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A $3\times 3$ matrix with $2$ independent vectors will span a $2$ dimensional plane in $\Bbb R^3$ but that plane is not $\Bbb R^2$. Is it just nomenclature or does $\Bbb R^2$ have some additional properties that other planes don't?

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Facts:

  • There are many two-dimensional planes in $\mathbb{R}^3$.

  • The planes that include the origin are two-dimensional subspaces of $\mathbb{R}^3$.

  • All two-dimensional subspaces of $\mathbb{R}^3$ are isomorphic to $\mathbb{R}^2$.

  • None of the two-dimensional subspaces of $\mathbb{R}^3$ are equal to $\mathbb{R}^2$ because $\mathbb{R}^2$ is defined to be points with two coordinates while points in $\mathbb{R}^3$ must have three coordinates.

  • One of the natural isomorphisms, is the map $$\begin{bmatrix}x\\y\\0\end{bmatrix}\mapsto\begin{bmatrix}x\\y\end{bmatrix}.$$ Since the points on the left and right aren't exactly the same, they aren't equal (you only get that the two objects have the same shape).

  • In linear algebra, properties are preserved under isomorphism, so two dimensional subspaces isomorphic to $\mathbb{R}^3$ look (from a linear algebra perspective) to be the same.

  • The difference is that the sets are different, even though the sets look the same from linear algebra's perspective, they are different as sets because the underlying objects are different.

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This is an interesting question. The usual definition of $\mathbb{R}^2$ is the set of all ordered pairs of real numbers, with pointwise operations. With this definition, no subspace of $\mathbb{R}^3$ is literally $\mathbb{R}^2$, although the $xy$ plane is quite similar, as we only need to adjoin an extra $0$ to the end of each sequence in $\mathbb{R}^2$.

There is a different viewpoint, however, which is more structural, and has a good analogy with category-theoretic foundations. Some mathematicians are uncomfortable with "extra" set theoretic baggage in the definitions of mathematical objects. For example, some dislike the fact that, in the usual definition of the natural numbers, we have $1 \in 5$. This is because the $\in$ (set membership) relation does not seem like an appropriate relation for natural numbers.

Similarly, the "take the first coordinate of the vector $v$" operation is not on its own a natural operation on a vector. The axioms of a vector space do not include anything about coordinates. Instead, thinking structurally, we should say "given an ordered basis $b_1, \ldots, b_n$, take the component of the vector $v$ in the direction of $b_1$". From this viewpoint, all isomorphic copies of $\mathbb{R}^2$ are equally good. Only by "looking inside" could we distinguish the "usual" $\mathbb{R}^2$ from its isomorphic copies. So from this structural viewpoint, any plane in $\mathbb{R}^3$ through the origin is a perfectly good copy of $\mathbb{R}^2$.

I should emphasize that the structural viewpoint is not commonly presented in elementary linear algebra texts. This leads to some challenges in seeing the structural viewpoint. For example, the dot product of two vectors in $\mathbb{R}^2$ should also only be defined relative to a basis - you know you have understood the structural viewpoint when you can grok the sentence "the dot product is an operation on pairs of finite sequences of real numbers, not on pairs of vectors".

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Start with Michael Burr's answer.

Here's another way to state the situation: Generally, a $2$-dimensional plane $P$ in $\mathbb R^3$ is isomorphic to $\mathbb R^2$ but not canonically isomorphic. That is, although there does exist a vector space isomorphism between $P$ and $\mathbb R^2$, there are also infinitely many other isomorphisms, and no particular isomorphism is privileged over the others.

Yes, there is also a matter of nomenclature. What if you need to work with two such planes at the same time? For example, let $P$ be the plane orthogonal to $(1,1,1)$ and let $Q$ be the plane orthogonal to $(1,1,-1)$. Since we've named them with separate letters, we can speak about things like their intersection, $P\cap Q$. But if we tried to write $P=\mathbb R^2$ and $Q=\mathbb R^2$, even though $P\neq Q$, we'd be in real trouble!

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$\mathbb{R}^2$ is a vector space. A two-dimensional plane in $\mathbb{R}^3$ is an affine space https://en.wikipedia.org/wiki/Affine_space.

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    $\begingroup$ A 2D plane through the origin is also a vector space, of course, because it contains 0. $\endgroup$ – Carl Mummert May 15 '17 at 23:10
  • $\begingroup$ Not by formal definition as $\mathbb{E}^2$ is not the same as $\mathbb{R}^2$ but it I know what you mean :) $\endgroup$ – JJR May 15 '17 at 23:11
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    $\begingroup$ From the OP's original question, this distinction unlikely to be known to the OP. It would make for a good answer to fully explain the difference. $\endgroup$ – Michael Burr May 15 '17 at 23:18
  • $\begingroup$ @MichaelBurr agreed and understood $\endgroup$ – JJR May 15 '17 at 23:20
  • $\begingroup$ @JJR - I am confused. Is a plane through the origin not a subspace of $\mathbb{R}^3$, and thus a vector space in its own right? I am not saying that a plane is the same set as $\mathbb{R}^2$. $\endgroup$ – Carl Mummert May 15 '17 at 23:21
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If $\mathbb{R}$ is the set of all real numbers, $\mathbb{R}^2$ is the set of all ordered pairs of real numbers. A point on a plane in $\mathbb{R}^3$ may be, for example, $(1,2,3)$. This is an ordered triple since there's 3 numbers, so it's not an element of $\mathbb{R}^2$.

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