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

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 $\mathbb{C}^{3}$?

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As you can see from the two answers below, the answer depends on the definition of the word "is" :-). Up to vector-space isomorphism, $\mathbb{C}^2$ can be identified (as Ekuurh wrote) with a (in fact many) subspace of $\mathbb{C}^3$. But under a strict(er) interpretation of your question statement, as Chris Eagle wrote, comparing $\mathbb{C}^2$ and (subspaces of) $\mathbb{C}^3$ is like comparing apples and oranges. (As an aside, why "that is not empty"? The set $\mathbb{C}^2$ of ordered pairs is manifestly non-empty.) – Willie Wong Apr 30 '12 at 11:59
But $\mathbb{C}^{2}$ does not necessarily equate to 2-dimensional space, does it? Otherwise, if a plane is a euclidean subspace of $\mathbb{R}^3$, $\mathbb{R}^2$ would be a subspace of $\mathbb{R}^3$. At least that's how it appears to me.. – mercurial Apr 30 '12 at 13:04
The Math Department in my university is housed in a 4-story building; the offices of the math department are located in the 2nd and 4th floor, exclusively. Does that mean that the Math Department is a 2-story building? – Arturo Magidin Apr 30 '12 at 16:57

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 $\mathbb{C}^2$.

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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 $\mathbb{C}^3$.

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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.

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Of course, you can take any two linearly independent vectors as an example, I just took the standard basis' first two vectors. Saying there are other subspaces of C3 isomorphic to C2 is trivial, true, but really irrelevant. If you must, map (a,b) to (a, a+b, 13.5*a + pi*b). – Ekuurh May 1 '12 at 11:55

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.

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You must also point out that the mapping is injective to properly be able to consider the first space as a subspace of the second one. – Najib Idrissi Apr 30 '12 at 12:28
That's trivial. – Ekuurh Apr 30 '12 at 12:36
Just because it's trivial doesn't mean you shouldn't say it when proving your result. Otherwise, since the result is trivial too, you shouldn't even have bothered with a proof. – Najib Idrissi Apr 30 '12 at 14:58
Every mathematician needs to find a balance of formality, as showing all "trivial" things would take me about 6-10 pages of modus ponens just to prove this easy theorem. – Ekuurh Apr 30 '12 at 15:27
@Ekuurh: Of course. But for someone who feels the need to ask this question, they're more likely to be helped by seeing the details, or at least having it pointed out that there are details. – Nate Eldredge Apr 30 '12 at 17:27