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I am currently teaching myself Linear Algebra. I have come across the closure of a vector space under addition. Now, I understand that 10 axioms must be satisfied before a set is considered a vector space.

However, suppose there is a set that contains all 2-tuples and 3-tuples. Why is this not considered a vector space since adding a vector such as $(1,2)$ and $(1, 2, 3)$ can be seen as giving $(2, 4, 3)$ and would therefore make the set closed under addition?

I have heard that vector addition requires vectors to be in the same dimension, but why is that? I have not come across an explicit definition in my textbook that states this.

Can someone please explain.

Thank you.

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    $\begingroup$ Why should $(1,2)+(1,2,3)$ equal $(2,4,3)$? If you insist on this being the case, then what is the difference between $(1,2)$ and $(1,2,0)$? Are those the same element or not? If not, then are you certain that inverses are unique? If inverses are not unique are you certain you have a valid group structure on your hands? $\endgroup$
    – JMoravitz
    Mar 17 at 12:11
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    $\begingroup$ If they are the same element, then write it the way everyone else does... where every element is a 3-tuple, and don't do something confusing like trying to talk about 2-tuples being there as well. $\endgroup$
    – JMoravitz
    Mar 17 at 12:11
  • $\begingroup$ I don't understand your rule for addition here. In any case, you need more than addition to make a vector space. You also need scalar multiplication and inverses. $\endgroup$
    – lulu
    Mar 17 at 12:12
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    $\begingroup$ Right, but that would make if treating 2-tuples as distinct from 3-tuples and using the rules I suggest that $(0,0)$ would be the additive identity but then you don't have any method of going from a 3-tuple back to a 2-tuple and so in particular no additive inverses exist for any 3-tuple. You could get from a 3-tuple to the other almost-additive inverse $(0,0,0)$ but this fails to be an actual additive inverse with these rules since $(1,0)+(0,0,0)=(1,0,0)\neq (1,0)$ for instance. That... or as alluded to previously... would imply that $(1,0)$ should not be considered different to $(1,0,0)$ $\endgroup$
    – JMoravitz
    Mar 17 at 12:19
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    $\begingroup$ But $(1,0,0)+(-1,0,0)$ is not equal to $(0,0)$... which this is the correct zero vector here. $(0,0,0)$ is not the zero vector here. Yes, both vectors are made up entirely of zeroes, but that is not how "a zero vector" is explicitly defined. It is defined as a vector who when added to any other will not change it. $(1,0)+(0,0,0)$ changed it to $(1,0,0)$. For an additive inverse to $(1,0,0)$ you need to find something, I'll call $v$, to add to it so that $(1,0,0)+v=(0,0)$ $\endgroup$
    – JMoravitz
    Mar 17 at 12:33

1 Answer 1

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Collating comments:

Your vector addition as I understand it is coordinate-wise addition whose output is a vector of length equal to the max length of the two inputs and whose entries are the sums of the corresponding entries with the caveat that if the entry is missing it is treated as zero.

Although addition can indeed be defined this way (and is done so in certain programming languages) it does not lend itself nicely to a group structure.

Recall that a group must have a "zero vector" $0$ such that $v+0=0+v=v$, that is to say adding by a zero vector will not change the object. Specifically note that $(0,0)$ is the proper "zero vector" here while $(0,0,0)$ is not since for example $(1,0)+(0,0,0)=(1,0,0)$ which is different (presumably) from $(1,0)$.

Recall that a group must have additive inverses for each element. That is, for any element $v$ you must have a $(-v)$ such that $v+(-v)=0$. But here, since any time you add with a 3-tuple you will get back another 3-tuple there is no way for us to have an additive inverse for a vector like $(1,0,0)$. At best we could have $(1,0,0)+(-1,0,0)=(0,0,0)$ but as mentioned in the previous paragraph this is not the zero vector in this context.

Since your set fails to have a group structure under the defined addition, your set does not qualify as a vector space.

As an alternative, if we were to amend our definition of what it means to be "equal" so as to allow $(1,0)=(1,0,0)$, letting two vectors be "equal" so long as every corresponding entry is either equal or if any missing entries exist that are missing in one but not the other that the corresponding entry is zero... then you will in fact have a vector space, but it is just the vectorspace commonly known as $\Bbb R^3$ and could have been written so as to contain only 3-tuples without having missed any elements.

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  • $\begingroup$ Thank you so much for your help. I completely get it now. You’re a legend $\endgroup$
    – Oofy2000
    Mar 17 at 13:03

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