]Subtracting $w$ from both sides, we obtain
\[i = \vec{0}.\]$\textbf{Q.E.D.}$
Commentary
The cancellation property in a vector space states that if $u + v = u + w$, then $v = w$. Applying this property to the equation $i + w = \vec{0} + w$, we cancel $w$ from both sides to conclude that $i = \vec{0}$.
Let $V$ be a vector space over the field $\mathbb{R}$, equipped with the operations of vector addition $+$ and scalar multiplication $\cdot$. Denote by $\vec{0}$ the additive identity in $V$, satisfying
\[\forall v \in V, \quad \vec{0} + v = v.\]Theorem 1.d.1. In any vector space $V$ over $\mathbb{R}$, the scalar multiplication of any real number $s$ with the additive identity $\vec{0}$ satisfies $s \cdot \vec{0} = \vec{0}$.
Proof. Let $s \in \mathbb{R}$ be an arbitrary scalar. Consider the scalar multiplication $s \cdot \vec{0}$.
By the distributive property of scalar multiplication over vector addition (Axiom 5), we have:
\[s \cdot (\vec{0} + \vec{0}) = s \cdot \vec{0} + s \cdot \vec{0}.\]Since $\vec{0}$ is the additive identity, it satisfies:
\[\vec{0} + \vec{0} = \vec{0}.\]Substituting this into the previous equation yields:
\[s \cdot \vec{0} = s \cdot \vec{0} + s \cdot \vec{0}.\]Let $x = s \cdot \vec{0}$. Then the equation becomes:
\[x = x + x.\]Subtracting $x$ from both sides (utilizing the existence of additive inverses, Axiom 4), we obtain:
\[x - x = (x + x) - x.\]Simplifying both sides:
\[\vec{0} = x.\]Thus,
\[s \cdot \vec{0} = \vec{0}.\]Since $s \in \mathbb{R}$ was arbitrary, the equality holds for all real scalars $s$. $\textbf{Q.E.D.}$
Let $V$ be a vector space over the field $\mathbb{R}$, endowed with the operations of vector addition $+$ and scalar multiplication $\cdot$. Denote by $\vec{0}$ the additive identity in $V$, satisfying
\[\forall v \in V, \quad \vec{0} + v = v.\]Theorem 1.e.1. In any vector space $V$ over $\mathbb{R}$, the scalar zero acts as the additive identity under scalar multiplication; that is, for all $w \in V$, $0 \cdot w = \vec{0}$.
Proof. Let $w \in V$ be arbitrary. Consider the scalar multiplication of $0$ with $w$:
\[0 \cdot w.\]By the distributive property of scalar multiplication over scalar addition (Axiom 6), we have:
\[(0 + 0) \cdot w = 0 \cdot w + 0 \cdot w.\]Calculating the left-hand side using the fact that $0 + 0 = 0$ in $\mathbb{R}$:
\[0 \cdot w = 0 \cdot w + 0 \cdot w.\]Let $x = 0 \cdot w$. Substituting, we obtain:
\[x = x + x.\]To isolate $x$, we subtract $x$ from both sides of the equation (utilizing the existence of additive inverses, Axiom 4):
\[x - x = (x + x) - x.\]Simplifying both sides:
\[\vec{0} = x.\]Thus,
\[0 \cdot w = \vec{0}.\]Since $w \in V$ was arbitrary, the equality holds for all $w \in V$. $\textbf{Q.E.D.}$
Commentary Through the application of the distributive property of scalar multiplication over scalar addition and the properties of additive inverses within the vector space $V$, it has been rigorously established that for the scalar zero $0 \in \mathbb{R}$ and for any vector $w \in V$, the scalar multiplication $0 \cdot w$ yields the additive identity $\vec{0}$. Therefore, $0 \cdot w = \vec{0}$ holds universally within the structure of the vector space $V$.
Commentary
Let $V$ be a vector space over the field $\mathbb{R}$, endowed with the operations of vector addition $+$ and scalar multiplication $\cdot$. In this context, we consider a modification of the standard vector space axioms by omitting the requirement for the existence of additive inverses. Specifically, we investigate the uniqueness of the additive identity under this relaxed structure.
Theorem 1.f.1. In a vector space $V$ over $\mathbb{R}$, where the existence of additive inverses is not assumed, if an element $i \in V$ satisfies $i + w = w$ for all $w \in V$, then $i$ coincides with the additive identity $\vec{0}$.
