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 $\mathbb{C}^{3}$?
Linear Subspace - is $\mathbb{C}^2$ a linear subspace of $\mathbb{C}^3$?
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2The 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? – 2012-04-30
4 Answers
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:
- $\iota$ is one-to-one;
- $\iota$ respects vector addition: $\iota\bigl((a,b)+(c,d)\bigr) = \iota(a,b)+\iota(c,d)$ for all $(a,b),(c,d)\in\mathbb{C}^2$;
- $\iota$ respects scalar multiplication: $\iota\bigl(\alpha(a,b)\bigr) = \alpha\iota(a,b)$ for all $\alpha\in\mathbb{C}$, $(a,b)\in\mathbb{C}$;
- 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 $\mathbb{C}^2$.
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 $\mathbb{C}^3$.
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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0Of 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). – 2012-05-01
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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2@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. – 2012-04-30