Dimension

So we know what a basis is now. How big can a basis be? If you look at some examples, you’ll notice that for a given vector space, every basis you come up with has the same size. This is no coincidence, and is one of the things that make vector spaces so nice to work with.

Theorem: Let V be a vector space. Then every basis for V must have the same size.

This is true, even for vector spaces with infinite bases, but we won’t prove it here. Let \mathcal B and \mathcal C be arbitrary bases for for V. We want to show that |\mathcal B|=|\mathcal C|. Instead, I’m going to prove that |\mathcal B|\leq|\mathcal C|. Since I chose the bases arbitrarily, I just as well could have chosen them in the other order. But if both directions are true, then equality must hold. If you take nothing else away from this post, take away the fact that this trick is really cool.

Take \mathcal B = \{b_1,b_2,\dots,b_n\} and remove a vector b_n. It its definitely still linearly independent, but it no longer spans all of V. Then there have to be some vectors from \mathcal C=\{c_1,c_2,\dots,c_m\} which I can add to \{b_1,b_2,\dots,b_{n-1}\}. Why? Well, if each c_i was linearly dependent on just the b_i, for 1\leq i\leq n-1, then anything we could write as a linear combination with the c_i we could write with b_i. This means that

V= \mbox{span}(\mathcal C)\subseteq \mbox{span}(\{b_1,\dots,b_{n-1}\})\subsetneq V,

which is a contradiction.

We add in linearly independent vectors from C until they span all of V. This must happen, since all of \mathcal C spans V. We now have a new basis for V. If there are any vectors left taken from \mathcal B, we should remove one, and add in vectors of \mathcal C until it forms a basis again.

Eventually we will have removed all of \mathcal B and will be left with \mathcal C. But in each step we removed one vector, and added at least one vector, possibly more. So we have proved that |\mathcal B|\leq|\mathcal C|. By the symmetry argument described above, |\mathcal B|=|\mathcal C|.

\square

What does this mean? It means that the size of a basis is a property of the vector space itself, not the specific basis chosen. Define the dimension of a vector space to be the size of any basis If V is a vector space over a field k, of dimension n, we write

\dim_kV=n.

Here are some examples. You should be able to come up with a basis for each each vector space. Remember, you only need one, since they are all the same size.

  • \dim_{\mathbb R}\mathbb R^n=n
  • \dim_{\mathbb R}\mathbb C=2
  • \dim_{k}\{0\}=0 for any field k
Advertisements

8 Responses to Dimension

  1. Pingback: More about Dimension « Andy Soffer

  2. Pingback: Linear Transformations « Andy Soffer

  3. Pingback: Linear transformations as matrices « Andy Soffer

  4. Pingback: Kernels and Images « Andy Soffer

  5. Pingback: The field of constructible lengths, part 3 « Andy Soffer

  6. Pingback: Rationals and the Square Root of 17 « Andy Soffer

  7. Pingback: The Tower Rule « Andy Soffer

  8. Pingback: The classical Greek questions « Andy Soffer

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s