Proof. Assume $V$ is a vector space over $\mathbb{R}$ with the operations $+$ and $\cdot$ satisfying all vector space axioms except the existence of additive inverses. Let $\vec{0} \in V$ denote the additive identity, satisfying
\[\forall w \in V, \quad \vec{0} + w = w.\]Suppose $i \in V$ is an element such that
\[\forall w \in V, \quad i + w = w.\]Consider an arbitrary element $w \in V$. By the given condition, we have
\[i + w = w.\]Subtracting $\vec{0}$ from both sides (noting that $\vec{0}$ acts as the additive identity), we obtain
\[i + w + \vec{0} = w + \vec{0}.\]However, since $\vec{0}$ is the additive identity, it satisfies
\[\vec{0} + v = v \quad \text{for all } v \in V.\]Thus, the equation simplifies to
\[i + w = w.\]Given that $i + w = w$ holds for all $w \in V$, we can deduce the following by substituting $w = \vec{0}$:
\[i + \vec{0} = \vec{0}.\]But since $\vec{0}$ is the additive identity,
\[i + \vec{0} = i.\]Combining the two results, we obtain
\[i = \vec{0}.\]Hence, the element $i$ satisfying $i + w = w$ for all $w \in V$ must be equal to the additive identity $\vec{0}$. $\textbf{Q.E.D.}$
Commentary
The result obtained herein is notably weaker than the corresponding assertion in Problem c, where the existence of additive inverses is assumed. In Problem c, the proof leverages the existence of additive inverses to facilitate the cancellation of terms, thereby directly establishing the equality $i = \vec{0}$ through subtraction. Specifically, the presence of additive inverses allows for the manipulation:
\[i + w = \vec{0} + w \implies i = \vec{0},\]by subtracting $w$ from both sides.
In contrast, Problem e circumvents the necessity of additive inverses by employing a more restrictive approach, relying solely on the properties of the additive identity. Consequently, while both proofs culminate in the uniqueness of the additive identity, the proof in Problem e is constrained by the absence of additive inverses and thus offers a less general mechanism for establishing $i = \vec{0}$.
Let $V$ be a vector space over the field $\mathbb{R}$, equipped with the operations of vector addition $+$ and scalar multiplication $\cdot$. Within this framework, for each vector $v \in V$, there exists an additive inverse $w \in V$ such that
\[v + w = \vec{0},\]where $\vec{0}$ denotes the additive identity in $V$.
Theorem 1.g.1. In any vector space $V$ over $\mathbb{R}$, the additive inverse of a vector $v \in V$ is unique. That is, if $w$ and $w’$ are both additive inverses of $v$, then $w = w’$.
Proof. Let $V$ be a vector space over $\mathbb{R}$ satisfying the standard vector space axioms, including the existence of additive inverses. Suppose $w, w’ \in V$ are both additive inverses of a vector $v \in V$. By definition of additive inverses, we have
\[v + w = \vec{0}, \quad \text{and} \quad v + w' = \vec{0}.\]Consider the following sequence of equalities:
\[v + w = \vec{0}, \quad \text{and} \quad v + w' = \vec{0}.\]Subtracting the first equation from the second yields
\[(v + w') - (v + w) = \vec{0} - \vec{0}.\]Applying the associativity and commutativity of vector addition, we have
\[v + w' - v - w = \vec{0}.\]Rearranging terms, this simplifies to
\[(v - v) + (w' - w) = \vec{0}.\]Since $v - v = \vec{0}$ (by the definition of additive inverses and the additive identity), the equation becomes
\[\vec{0} + (w' - w) = \vec{0}.\]By the property of the additive identity, $\vec{0} + x = x$ for any $x \in V$. Therefore,
\[w' - w = \vec{0}.\]Adding $w$ to both sides yields
\[w' = w.\]Thus, $w = w’$ establishing that the additive inverse of $v$ is unique. $\textbf{Q.E.D.}$
Commentray The uniqueness of additive inverses is a pivotal property in the theory of vector spaces. It ensures that each vector possesses exactly one inverse, thereby preventing ambiguity in vector operations and facilitating the construction of linear combinations and subspaces.
Let $V$ be a vector space over the field $\mathbb{R}$, equipped with the operations of vector addition $+$ and scalar multiplication $\cdot$. Within this framework, for each vector $v \in V$, there exists a unique additive inverse $v’ \in V$ such that
\[v + v' = \vec{0},\]where $\vec{0}$ denotes the additive identity in $V$.
Theorem 1.h.1. In any vector space $V$ over $\mathbb{R}$, if $v’$ is the additive inverse of $v \in V$, then for any $w \in V$ satisfying $w \neq v$, the equation $w + v’ \neq \vec{0}$ holds.
