1
$\begingroup$

theorem: "Given finite dimensional subspaces $U_1$ and $U_2$ in a vector space V then: $$dim(U_1+U_2)=dim(U_1)+dim(U_2)-dim(U_1 \cap U_2)$$ "

The first step of the proof establishes that $U_1 \cap U_2$ is finite dimensional.

My question is why exactly is this important? Is it because the dimension of a non-finite dimensional space is undefined??

(I gather that this is of course important, just want to know the reason)

$\endgroup$
1
  • 1
    $\begingroup$ The dimension of an infinite-dimensional space is defined (as an infinite cardinal number, a concept that you may not have come across yet). Apart from that, you need to tell us more about the proof (or maybe give a link or a reference) to help us help you: we can't say why a step in a proof is needed without seeing the proof. $\endgroup$ – Rob Arthan Apr 1 '19 at 20:54
0
$\begingroup$

Dimension is meaningful even when it is not finite. In general, the dimension of a vector space is defined to be the cardinality of its basis (one can show that any two bases have to share the same cardinality).

Cardinality essentially means the "size" of a set. For example: the cardinality of $\{a,b,c\}$ is 3, whereas the cardinality of $\mathbb{N} = \{ 0,1,2,\dots \}$ is $\aleph_0$ (i.e., countable infinity). The cardinality of $\mathbb{R}$ is $2^\aleph_0$ (i.e., continuum). We have (this can be made precise) that $3<\aleph_0<2^\aleph_0$. So that each of my example sets is strictly bigger than the previous one.

If you take polynomials $\mathbb{R}[x] = \{ n \geq 0 \text{ and } a_nx^n +\cdots+a_1x+a_0 \;|\; a_0,\dots,a_n \in\mathbb{R}\}$, you get an infinite dimensional vector space (over the reals). So many texts write: $\dim(\mathbb{R}[x])=\infty$.

If we are being more careful, we will specify the kind of infinite dimension we have. Specifically, $\dim(\mathbb{R}[x])=\aleph_0$. The reason for this is that $\beta = \{1,x,x^2,\dots\}$ forms a basis for the space of polynomials and there is an invertible map between $\beta$ and $\mathbb{N}$ (i.e., $x^i \mapsto i$).

Now while subtraction is ok for finite quantities, it is no longer ok for cardinalities. This is much like the fact that $\infty-\infty$ is an indeterminant form.

For example: $\{1,2,3,\dots\}$ take away $\{3,4,\dots\}$ leaves $\{1,2\}$, so one might conclude that $\infty-\infty=2$. But if we took $\{1,2,3,\dots\}$ and removed $\{2,4,6,\dots\}$ we would have $\{1,3,5,\dots\}$ so now we should have $\infty-\infty=\infty$.

So subtracting cardnalities is problematic. Adding them is not. One says that if $A$ has cardinal number $\alpha$ and $B$ has cardinal number $\beta$ then the disjoint union of $A$ and $B$ (put these sets together without allowing overlapping) has cardinal number $\alpha+\beta$.

For example: $A=\{1,2,3\}$ and $B=\{1,2\}$ then make $B$ distinct: $B' = \{1',2'\}$ put them together and get $\{1,2,3,1',2'\}$ thus $2+3=5$.

One can make sense of this for infinite stuff as well.

So why did you text prove finite dimensional first? Possibly because the statement involved a subtraction.

Actually if you say: $\dim(U_1+U_2)+\dim(U_1\cap U_2) = \dim(U_1)+\dim(U_2)$, then this statement holds without the finite dimensional assumption!

Proof sketch: Let $\alpha$ be a basis for $U_1 \cap U_2$. Extend this to a basis $\beta = \alpha \cup \beta_0$ (disjoint union) for $U_1$ and extend also to a basis $\gamma = \alpha \cup \gamma_0$ (disjoint union) for $U_2$.

You can show that $\alpha \cup \beta_0 \cup \gamma_0$ (disjoint union) is a basis for $U_1+U_2$. So that:

$$\dim(U_1+U_2)+\dim(U_1\cap U_2)=|\alpha \cup \beta_0 \cup \gamma_0| + |\alpha|$$ $$= |\alpha|+|\beta_0|+|\gamma_0|+|\alpha| = |\alpha \cup \beta_0| + |\alpha\cup\gamma_0| = \dim(U_1)+\dim(U_2)$$

[where $|X|$ denotes the cardinality of $X$. We used the fact that $|X\cup Y| = |X|+|Y|$ for disjoint sets and that cardinal addition is commutative and associative.]

This proof works whether the dimensions are finite or infinite. :)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.