Why is $\mathbb{R}^2$ not a vector subspace of $\mathbb{R}^3$?

Question

Lets suppose we have the vector space $V=\mathbb{R}^3$ and $(x,y,z)\in V$ then $W=(x,y,0)$ is a subspace of $V$ but $\mathbb{R}^2$ is not one as $(x,y)\notin V$(with $(x,y)\in \mathbb{R}^2$.) But often we consider groups, rings, fields, or vector spaces up to isomorphism. Now as $W$ is isomorphic to $\mathbb{R}^2$, shouldn't $\mathbb{R}^2$ also be a subspace of $V$?

Answer 1

A subset of a vector space is a subspace iff (either by definition or by proof) it has the same zero and "closed" under the same addition and multiplication; but, clearly, the zero in $\Bbb{R}^{2}$ is $(0,0)$, which is $\neq (0,0,0)$, the zero of $\Bbb{R}^{3}$.

Answer 2

$V$ consists of $3$-tuples of real numbers $(x, y, z)$ $\mathbb{R}^2$ only consists of ordered pairs $(x, y)$

Answer 3

Yes, they are isomorph... as are many other subspaces :

Is $\Bbb R^2$

• $\{ (x,y,0) \in \Bbb R^3 \}$
• $\{ (x,0,z) \in \Bbb R^3 \}$
• $\{ (0,y,z) \in \Bbb R^3 \}$
• any other 2 dimensional space?

There is no "canonic" subspace of $\Bbb R^3$ that you can identify to $\Bbb R^2$

Answer 4

It's very close to a "semantic" distinction: namely, the difference between equality and isomorphism. The latter is an equivalence relation, and preserves all the algebraic qualities, but not necessarily the identity of the set-elements involved.

To a lay person, it seems intuitively obvious that the $xy$-plane (what we use as a model for $\Bbb R^2$) is a subset of the $xyz$-space (what we use as a model for $\Bbb R^3$), but this identification depends, in an essential way, on identifying certain distinguished basis vectors (what we choose to use as axes). This choice is not inherent in $\Bbb R^3$ as a vector space, and the "conventional" way is somewhat arbitrary.

Put another way, there are, in fact, an infinite number of ways to extend the Euclidean plane to Euclidean $3$-space, and the most we can say "universally" is that each way results in a $2$-dimensional subspace of a $3$-dimensional space-only the dimensions are invariant-the "coordinate systems" we use may differ radically (resulting in perhaps weird numerical "coordinates").

That said, if we assign our Euclidean plane an orthonormal basis (and this can always be done), there are then basically just "two" ways to extend this to an orthonormal basis for a Euclidean $3$-space it lives in: that is we have just two choice for a unit normal vector to this plane. Note that although this narrows the possiblities down, it still does not uniquely determine the result: two is not one.

The usual convention is to determine the "positive" unit normal by "the right-hand rule", this is the same as assigning a sign to the pseudo-vector $\mathbf{e_1} \times \mathbf{e_2}$, which determines which orientation we have.

As others have pointed out above, this corresponds to "embedding" the ordered basis $\{(1,0),(0,1)\}$ into the ordered basis $\{(1,0,0),(0,1,0),(0,0,1)\}$ via the linear isomorphism: $\Bbb R^2 \to \Bbb R^3$ given by $(x,y) \mapsto (x,y,0)$. Although the former is a pair, and the latter is a triple, in practice they behave and calculate no differently.