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I know that the sum of vectors of different spaces is not defined, but what about the multiplication of vectors of different spaces. For example, what about the multiplication of $v_1 = \left(\begin{matrix} 3 \\ 9\end{matrix}\right)\in \mathbb{R}^2$ and $v_2 = \left(\begin{matrix} 2 \\ 5 \\ 7\end{matrix}\right)\in \mathbb{R}^3$?

My intuitive answer would be also no, since they effectively belong to different worlds, but I would like to hear a correct answer. Is it possible to add a $3rd$ null ($0$) dimension to $v_1$ and therefore add them? So, is or not defined the product between them?

This problem came out when I had to find the angle between 2 vectors. The exercise explicitly says that we should find the angle, if it is defined.

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  • $\begingroup$ The first question you should ask yourself is what kind of properties you want this product to have, and only then ask whether it is actually possible to define a product which satisfies said properties. $\endgroup$ – zarathustra Mar 25 '15 at 17:03
  • $\begingroup$ @zarathustra This problem came out when I had to find the angle between those 2 vectors. The exercise explicitly says that we should find the angle, if it is defined. $\endgroup$ – nbro Mar 25 '15 at 17:08
  • $\begingroup$ Well, then it is not well-defined. There is no "canonical" way to imagine $\mathbb R^2$ as a subspace of $\mathbb R^3$! $\endgroup$ – zarathustra Mar 25 '15 at 17:11
  • $\begingroup$ @zarathustra What are the cases in which it could be defined, just for curiosity? $\endgroup$ – nbro Mar 25 '15 at 17:12
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If you are searching for a short answer, here it is: no, it is not defined. However, I think it is useful to think about the following answer:

Are we searching for kind of a inner product or are we searching for something else?

Well, if we are searching for a inner product, the answer is still no: of course you can add a $0$ as the third real coordinate of $v_1$, but it is not clear why to use $0$; one could use any $r\in \mathbb{R}$. Probably you are trying to add $0$ because you are thinking $\mathbb{R}^2$ as the plane $z=0$ inside $\mathbb{R}^3$. Of course the plane $z=0$ is isomorphic to $\mathbb{R}^2$, but inside $\mathbb{R}^3$ there are infinitely many isomorphic copies of $\mathbb{R}^2$, not only that one.

In a different fashion, there are infinitely many monomorphism (injective linear maps) $$ \mathbb{R}^2 \to \mathbb{R}^3$$ and among these there is not a "natural", "canonical" one.


A different question is: there is a way to "multiply" two vector spaces, in order to obtain another vector space? This time we are not asking a inner product, but a sort of a "product" which takes vectors and gives vectors.

The answer is yes, there is. The construction is called tensor product of two vector spaces, and is performed taking a particular quotient of the cartesian product. If you are interested, you can find further informations about this in a lot of linear algebra books.

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