MATH 323 Lecture 10

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Vector Space

Closure criteria:

1. $\overrightarrow{x}+\overrightarrow{y}\in V\mathrm{\forall }\overrightarrow{x},\overrightarrow{y}\in V$
2. $\alpha \overrightarrow{x}\in V\mathrm{\forall }\overrightarrow{x}\in V\mathrm{\forall }\alpha \in \mathbb{ℝ}$

Subspace

A vector space $V$ automatically has two subspaces:

1. $S_1=\{0\}$
2. $S_2=V$

When $S\ne \{0\}$ and $S\ne V$, then $S$ is called a proper subspace.

Null Space of Matrix

Let $A$ be a $m\times n$ matrix, and let $N(A)$ be the set of all solutions to the system $A\overrightarrow{x}=\overrightarrow{0}$:

$N(A)=\{\overrightarrow{x}\in {\mathbb{ℝ}}^n\mid A\overrightarrow{x}=\overrightarrow{0}\}$

$N(A)$ is clalled the null space (also called nullspace or kernel) of $A$.

$N(A)$ is a vector space since $A(\overrightarrow{x}+\overrightarrow{y})=A\overrightarrow{x}+A\overrightarrow{y}=\overrightarrow{0}$ and $A(\alpha \overrightarrow{x})=\alpha A\overrightarrow{x}=\alpha \overrightarrow{0}=\overrightarrow{0}$

Example

Determine $N(A)$ if $A=\left[\begin{array}{cccc}1& 1& 1& 0\\ 2& 1& 0& 1\end{array}\right]$

Gauss-Jordan reduction of $A$ gives $\left[\begin{array}{ccccc}1& 0& -1& 1& 0\\ 0& 1& 2& -1& 0\end{array}\right]$, so the solutions of $A\overrightarrow{x}=\overrightarrow{0}$ are

$\begin{array}{rl}\overrightarrow{x}& =⟨\alpha -\beta ,-2\alpha +\beta ,\alpha ,\beta ⟩\mathrm{\forall }\alpha ,\beta \in \mathbb{ℝ}\\ & =\alpha \left(\begin{array}{c}1\\ -2\\ 1\\ 0\end{array}\right)+\beta \left(\begin{array}{c}-1\\ 1\\ 0\\ 1\end{array}\right)\end{array}$

Linear Combinations and Span

Let $\overrightarrow{v}=v_1,\dots ,v_n\in V$.

The sum ${\alpha }_1{v}_1+{\alpha }_2{v}_2+\dots +{\alpha }_n{v}_n$, where $\overrightarrow{\alpha }\in {\mathbb{ℝ}}^n$, is called a linear combination of $\overrightarrow{v}$.

The set of all such linear combinations is called the span of $\overrightarrow{v}$:

$\mathrm{S}\mathrm{p}\mathrm{a}\mathrm{n}(v_1,\dots ,v_n)=\{v\in V\mid V={\alpha }_1{v}_1+\dots +{\alpha }_n{v}_n\}$

Example

${\overrightarrow{b}}_1=⟨1,0,0⟩$ and ${\overrightarrow{b}}_2=⟨0,1,0⟩$ are in ${\mathbb{ℝ}}^3$

The span of ${\overrightarrow{b}}_1$ and ${\overrightarrow{b}}_2$ is the set of all vectors $\overrightarrow{x}=\alpha {\overrightarrow{b}}_1+\beta {\overrightarrow{b}}_2=⟨\alpha ,\beta ,0⟩$.

These two vectors form the standard basis in 2D space.

Theorem

$\mathrm{S}\mathrm{p}\mathrm{a}\mathrm{n}(v_1,\dots ,v_n)$ is a subspace of $V$

$\begin{array}{rl}\overrightarrow{x}& ={\alpha }_1{v}_1+\dots +{\alpha }_n{v}_n\\ \overrightarrow{y}& ={\beta }_1{v}_1+\dots +{\beta }_n{v}_n\\ \overrightarrow{x}+\overrightarrow{y}& =({\alpha }_1+{\beta }_1)v_1+\dots +({\alpha }_n+{\beta }_n)v_n\in \mathrm{S}\mathrm{p}\mathrm{a}\mathrm{n}(v_1,\dots ,v_n)\\ \lambda \overrightarrow{x}& =(\lambda {\alpha }_1)v_1+\dots +(\lambda {\alpha }_n)v_n\in \mathrm{S}\mathrm{p}\mathrm{a}\mathrm{n}(v_1,\dots ,v_n)\end{array}$

Spanning Set

The set $S=\{v_1,\dots ,v_n\}$ is a spanning set for $V$ if $V=\mathrm{S}\mathrm{p}\mathrm{a}\mathrm{n}(v_1,\dots ,v_n)$

If a subset of $S$ is a spanning set of $V$, then $S$ itself is a spanning set of $V$ (set linear coefficient of other terms in set to 0).

Example

$\mathrm{S}\mathrm{p}\mathrm{a}\mathrm{n}(\hat{ı},\hat{ȷ},\hat{k})={\mathbb{ℝ}}^3$

These are the standard basis vectors for 3D space.

Example

Is $\{\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right),\left(\begin{array}{c}1\\ 1\\ 0\end{array}\right),\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)\}$ a spanning set of ${\mathbb{ℝ}}^3$?

$\begin{array}{rl}{\alpha }_1⟨1,1,1⟩+{\alpha }_2⟨1,1,0⟩+{\alpha }_3⟨1,0,0⟩& =⟨x_1,x_2,x_3⟩\\ {\alpha }_1+{\alpha }_2+{\alpha }_3& =x_1\\ {\alpha }_1+{\alpha }_2& =x_2\\ {\alpha }_1& =x_3\\ \end{array}$

This has a solution for any $\overrightarrow{x}\in {\mathbb{ℝ}}^3$, so it is indeed a spanning set (the vectors in the original problem are noncoplanar, so that's easier to visualize)

Example

$P_3$ is the set of polynomials of degree < 3.

$P_3=\mathrm{S}\mathrm{p}\mathrm{a}\mathrm{n}(1,x,x^2)$

$c{v}_1+b{v}_2+a{v}_3=a{x}^2+bx+c\in P_3$

Linear Independence

Given ${\overrightarrow{x}}_1=⟨1,-1,2⟩$, ${\overrightarrow{x}}_2=⟨-2,3,1⟩$, and $\overrightarrow{x_3}=⟨-1,3,8⟩$,

$S=\mathrm{S}\mathrm{p}\mathrm{a}\mathrm{n}({\overrightarrow{x}}_1,{\overrightarrow{x}}_2,{\overrightarrow{x}}_3)=\mathrm{S}\mathrm{p}\mathrm{a}\mathrm{n}({\overrightarrow{x}}_1,{\overrightarrow{x}}_2)$

This is true since ${\overrightarrow{x}}_3=3{\overrightarrow{x}}_1+2{\overrightarrow{x}}_2$

The set $\{{\overrightarrow{x}}_1,{\overrightarrow{x}}_2,{\overrightarrow{x}}_3\}$ is called linearly dependent

1. If $v_1,\dots ,v_n$ span $V$ and one of these vectors can be written as a linear combination of the $n-1$ others, then those $n-1$ vectors span $V$.
2. Given $n$ vectors $v_1,\dots ,v_n$, it is possible to write one of the vectors as a linear combination of the other $n-1$ vectors iff there exist scalars $c_1,\dots ,c_n$ (not all zero!) such that $c_1{v}_1+\dots +c_n{v}_n=0$.