Proof. Assume $V$ is a vector space over $\mathbb{R}$ satisfying all vector space axioms, including the existence of additive inverses. Let $v \in V$ be an arbitrary vector, and let $v’ \in V$ be its additive inverse, satisfying
\[v + v' = \vec{0}.\]Suppose $w \in V$ is another vector such that $w \neq v$.
Assume for the sake of contradiction that
\[w + v' = \vec{0}.\]Then adding $v$ to both sides of the equation yield \(v + (w + v') = v + \vec{0}.\)
By the associativity of vector addition (Axiom 3), this can be rewritten as
\[(v + w) + v' = v.\]Given that $v + v’ = \vec{0}$, substitute to obtain
\[\vec{0} + w = v.\]By the definition of the additive identity (Axiom 2),
\[w = v.\]This contradicts the initial assumption that $w \neq v$. Therefore, the assumption that $w + v’ = \vec{0}$ must be false. Consequently,
\[w + v' \neq \vec{0}.\]Thus, for any $w \in V$ with $w \neq v$, it holds that $w + v’ \neq \vec{0}$. $\textbf{Q.E.D.}$
Commentray
Through a contradiction argument grounded in the fundamental axioms of vector spaces, it has been rigorously demonstrated that if $v’$ is the additive inverse of $v \in V$, then for any $w \in V$ distinct from $v$, the sum $w + v’$ cannot equal the additive identity $\vec{0}$.
The result established herein is a weaker assertion compared to the conclusion derived in Question (c), where the uniqueness of the additive identity itself was proven. Specifically, Question (c) demonstrated that any element $i \in V$ satisfying $i + w = w$ for some $w \in V$ must coincide with the additive identity $\vec{0}$. This directly leveraged the existence of additive inverses to facilitate the cancellation of terms, thereby establishing the equality $i = \vec{0}$ unequivocally.
In contrast, Question (h) addresses a more nuanced scenario wherein the addition of an additive inverse $v’$ to a vector $w \neq v$ does not result in the additive identity $\vec{0}$. This conclusion does not rely on the existence of additive inverses for vectors other than $v$ and does not assert the uniqueness of additive inverses per se. Instead, it affirms that distinct vectors retain their distinctiveness when combined with the additive inverse of a specific vector, provided they are not identical to that vector.
Thus, while Question (c) establishes a fundamental uniqueness property of the additive identity within the vector space, Question (h) extends this uniqueness to the interaction between distinct vectors and the additive inverse of a particular vector, albeit in a less general context. The latter does not encompass the full breadth of additive inverse uniqueness across the entire vector space but rather focuses on the non-vanishing of specific vector combinations.
Theorem 1.i.1. In any vector space $V$ over $\mathbb{R}$, the scalar multiplication by $-1$ yields the additive inverse of any vector $v \in V$. That is, for all $v \in V$, \((-1) \cdot v = -v.\)
Proof. Let $V$ be a vector space over $\mathbb{R}$ satisfying all vector space axioms, including the existence of additive inverses. Let $v \in V$ be an arbitrary vector, and let $v’ \in V$ denote its additive inverse, satisfying
\[v + v' = \vec{0}.\]Consider the scalar multiplication of $-1$ with $v$:
\[(-1) \cdot v.\]We analyze the sum of $v$ and $(-1) \cdot v$:
\[v + [(-1) \cdot v].\]By the distributive property of scalar multiplication over vector addition (Axiom 5), we have
\[v + [(-1) \cdot v] = 1 \cdot v + (-1) \cdot v.\]Applying the distributive property of scalar addition over scalar multiplication (Axiom 6), we obtain
\[1 \cdot v + (-1) \cdot v = (1 + (-1)) \cdot v.\]Simplifying the scalar sum:
\[1 + (-1) = 0.\]Thus,
\[(1 + (-1)) \cdot v = 0 \cdot v.\]By the previously established property that $0 \cdot v = \vec{0}$ for all $v \in V$ (Theorem 1.e.1), it follows that
\[0 \cdot v = \vec{0}.\]Substituting back, we have
\[v + [(-1) \cdot v] = \vec{0}.\]By the definition of the additive inverse, $v’$ is the unique element in $V$ satisfying
\[v + v' = \vec{0}.\]Comparing the two expressions, we observe that
\[v + [(-1) \cdot v] = v + v' = \vec{0}.\]By the uniqueness of additive inverses (Theorem 1.g.1), it must be that
\[(-1) \cdot v = v'.\]Therefore, scalar multiplication by $-1$ yields the additive inverse of $v$, establishing that
\[(-1) \cdot v = -v.\]This equivalence justifies the conventional notation $-v$ to represent the additive inverse of $v$ in $V$. $\textbf{Q.E.D.}$
Theorem 1.j.1. In any vector space $V$ over $\mathbb{R}$, if $v \neq \vec{0}$, then for all scalars $s \neq 0$, the scalar multiplication $s \cdot v$ satisfies $s \cdot v \neq \vec{0}$.
