MATH 240

Wed. October 23rd, 2019

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

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

Proof. Let $V$ be a vector space, and let $B$ be a basis for $V$ with $n$ vectors. Let $B'$ be another basis for $V$.

• If $B'$ had more than $n$ vectors, then $B'$ would be linearly dependent; however, since it is a basis it must be linearly independent. Thus $B'$ has at most $n$ elements.
• If $B'$ had less than $n$ vectors, then we couldn’t write down the elements of $B$ as a linear combination of elements of $B'$; however, since it is a basis this must be true. Thus $B'$ has at least $n$ elements.

Thus $B'$ has exactly $n$ 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.

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Dimension

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

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

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

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Subspaces of a finite dimensional vector space

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