MATH 240

Wed. October 23rd, 2019


Theorem. If a vector space has a basis of nn vectors, then any collection of pp vectors, where p>np>n, is going to be linearly dependent.
(We get this theorem "for free" because of the isomorphism between VV and Rn\R^n that we previously showed exists.)


Theorem. If a vector space has a basis of nn vectors, then any other basis of that vector space will also have nn vectors.

Proof. Let VV be a vector space, and let BB be a basis for VV with nn vectors. Let BB' be another basis for VV.

Thus BB' has exactly nn elements.

This theorem indicates that the number of elements in a basis is invariant for a vector space – it’s the same for every basis of that vector space.


Dimension

Definition. If VV is spanned by a finite set, we say that VV is finite dimensional, and the dimension of VV, or dimV\text{dim}V, is the number of vectors in a basis of VV. The dimension of {0}\{0\} is defined to be 00.
If VV is spanned by an infinite set, we say that VV is infinite dimensional.


Ex. Let M2×2M_{2\times 2} be the space of 2×22\times 2 matrices over R\R (that is, the 2×22\times 2 matrices whose entries are real numbers). M2×2M_{2\times 2} is a vector space over the real numbers. What is dimM2×2\text{dim}M_{2\times 2}?

If we can find a basis for M2×2M_{2\times 2}, then all we have to do is count the number of vectors in that basis to determine dimM2×2\text{dim}M_{2\times 2}. (abcd)=a(1000)+b(0100)+c(0010)+d(0001)B={(1000),(0100),(0010),(0001)}SpanB=M2×2\begin{gathered} \begin{pmatrix} a&b\\c&d \end{pmatrix}=a\begin{pmatrix} 1&0\\0&0 \end{pmatrix}+b\begin{pmatrix} 0&1\\0&0 \end{pmatrix}+c\begin{pmatrix} 0&0\\1&0 \end{pmatrix}+d\begin{pmatrix} 0&0\\0&1 \end{pmatrix}\\ B=\{\begin{pmatrix} 1&0\\0&0 \end{pmatrix},\begin{pmatrix} 0&1\\0&0 \end{pmatrix},\begin{pmatrix} 0&0\\1&0 \end{pmatrix},\begin{pmatrix} 0&0\\0&1 \end{pmatrix} \}\\ \text{Span}B=M_{2\times 2}\\ \end{gathered}
BB spans M2×2M_{2\times 2} and is linearly independent, so BB is a basis for M2×2M_{2\times 2}.
BB contains four elements, thus dimM2×2=4\text{dim}M_{2\times 2}=4.


Subspaces of a finite dimensional vector space

Theorem. Let HH be a subspace of a finite dimensional vector space VV. Any linearly independent set in HH can be expanded if needed to be a basis of HH. (You can keep adding linearly independent vectors to the set to eventually get a vector of HH.) Also, HH is finite dimensional and dimHdimV\text{dim}H\leq \text{dim}V.