Theorem 1.2.1.
In any vector space \(V\text{,}\) we have \(0\vv = \zer\) for all \(\vv\in V\text{.}\)
Section 1.2 Properties
There are a number of other algebraic properties that are common to all vector spaces; for example, it is true that \(0\vv = \zer\) for all vectors \(\vv\) in any vector space \(V\text{.}\) The reason these are not included is that the ten axioms in Definition 1.1.3 are the ones deemed “essential” – all other properties can be deduced from the axioms. To demonstrate, we next give the proof that \(0\vv = \zer\text{.}\)
The focus on proofs may be one place where your second course in linear algebra differs from your first. Learning to write proofs (and to know when a proof you have written is valid) is a difficult skill that takes time to develop. Some of the proofs in this section are “clever”, in the sense that they require you to apply vector space axioms in ways that may not seem obvious. Proofs in later sections will more often be more straightforward “direct” proofs of conditional (if … then) statements, although they may not feel straightforward on your first encounter.
Strategy.
The goal is to show that multiplying by the scalar \(0\) produces the vector \(\zer\text{.}\) We all learn early on that “anything times zero is zero”, but why is this true? A few strategies that show up frequently when working with the axioms are:
- Adding zero (the scalar, or the vector) does nothing, including when you add it to itself.
- We can always add the same thing to both sides of an equation.
- Liberal use of the distributive property!
What we do here is to start with a very simple statement: \(0+0=0\text{.}\) The reason for doing so is that when we multiply by \(0+0\text{,}\) we have an opportunity to use the distributive property.
Proof.
Since \(0+0=0\text{,}\) we have \(0\vv=(0+0)\vv\text{.}\) Using the distributive axiom S3, this becomes
\begin{equation*}
0\vv + 0\vv = 0\vv\text{.}
\end{equation*}
By axiom A5, there is an element \(-0\vv\in V\) such that \(0\vv+(-0\vv)=\zer\text{.}\) Adding this to both sides of the equation above, we get:
\begin{equation*}
(0\vv+0\vv)+(-0\vv) = 0\vv+(-0\vv)\text{.}
\end{equation*}
Now, apply the associative property (A3) on the left, and A5 on the right, to get
\begin{equation*}
0\vv + (0\vv+(-0\vv)) = \zer\text{.}
\end{equation*}
Using A5 again on the left, we get \(0\vv+\zer = \zer\text{.}\) Finally, axiom A4 guarantees \(0\vv = 0\vv+\zer = \zer\text{.}\)
Exercise 1.2.2.
Tactics similar to the ones used in Theorem 1.2.1 can be used to establish the following results, which we leave as an exercise. Solutions are included in at the end of the book, but it will be worth your while in the long run to wrestle with these.
Show that the following properties are valid in any vector space \(V\text{:}\)
(a)
If \(\uu+\vv=\uu+\ww\text{,}\) then \(\vv=\ww\text{.}\)
Hint.
Remember that every vector \(\uu\) in a vector space has an additive inverse \(-\uu\text{.}\)
(b)
For any scalar \(c\text{,}\) \(c\zer=\zer\text{.}\)
Hint.
Your approach should be quite similar to the one used in Theorem 1.2.1.
(c)
The zero vector is the unique vector such that \(\vv+\zer=\vv\) for all \(\vv\in V\text{.}\)
Hint.
If you want to prove something is unique, try assuming that you have more than one! If any two different elements with the same property have to be equal, then all such elements must, in fact, be the same element.
(d)
The negative \(-\vv\) of any vector \(\vv\) is unique.
Example 1.2.3.
Prove the following statement:
Let \(V\) be a vector space. For any \(\vv\in V\text{,}\) \((-1)\vv=-\vv\text{.}\)
Indicate all axioms needed in the proof.
Solution.
This proof will be far more detailed than what you’ll see in later sections. (We usually do not bother to take note of the axioms, one by one.) We will also try to explain our reasoning as we go, to help you get used to the sort of careful reasoning involved in a proof. Lines that should actually be included in the proof will be set aside in block quotes.
