Suppose that $V$ is a vector space and $\set{\vectorlist{v}{t}}$ is a basis of $V$. Then the dimension of $V$ is defined by $\dimension{V}=t$. If $V$ has no finite bases, we say $V$ has infinite dimension.
This is a very simple definition, which belies its power. Grab a basis, any basis, and count up the number of vectors it contains. That is the dimension. However, this simplicity causes a problem. Given a vector space, you and I could each construct different bases — remember that a vector space might have many bases. And what if your basis and my basis had different sizes? Applying Definition D we would arrive at different numbers! With our current knowledge about vector spaces, we would have to say that dimension is not “well-defined.” Fortunately, there is a theorem that will correct this problem.
In a strictly logical progression, the next two theorems would precede the definition of dimension. Many subsequent theorems will trace their lineage back to the following fundamental result.
Suppose that $S=\set{\vectorlist{v}{t}}$ is a finite set of vectors which spans the vector space $V$. Then any set of $t+1$ or more vectors from $V$ is linearly dependent.
Notice how the swap of the two summations is so much easier in the third step above, as opposed to all the rearranging and regrouping that takes place in the previous proof. And using only about half the space. And there are no ellipses (…).
Theorem SSLD can be viewed as a generalization of Theorem MVSLD. We know that $\complex{m}$ has a basis with $m$ vectors in it (Theorem SUVB), so it is a set of $m$ vectors that spans $\complex{m}$. By Theorem SSLD, any set of more than $m$ vectors from $\complex{m}$ will be linearly dependent. But this is exactly the conclusion we have in Theorem MVSLD. Maybe this is not a total shock, as the proofs of both theorems rely heavily on Theorem HMVEI. The beauty of Theorem SSLD is that it applies in any vector space. We illustrate the generality of this theorem, and hint at its power, in the next example.
Theorem SSLD is indeed powerful, but our main purpose in proving it right now was to make sure that our definition of dimension (Definition D) is well-defined. Here is the theorem.
Suppose that $V$ is a vector space with a finite basis $B$ and a second basis $C$. Then $B$ and $C$ have the same size.
Theorem BIS tells us that if we find one finite basis in a vector space, then they all have the same size. This (finally) makes Definition D unambiguous.