Axiom number 4 for vector space states that "there exists a zero vector 0 in V, called a zero vector for V, such that 0 + u = u + 0 = u, for all u in V.

My questions is, must all the elements in the zero vector be zeroes?

For this question: "Given the set of all pairs of real numbers of the form (x, 1) with the operations (x, 1) + (x', 1) = (x + x', 1)", what is the zero vector for this case, if it exists? Is it (0, 0) or (0, 1)? Also, is it necessary for the zero vector to be contained in V?

I think that the zero vector exists, which is (0, 1), and all zero vectors for any vector spaces must be contained in V. But then again, the definition for zero vector is that all the elements / components must be zeroes.

  • 4
    $\begingroup$ This is a good opportunity for you to stop thinking of vectors only as tuples of numbers. The zero vector is whatever object from the set you’re working with satisfies the corresponding axioms. $\endgroup$
    – amd
    Apr 11, 2017 at 0:23
  • $\begingroup$ Thanks. That explains as well the zero vector should be inside V. $\endgroup$
    – Matthew07
    Apr 11, 2017 at 2:07

1 Answer 1


Perhaps it might be instructive to think of the properties of the $0 \in \mathbb{R}$ that you are familiar with. If you add any other number $x \in \mathbb{R}$ to $0_{\mathbb{R}}$, you get back the number $x$, i.e. \begin{equation*} x + 0_{\mathbb{R}} = x \end{equation*} The "zero vector" that you are referring to has the same property as $0_\mathbb{R}$. The difference is that this equation takes place in some vector space $V$, so you get \begin{equation*} v + 0_V = v \end{equation*}

Note that the second equation will hold for any vector space $V$. Depending on the specific vector space you are working with, the vector $0_V$ will look different.

For example, if $V$ is \begin{equation*} V = \{(x, 1): \text{ x is in } \mathbb{R}\} \end{equation*} then the zero vector here is $(0, 1)$, i.e. \begin{equation*} 0_V = (0, 1) \end{equation*} Note that the zero vector cannot possibly be $(0, 0)$, since $(0,0)$ is not even in the set $V$!!!!

  • $\begingroup$ You didn't define an addition for your $V$ (the standard one, element-wise addition, isn't closed on $V$). You probably thought of addition on the first element of the tuple only, but you didn't state that. You could just as well introduce the addition $(x,1)+(y,1)=(x+y-1,1)$, in which case the zero vector is $(1,1)$. $\endgroup$
    – celtschk
    Apr 11, 2017 at 20:08
  • $\begingroup$ @celtschk You are right, I thought of $+$ in $V$ as being defined by $(x, 1) +_V (y, 1) = (x + y, 1)$. In this case the zero vector is $(0, 1)$...do you agree? $\endgroup$
    – Sam Y.
    Apr 11, 2017 at 23:06
  • $\begingroup$ Of course I agree. My point was that if you use a non-standard addition, you have to state that addition explicitly. $\endgroup$
    – celtschk
    Apr 12, 2017 at 7:03

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