Linear Subspace - is $\mathbb{C}^2$ a linear subspace of $\mathbb{C}^3$?

Question

Assuming a vector space $\mathbb{C}^{2}$ that is not empty and fulfills the addition and multiplication conditions, is $\mathbb{C}^{2}$ a linear subspace of $\mathbb{C}^{3}$?

Answer 1

By definition, a vector space is a set together with some operations. If $V$ is a vector space and $W$ is a subspace, then the set that is $W$ must be, among other things, a subset of the set that is $V$.

$\mathbb{C}^2$, as a set, consists of the ordered pairs $(a,b)$ with $a,b\in\mathbb{C}$. On the other hand, $\mathbb{C}^3$, as a set, consists of the ordered triples $(a,b,c)$ with $a,b,c\in\mathbb{C}$.

Since the collection of ordered pairs is not a subset of the collection of ordered triples, then $\mathbb{C}^2$ is not a subspace of $\mathbb{C}^3$.

That said: there is a natural way of finding an "isomorphic copy" of $\mathbb{C}^2$ inside of $\mathbb{C}^3$ (or more generally, of $F^n$ inside of $F^{n+k}$ for any positive integers $n$ and $k$): define $\iota\colon\mathbb{C}^2\to\mathbb{C}^3$ to be the map $$\iota(a,b) = (a,b,0)$$ for all $(a,b)\in\mathbb{C}^2$. Then you can verify that:

1. $\iota$ is one-to-one;
2. $\iota$ respects vector addition: $\iota\bigl((a,b)+(c,d)\bigr) = \iota(a,b)+\iota(c,d)$ for all $(a,b),(c,d)\in\mathbb{C}^2$;
3. $\iota$ respects scalar multiplication: $\iota\bigl(\alpha(a,b)\bigr) = \alpha\iota(a,b)$ for all $\alpha\in\mathbb{C}$, $(a,b)\in\mathbb{C}$;
4. The image of $\iota$ is a subspace of $\mathbb{C}^3$.

That is, there is a "copy" of $\mathbb{C}^2$ inside of $\mathbb{C}^3$. It is common to identify $\mathbb{C}^2$ with its copy sitting inside $\mathbb{C}^3$ (just like we usually identify the plane $\mathbb{R}^2$ with the $xy$-plane in $\mathbb{R}^3$: but note that we specify "$xy$-plane", not just calling it "the plane").

An analogy might be: if you take the first two floors of a 3-story building, then they are, functionally, pretty much the same as a 2-story building. However, they are not a 2-story building. Similarly, by considering the vectors $(a,b,0)$ in $\mathbb{C}^3$ we obtain something which is, functionally, pretty much the same as $\mathbb{C}^2$, but it is not actually $\mathbb{C}^2$.

Answer 2

No. $\mathbb{C}^2$ is the space of all ordered pairs of complex numbers. $\mathbb{C}^3$ is the space of all ordered triples of complex numbers. No ordered pair is an ordered triple, so no element of $\mathbb{C}^2$ is an element of $\mathbb{C}^3$, so $\mathbb{C}^2$ isn't even a subset of $\mathbb{C}^3$.

Answer 3

Maybe it's easier to look at things the other way around. Inside $\Bbb C^3$ you can certainly find (many!) linear subspaces that are isomorphic to $\Bbb C^2$. Ekuurh's answer gives just an example.

Answer 4

Yea, if you do this little trick: Say (a,b) = (a,b,0). This mapping is obviously linear, and the image is clearly closed under addition and multiplication by a (complex)y scalar.