MATH 323 Lecture 12

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Substitute lecture

Chapter 3.4: Basis and Dimension

Definitions

We say that vectors ${\overrightarrow{v}}_1,\dots ,{\overrightarrow{v}}_k$ are linearly independendent if $a_1{\overrightarrow{v}}_1+\dots +a_k{\overrightarrow{v}}_k=\overrightarrow{0}$

We say that ${\overrightarrow{w}}_1,\dots {\overrightarrow{w}}_k\in V$ span $V$ if any $\overrightarrow{u}\in V$ can be written as a linear combination of ${\overrightarrow{w}}_1,\dots ,{\overrightarrow{w}}_k$.

Let $V$ be a vector space.

The set ${\overrightarrow{e}}_1,\dots ,{\overrightarrow{e}}_n\in V$ is called a basis in $V$ if

1. ${\overrightarrow{e}}_1,\dots ,{\overrightarrow{e}}_n$ are linearly independent
2. They span $V$

The number of vectors in the basis is called the dimension of $V$

Theorem

All bases in $V$ have the same number of elements called the dimension of $V$.

Note: it is possible for a "finitely defined" Vector space to have infinite dimension.

Theorem 3.4.1

If $\{v_1,\dots ,v_n\}$ is a spanning set of $V$, $m>n$, then any vectors $w_1,\dots ,w_m$ are linearly independent.

Proof. (substitution/elimination process).

If $w_j$, $1\le j\le m$ is a 0 vector, then there is nothing to prove:

$0{\overrightarrow{w}}_1+\dots +\alpha {\overrightarrow{w}}_j+\dots +0{\overrightarrow{w}}_m=\overrightarrow{0}$

If ${\overrightarrow{w}}_1\ne \overrightarrow{0}$, then ${\overrightarrow{w}}_1=a_1{\overrightarrow{v}}_1+\dots +a_n{\overrightarrow{v}}_n$. At least one of the $a$'s is nonzero. Without loss of generality, we assume that it is $a_1$ which is nonzero.

Therefore ${\overrightarrow{v}}_1=\frac{{\overrightarrow{w}}_1-a_2{\overrightarrow{v}}_2-\dots -a_n{\overrightarrow{v}}_n}{a_1}$, so $\{{\overrightarrow{w}}_1,{\overrightarrow{v}}_2,\dots ,{\overrightarrow{v}}_n\}$ is a spanning set for $V$.

Repeat the process with the above spanning set and ${\overrightarrow{w}}_2,\dots ,{\overrightarrow{w}}_n$. When we can no longer continue, we end up with $\{{\overrightarrow{w}}_1,\dots ,{\overrightarrow{w}}_n\}$ as a spanning set for $V$, and $\{{\overrightarrow{w}}_{n+1}=k_1{\overrightarrow{w}}_1+\dots +k_n{\overrightarrow{w}}_n\}$

Example

${\mathbb{ℝ}}^2$ has bases ${\overrightarrow{e}}_1=⟨1,0⟩$, ${\overrightarrow{e}}_2=⟨0,1⟩$ and ${\overrightarrow{v}}_1=⟨1,1⟩$, ${\overrightarrow{v}}_2=⟨1,0⟩$.

Let ${\overrightarrow{w}}_1=⟨1,2⟩$, ${\overrightarrow{w}}_2=⟨2,3⟩$, and ${\overrightarrow{w}}_3=⟨3,4⟩$

Note: $\{{\overrightarrow{e}}_1,{\overrightarrow{e}}_2\}$ is a spanning set of ${\mathbb{ℝ}}^2$.

$w_1=1{\overrightarrow{e}}_1+2{\overrightarrow{e}}_2\to {\overrightarrow{e}}_1={\overrightarrow{w}}_1-2\overrightarrow{e}$

…

Corollary 3.4.2

If $\{v_1,\dots ,v_n\}$ and $\{w_1,\dots ,w_m\}$ are two bases in $V$, then $m=n$.

Proof. Assume $v$'s are spanning set and $m>n$, then from Thm. 3.4.1, we know that $w_1,\dots ,w_m$ are linearly dependent. Not a basis (contradiction)

Flip $n>m$ and by same logic → contradiction. Therefore $m=n$.

Infinite Dimension

If the vector space $V$ has no basis of finitely many vectors, we say that $V$ has infinite dimension.

If $V=\{\overrightarrow{0}\}$, we say that $\dim V=0$

Example

Find the dimension of the space of solutions to the system

$\begin{array}{rl}x+y+z& =0\\ x-2y+3z& =0\end{array}$

Three variables, two constraints: $3-2=1$

Note that the solutions form a vector space, and the set of solutions will be the nullspace of the coefficient matrix $\left[\begin{array}{ccc}1& 1& 1\\ 1& -2& 3\end{array}\right]$.

The solution is of the form $\alpha ⟨-5,2,3⟩$, so the basis has only one vector. Therefore $\dim N=1$

Example

Find the dimension of the nullspace for the operation $A=(x\frac{\mathrm{d}}{\mathrm{d}x}-1)$ over $P_2$

In other words, $A$ does $p(x)\mapsto x\frac{\mathrm{d}p(x)}{\mathrm{d}x}-p(x)$. For a polynomial of degree ≤ 1, $a+bx\mapsto x(b)-(a+bx)=-a$. If $A(a+bx)=0$, then $a=0$, and $b$ is anything. Therefore, $N(A)=\{bx\mid b\in \mathbb{ℝ}\}$, the basis is $\{x\}$, and $\dim N(A)=1$.

Example

${\mathbb{ℝ}}^{2\times 2}$

A basis in ${\mathbb{ℝ}}^{2\times 2}$ would be:

$\begin{array}{rlrlr}{\overrightarrow{e}}_1& =\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]{\overrightarrow{e}}_2& =\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]{\overrightarrow{e}}_3& =\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]{\overrightarrow{e}}_4& =\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]\end{array}$

Theorem 3.4.3

Let $V$ be a vector space, and $\dim V=n>0$. Then

1. any $n$ linearly independent vectors $f_1,\dots ,f_n$ span $V$
2. any $n$ vectors that span $V$ are linearly independent.

Proof.

1. assume they do not span $V$: there is a vector $\overrightarrow{g}$ which is not a linear combination of $f$'s, then $\{f_1,\dots ,f_n,g\}$ would be linearly independent. this contradicts #Theorem 3.4.1 since there are $n+1$ vectors with only $n$ in a basis.
2. assume they are linearly dependent: then $c_1{f}_1+\dots +c_n{f}_n=\overrightarrow{0}$, where $c_1,\dots ,c_n$ are not all 0. This means that one vector could be written as a linear combination of other vectors, so $n-1$ vectors would span $V$. This contradicts the definition of a basis, stating that $n$ vectors must be linearly independent.

Theorem 3.4.4

$\dim V=n>0$

1. No set of fewer than $n$ vectors spans $V$
2. Any $m<n$ linearly independent vectors could be extended (by more vectors) to form a basis.
3. If $v_1,\dots ,v_N$, $N>n$, span $V$, we can pare them down to $n$ vectors, which would be a basis.