# Is it defined the product of vectors of different spaces?

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.

• 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. – zarathustra Mar 25 '15 at 17:03
• @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. – nbro Mar 25 '15 at 17:08
• Well, then it is not well-defined. There is no "canonical" way to imagine $\mathbb R^2$ as a subspace of $\mathbb R^3$! – zarathustra Mar 25 '15 at 17:11
• @zarathustra What are the cases in which it could be defined, just for curiosity? – nbro Mar 25 '15 at 17:12

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.