First, we are proving a “for all” (universally quantified) statement. This means we should be careful not to assume anything about the vector we choose, so that our argument can apply to any vector we want:
Let \(\vv\) be any vector in \(V\text{.}\)
Next, you might want to remind yourself of the goal: we want to show that \((-1)\vv=-\vv\text{.}\) You can state that this is what you want to show, but it’s not absolutely necessary to do this. A common trick that shows up in a lot of mathematical proofs is a simple bit of arithmetic: \(a=b\) is the same thing as \(a-b=0\text{!}\) So really, what we want to show is that \((-1)\vv+\vv=\zer\text{.}\)
Now, remember that \(1\vv=\vv\text{,}\) and \(0\vv=\zer\text{,}\) so what we want to show is equivalent to \((-1)\vv+1\vv=0\vv\text{.}\) Remove the \(\vv\text{,}\) and we’re left with \(-1+1=0\text{,}\) and that we definitely know is true! (Of course, we can’t just “remove \(\vv\)”, but we can use the distributive property!)
This is basically the proof, but we need to state all the axioms we use, and best practice in logical arguments is that we should begin with our assumptions, and statements we agree are true, and proceed from those to the desired conclusion.
Since we know that \(-1+1=0\text{,}\) it follows that\begin{equation*} (-1+1)\vv=0\vv\text{.} \end{equation*}On the left hand side, we can use the distributive property S3 to get \((-1+1)\vv=(-1)\vv+1\vv\text{.}\) On the right hand side, we can use Theorem 1.2.1 to get \(0\vv=\zer\text{.}\) Therefore,\begin{equation*} (-1)\vv+1\vv=\zer\text{.} \end{equation*}By axiom S5, \(1\vv=\vv\text{,}\) so we have \((-1)\vv+\vv=\zer\text{.}\)
OK, that’s more or less where we said we wanted to get to, and then we can just move \(\vv\) to the other side as \(-\vv\text{,}\) and we’re done. But we want to be careful to state all axioms! The rest of the proof involves carefully stepping through this process.
Another way to proceed, which shortcuts this whole process, is to use Part d of Exercise 1.2.2: since the additive inverse of \(\vv\) is the unique vector \(-\vv\) such that \(-\vv+\vv=\zer\text{,}\) and \((-1)\vv+\vv=\zer\text{,}\) it must be the case that \((-1)\vv=-\vv\text{.}\)
This approach is completely valid, and you are free to use it, but we will take the long route to demonstrate further use of the axioms.
Since \((-1)\vv+\vv=\zer\text{,}\) we can add \(-\vv\) to both sides of the equation, giving\begin{equation*} ((-1)\vv+\vv)+(-\vv)=\zer+(-\vv)\text{.} \end{equation*}By the associative property (axiom A3), \(((-1)\vv+\vv)+(-\vv)=(-1)\vv+(\vv+(-\vv))\text{,}\) and by the identity axiom A5, \(\zer+(-\vv)=-\vv\text{.}\) This gives us\begin{equation*} (-1)\vv+(\vv+(-\vv))=-\vv\text{.} \end{equation*}By axiom A5, \(\vv+(-\vv)=\zer\text{,}\) so \((-1)\vv+\zer=-\vv\text{.}\) Finally, we use axiom A4 one last time, and we have our result: \((-1)\vv=-\vv\text{.}\)
Note that in the above example, we could have shortened the proof: In Exercise 1.2.2 we showed that additive inverses are unique. So once we reach the step where \(-1\vv+\vv=\zer\text{,}\) we can conclude that \(-1\vv=-\vv\text{,}\) since \(-\vv\) is the unique vector that satisfies this equation.
To finish off this section, here is one more problem similar to the one above. This result will be useful in the future, and students often find the logic tricky, so it is worth your time to ensure you understand it.
Exercise 1.2.4.
Rearrange the blocks to create a valid proof of the following statement:
If \(c\vv=\zer\text{,}\) then either \(c=0\) or \(\vv=\zer\text{.}\)
Hint.
The main difficulty with this problem is getting the logic of the statement correct, and making sure you know what it is you’re trying to prove. Consider a proof by cases: either \(c=0\) or \(c\neq 0\text{.}\) In the first case, there is nothing to prove (why?). In the second case, use the fact (see Definition 1.1.4) that for any nonzero scalar \(c\text{,}\) there exists a scalar \(1/c\) such that \(c\cdot 1/c=1\text{.}\)
