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Problem 48: Prove Vector Space Properties Using Vector Space Axioms

Problem 48: Prove Vector Space Properties Using Vector Space Axioms

Let $\V$ be a vector space over $\R$. Let $ u,v,w \in \V$.
$(a)$ If $u+v=u+w$, then $v=w$.
$(b)$ If $v+u=w+u$, then $v=w$.
$(c)$ The zero vector $0$ is unique.
$(d)$ For each $v \in \V$, the additive inverse $−v$ is unique.
$(e) 0v=\mathbf{0}$ for every $v \in \V$, where $0 \in \R$ is the zero scalar.
$(f) a\mathbf{0}=\mathbf{0}$ for every scalar $a$.
$(g)$ If $av=\mathbf{0}$, then $a=0$ or $v=\mathbf{0}$.
$(h) (−1)v=−v$.

Solution:
The Axioms of a Vector Space
The following properties must hold for all $ u,v,w \in \V$ and $a,b \in \R$:
Closure Properties
$(A1) u+v \in \V$.
$(A2) av \in \V$.
Properties of Addition
$(A3) u+v=v+u$.
$(A4) u+(v+w)=(u+v)+w$.
$(A5)$ There is an element $\mathbf{0} \in \V$ such that $\mathbf{0}+v=v$ for all $v \in \V$.
$(A6)$ Given an element $v \in \V$, there is an element $−v \in \V$ such that $v+(−v)=\mathbf{0}$.
Properties of Scalar Multiplication
$(A7) a(bv)=(ab)v$.
$(A8) a(u+v)=au+av$.
$(A9) (a+b)v=av+bv$.
$(A10) 1v=v$ for all $v \in \V$.

$(a)$ If $u+v=u+w$, then $v=w$.
we know by $(A6)$ that there is an additive inverse $-u \in \V$. Then \begin{align*} u+v=u+w \Rightarrow -u+(u+v) = -u+(u+w)\\ \overset{(A4)}\Rightarrow (-u+u)+v = (-u+u)+w\\ \overset{(A3)}\Rightarrow (u+(-u))+v = (u+(-u))+w\\ \overset{(A6)}\Rightarrow \mathbf{0}+v = \mathbf{0}+w\\ \overset{(A5)}\Rightarrow v = w\\ \end{align*}

$(b)$ If $v+u=w+u$, then $v=w$.
we know by $(A6)$ that there is an additive inverse $-u \in \V$. Then \begin{align*} v+u=w+u \Rightarrow -u+(v+u) = -u+(w+u)\\ \overset{(A3)}\Rightarrow -u+(u+v) = -u+(u+w)\\ \overset{(A4)}\Rightarrow (-u+u)+v= (-u+u)+w\\ \overset{(A3)}\Rightarrow (u+(-u))+v = (u+(-u))+w\\ \overset{(A6)}\Rightarrow \mathbf{0}+v = \mathbf{0}+w\\ \overset{(A5)}\Rightarrow v = w\\ \end{align*}

$(c)$ The zero vector $0$ is unique.
Suppose that $0'$ is another vector satisfying axiom $(A5)$. That is, we have $0'+v=v$ for any $v \in \V$.
Since $0$ is also satisfy $0+v=v$, we have \begin{align*} 0'+v=v=0+v \end{align*} where $v$ is any fixed vector. For example $v=0$ is enough. \begin{align*} 0'+v=0+v \Rightarrow -v+(0'+v) = -v+(0+v)\\ \overset{(A3)}\Rightarrow -v+(v+0') = -v+(v+0)\\ \overset{(A4)}\Rightarrow (-v+v)+0'= (-v+v)+0\\ \overset{(A3)}\Rightarrow (v+(-v))+0' = (v+(-v))+0\\ \overset{(A6)}\Rightarrow \mathbf{0}+0' = \mathbf{0}+0\\ \overset{(A5)}\Rightarrow 0' = 0\\ \end{align*} Thus, there is only one zero vector $0$.