Proof. Let $V$ be a vector space over $\mathbb{R}$ satisfying all vector space axioms, including the existence of additive inverses and the distributive properties of scalar and vector addition. Let $v \in V$ be an arbitrary vector such that $v \neq \vec{0}$, and let $s \in \mathbb{R}$ be an arbitrary scalar with $s \neq 0$.
For the sake of contradiction, assume that:
\[s \cdot v = \vec{0}.\]Consider the scalar $\frac{1}{s} \in \mathbb{R}$, which exists since $s \neq 0$. Perform scalar multiplication of both sides of the equation $s \cdot v = \vec{0}$ by $\frac{1}{s}$:
\[\frac{1}{s} \cdot (s \cdot v) = \frac{1}{s} \cdot \vec{0}.\]By the associativity of scalar multiplication (Axiom 7),
\[\left( \frac{1}{s} \cdot s \right) \cdot v = \vec{0},\]which simplifies to
\[1 \cdot v = \vec{0},\]since $\frac{1}{s} \cdot s = 1$.
By the property of the scalar multiplicative identity (Axiom 8),
\[1 \cdot v = v.\]Therefore, we have
\[v = \vec{0}.\]This conclusion directly contradicts our initial assumption that $v \neq \vec{0}$. Hence, the assumption that $s \cdot v = \vec{0}$ must be false. Consequently, it must hold that
\[s \cdot v \neq \vec{0}.\]Since $v \in V$ and $s \in \mathbb{R}$ were arbitrary, with the conditions $v \neq \vec{0}$ and $s \neq 0$, the result follows universally. $\textbf{Q.E.D.}$
Commentary
This result underscores the injectivity of scalar multiplication by non-zero scalars within vector spaces, ensuring that non-zero vectors remain non-trivial under such operations. Specifically, Theorem 1.j.1 complements earlier established properties of vector spaces, particularly regarding the uniqueness of additive inverses and the behavior of the additive identity under scalar multiplication. While Question (c) addressed the uniqueness of the additive identity itself, and Question (g) affirmed the uniqueness of additive inverses, this result emphasizes the preservation of vector non-triviality under non-zero scalar transformations.
Teorem 1.k.1. In any vector space $V$ over $\mathbb{R}$, if a vector $v \in V$ satisfies $v = -v$, then $v$ is the additive identity $\vec{0}$.
Proof. Consider the vector addition of $v$ with itself:
\[v + v.\]Substituting $v = -v$ into the above expression, we obtain
\[v + v = (-v) + v.\]By the commutativity of vector addition (Axiom 1),
\[(-v) + v = v + (-v).\]By the definition of additive inverses, $v + (-v) = \vec{0}$. Therefore,
\[v + v = \vec{0}.\]Thus, we have established that
\[2v = \vec{0},\]where $2v$ denotes the scalar multiplication of $v$ by $2$, i.e.,
\[2v = 2 \cdot v.\]Next, consider scalar multiplication by $\frac{1}{2}$, which is permissible since $\frac{1}{2} \in \mathbb{R}$ and $2 \neq 0$. Multiply both sides of the equation $2v = \vec{0}$ by $\frac{1}{2}$:
\[\frac{1}{2} \cdot (2v) = \frac{1}{2} \cdot \vec{0}.\]By the associativity of scalar multiplication (Axiom 7),
\[\left( \frac{1}{2} \cdot 2 \right) \cdot v = \frac{1}{2} \cdot \vec{0}.\]Simplifying the scalar product,
\[1 \cdot v = \frac{1}{2} \cdot \vec{0}.\]By the property of the scalar multiplicative identity (Axiom 8),
\[1 \cdot v = v.\]Additionally, by Theorem 1.e.1, scalar multiplication by zero yields the additive identity:
\[\frac{1}{2} \cdot \vec{0} = \vec{0}.\]Therefore, we have
\[v = \vec{0}.\]Thus, any vector $v \in V$ satisfying $v = -v$ must be the additive identity $\vec{0}$. $\textbf{Q.E.D.}$
Commentary
The theorem precludes the existence of non-trivial vectors that annihilate themselves under addition.