18 - Span

Linear Combination

Let $u_1,u_2,\cdots ,u_m\in V$ for some vector space over $\mathbf{F}$
A linear combination of $u_1,u_2,\cdots ,u_m$ is

$${\alpha }_1{u}_1+{\alpha }_2{u}_2+\cdots +{\alpha }_m{u}_m\text{ for some }{\alpha }_1,{\alpha }_2,\cdots ,{\alpha }_m\in \mathbf{F}$$

Span

Let $u_1,u_2,\cdots ,u_m\in V$ for some vector space over $\mathbf{F}$
The span of $u_1,u_2,\cdots ,u_m$ is defined as

$$⟨u_1,\cdots ,u_m⟩:=\{{\alpha }_1{u}_1+\cdots +{\alpha }_m{u}_m:{\alpha }_1,\cdots ,{\alpha }_m\in \mathbf{F}\}$$

Also the smallest subspace of $V$ that contains $u_1,\cdots u_m$

AKA all the possible linear combinations of $u_1,u_2,\cdots ,u_m$

Spans are a subspace of the Vector Space lemma

Let $u_1,u_2,\cdots ,u_m\in V$ for some vector space over $\mathbf{F}$ then

$$⟨u_1,u_2,\cdots ,u_m⟩\le V$$

Proof of Subspace Property

1. $0_V\in ⟨u_1,u_2,\cdots ,u_m⟩$
   Setting ${\alpha }_1={\alpha }_2=\cdots ={\alpha }_m=0$ we get

$$0_V=0u_1+0u_2+\cdots +0u_m$$

So $0_V\in ⟨u_1,u_2,,\cdots ,u_m⟩$
2) $\lambda v_1+v_2\in U$
Take $v_1,v_2\in U$ so suppose

$$v_1={\alpha }_1{u}_1+{\alpha }_2{u}_2+\cdots +{\alpha }_m{u}_m$$ $$v_2={\beta }_1{u}_1+{\beta }_2{u}_2+\cdots +{\beta }_m{u}_m$$

Then for $\lambda \in \mathbf{F}$

$$\lambda v_1+v_2=(\lambda {\alpha }_1+{\beta }_1)u_1+\cdots +(\lambda {\alpha }_m+{\beta }_m)u_m\in U$$

So by the subspace test $U\le V$

Spanning Set

Let $V$ be a vector space over $\mathbf{F}$
If $S\subseteq V$ and $V=⟨S⟩$ then

$$S\text{ spans }V$$

in other words

$$S\text{ is a spanning set for }V$$

Also means every vector $v\in V$ can be written as a linear combination of the elements in $S$

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Row Space ( $m\times n$ Matrix)

$\mathrm{Row}(A)$: Span of the rows of a matrix
With $\mathrm{Row}(A)\le R^n$

Column Space

$\mathrm{col}(A)$: Span of the columns of the matrix
With $\mathrm{col}(A)\le {\mathbb{R}}_{\mathrm{col}}^m$

Property of Row Space

Let $A=(a_{ij})$ be a $n\times n$ matrix and $B=(b_{ij})$ be a $k\times m$ matrix
Let $R=(r_{ij})$ be the RRE form of $A$ by EROs

1. The non-zero rows of $R$ are independent
2. Rows of $R$ are linear combinations of the rows of $A$
3. $\mathrm{Row}(BA)$ is contained in $\mathrm{Row}(A)$
4. If $k=m$ and $B$ is invertible then $\mathrm{Row}(BA)=\mathrm{Row}(A)$
5. $\mathrm{Row}(R)=\mathrm{Row}(A)$

Proof

• (1) Denote the non-zero rows of $R$ as ${\mathbf{r}}_1,\cdots ,{\mathbf{r}}_r$ and suppose that

$$c_1{\mathbf{r}}_1+c_2{\mathbf{r}}_2+\cdots +c_r{\mathbf{r}}_r=0$$

Suppose that the leading $1$ or $r_1$ appears in the $j$th column then

$$c_1+c_2{r}_{2j}+c_3{r}_{3j}+\cdots +c_r{r}_{rj}=0$$

However as $R$ is in $\mathrm{RRE}$ form then each of $r_{2j},r_{3j},\cdots ,r_{rj}$ are $0$ as they are entries under a leading $1$ by RRE Form Properties
Hence

$$c_1=0$$

Similarly by focusing on the column which contains the leading $1$ of ${\mathbf{r}}_2$ then likewise

$$c_2=0$$

Repeat this for the rest
Then as $c_i=0$ for each $1\le i\le r$ then the non-zero rows ${\mathbf{r}}_i$ are independent

• (3) Remember that

$$(i,j)\text{th entry of }BA=\sum _{s=1}^m{b}_{is}{a}_{sj}(1\le i\le k,1\le j\le n)$$

Hence the $i$th row of $BA$ is the row vector

$$\left(\sum _{s=1}^m{b}_{is}{a}_{s1},\sum _{s=1}^m{b}_{is}{a}_{s2},\cdots ,\sum _{s=1}^m{b}_{is}{a}_{sn}\right)=\sum _{s=1}^m{b}_{is}\underset{s\text{th row of }A}{\underbrace{(a_{s1},a_{s2},\cdots ,a_{sn})}}$$

Thus it is a linear combination of the rows of $A$ so every row of $BA$ is in $\mathrm{Row}(A)$
As a vector in the row space $\mathrm{Row}(BA)$ is a linear combination of the rows of $BA$ then they are a linear combination of the rows of $A$ hence
$\mathrm{Row}(BA)$ is contained in $\mathrm{Row}(A)$

• (2) By (3) as $R=BA$ with $B=E_k\cdots E_1$for some Elementary Matrices which are invertible
  Hence $\mathrm{Row}(R)$ is contained in $\mathrm{Row}(A)$
• (4) By (3) as $\mathrm{Row}(A)=\mathrm{Row}(B^{-1}(BA))$ is contained in $\mathrm{Row}(BA)$ and by $(3)$ $\mathrm{Row}(BA)$ is contained in $\mathrm{Row}(A)$ so $\mathrm{Row}(A)=\mathrm{Row}(BA)$

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Test for a Spanning Set corollary

Let $A$ be a $m\times n$ matrix.
Then the rows of $A$ span ${\mathbb{R}}^n$ if and only if

$$\mathrm{RRE}(A)=\left(\begin{array}{c}{I}_n\\ \\ 0_{(k-n)n}\end{array}\right)$$

Proof

1. Let ${\mathbf{r}}_1,{\mathbf{r}}_2,\cdots ,{\mathbf{r}}_m$ be the rows of $A$ in ${\mathbb{R}}^n$ and suppose they span ${\mathbb{R}}^n$
   By Property of Row Space then the row space is invariant under EROs
   If the $i$th column of $\mathrm{RRE}(A)$ does not contain a leading $1$ then ${\mathbf{e}}_i$ would not be in the row space
   Consequently every column contains a leading $1$ and so

$$\mathrm{RRE}(A)=\left(\begin{array}{c}{I}_n\\ \\ 0_{(k-n)n}\end{array}\right)$$

2. Conversely if $\mathrm{RRE}(A)$ has the above form then the rows of $\mathrm{RRE}(A)$ are spanning and hence so the original rows ${\mathbf{r}}_1,{\mathbf{r}}_2,\cdots ,{\mathbf{r}}_m$ are also spanning