$(d)$ For each $v \in \V$, the additive inverse $−v$ is unique.
Since $−v$ is the additive inverse of $v \in \V$, we have $v+(−v)=0$. This is just $(A6)$.
Now, suppose that we have a vector $w \in \V$ satisfying $v+w=0$. So, $w$ is another element satisfying axiom $(A6)$.
Then we have \begin{align*} v+(−v)=0=v+w \end{align*} \begin{align*} v+(−v)=v+w \Rightarrow -v+(v+(−v)) = -v+(v+w)\\ \overset{(A4)}\Rightarrow (-v+vu)+(-v) = (-v+v)+w\\ \overset{(A3)}\Rightarrow (v+(-v))+(-v) = (v+(-v))+w\\ \overset{(A6)}\Rightarrow \mathbf{0}+(-v) =\mathbf{0}+w\\ \overset{(A5)}\Rightarrow -v = w\\ \end{align*} we have $−v=w$. Thus, the additive inverse is unique.

$(e) 0v=0$ for every $v \in \V$, where $0 \in \R$ is the zero scalar.
Note that $0$ is a real number and $\mathbf{0}$ is the zero vector in $V$. For $v \in \V$, we have
\begin{align*} 0v=(0+0)v\\ \overset{(A9)}\Rightarrow 0v+0v. \end{align*} We also have \begin{align*} 0v\overset{(A5)}= \mathbf{0}+0v. \end{align*} Hence, combining these, we see that \begin{align*} 0v+0v=\mathbf{0}+0v, \end{align*} \begin{align*} 0v+0v=\mathbf{0}+0v, \end{align*} By cancellation law, we obtain $0v=\mathbf{0}$.

$(f) a\mathbf{0}=\mathbf{0}$ for every scalar $a$. Note that we have $\mathbf{0}+\mathbf{0}=\mathbf{0}$ by $(A5)$. Thus, we have \begin{align*} a\mathbf{0}=a(\mathbf{0}+\mathbf{0})\\ \overset{(A8)}\Rightarrow a\mathbf{0}+a\mathbf{0}. \end{align*} We also have \begin{align*} a\mathbf{0}=\mathbf{0}+a\mathbf{0} \end{align*} by $(A5)$. Combining these, we have \begin{align*} a\mathbf{0}+a\mathbf{0}=\mathbf{0}+a\mathbf{0}, \end{align*} and the cancellation law yields $a\mathbf{0}=\mathbf{0}$.

$(g)$ If $av=\mathbf{0}$, then $a=0$ or $v=\mathbf{0}$. For this problem, we use a little bit logic.
Our assumption is $av=\mathbf{0}$. From this assumption, we need to deduce that either $a=0$ or $v=\mathbf{0}$.
Note that if $a=0$, then we are done as this is one of the consequence we want.
So, let us assume that $a \ne 0$. Then we want to prove $v=\mathbf{0}$.
Since $a$ is a nonzero scalar, we have $a^{−1}$. Then we have \begin{align*} a^{−1}(av)=a^{−1}\mathbf{0}. \end{align*} The right hand side $a^{−1}\mathbf{0}$ is $\mathbf{0}$ by part $(f)$.
On the other hand, the left hand side can be computed as follows:
\begin{align*} a^{−1}(av)\overset{(A7)} =(a^{−1}a)v=1v\overset{(A10)} =v. \end{align*} Therefore, we have $v=\mathbf{0}$.
Thus, we conclude that if $av=\mathbf{0}$, then either $a=0$ or $v=\mathbf{0}$.

$(h) (−1)v=−v$.
Note that $(−1)v$ is the scalar product of $−1$ and $v$.
On the other hand, $−v$ is the additive inverse of $v$.
We show that $(−1)v$ is also the additive inverse of $v$:
\begin{align*} v+(−1)v\overset{(A10)} =1v+(−1)v\\ \overset{(A9)} =(1+(−1))v=0v=\mathbf{0} \end{align*} So $(−1)$v is the additive inverse of $v$.
we know that the additive inverse is unique, it follows that $(−1)v=−v$.

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