Exercises — Span — definition

WHAT we do: find a scalar $c$ with $c(2,1)=(0,0)$. The reasoning: $2c=0$ and $1c=0$ both force $c=0$. Choosing $c=0$ works. Answer: Yes, with $c=0$. This is the general fact from the parent: taking every scalar zero always lands on the origin, so $0$ is in every span.

False. The span is the set of all linear combinations $a(1,0)+b(0,1)=(a,b)$ with $a,b$ ranging over all of $\mathbb{R}$. That is the entire plane ${\mathbb{R}}^2$ — infinitely many points, not just the two generators.

Set every scalar to $0$: $0(3,7)+0(-1,2)+0(5,5)=(0,0)$. The all-zero combination is always legal, so $0$ is always reachable. Every span contains the origin — no matter how many generators or which ones.

WHAT/WHY: membership means "is there a scalar $c$ with $c(2,1)=w$?" For $w=(6,3)$: $2c=6\Rightarrow c=3$; check second coord $1\cdot 3=3$ ✔. Yes, $c=3$. For $u=(2,3)$: $2c=2\Rightarrow c=1$; check $1\cdot 1=1\ne 3$ ✗. No — no single scalar satisfies both coordinates.

A combination is $a(1,0,0)+b(0,1,0)=(a,b,0)$: the third coordinate is forced to $0$. $w=(3,-7,0)$: take $a=3,b=-7$; third coord is $0$ ✔. Yes. $u=(0,0,5)$: needs third coord $=5$, but it can only be $0$. No. Geometrically this span is the $xy$-plane sitting inside 3D space (see figure below).

Set up: $c_1(1,2)+c_2(3,1)=(4,5)$ gives the system $c_1+3c_2=4,2c_1+c_2=5.$ Solve: from the first, $c_1=4-3c_2$. Substitute: $2(4-3c_2)+c_2=5\Rightarrow 8-6c_2+c_2=5\Rightarrow -5c_2=-3\Rightarrow c_2=\frac{3}{5}$. Then $c_1=4-3\cdot \frac{3}{5}=4-\frac{9}{5}=\frac{11}{5}$. Answer: Yes, with $c_1=\frac{11}{5},c_2=\frac{3}{5}$. (Check: $\frac{11}{5}+3\cdot \frac{3}{5}=\frac{11+9}{5}=4$ ✔, $2\cdot \frac{11}{5}+\frac{3}{5}=\frac{22+3}{5}=5$ ✔.)

Observe: $v_2=2v_1$, so any combination $c_1{v}_1+c_2{v}_2=c_1{v}_1+2c_2{v}_1=(c_1+2c_2)v_1$. Renaming $t=c_1+2c_2$ (a single free real number), the span is $\{t(1,1):t\in \mathbb{R}\}$ — just the line through the origin in direction $(1,1)$. Why not ${\mathbb{R}}^2$: the two vectors point the same way (dependent), so the second adds no new direction. To fill the plane you need two genuinely different directions. (See figure: both arrows lie on one line.)

Stack the vectors as columns: $M=\left[\begin{array}{cc}1& 3\\ 2& 1\end{array}\right]$. Then $\det M=1\cdot 1-3\cdot 2=1-6=-5$. Since $\det M\ne 0$, $M$ is invertible, so $Mc=(x,y)$ has a (unique) solution $c$ for every target $(x,y)$. Every vector is reachable $\Rightarrow$ yes, they span ${\mathbb{R}}^2$.

The current span is the $x$-axis $\{(t,0)\}$. A new vector enlarges the span only if it is not already in it.

• $(3,0)$: already on the $x$-axis $\Rightarrow$ no change.
• $(-7,0)$: already on the $x$-axis $\Rightarrow$ no change.
• $(0,1)$: not on the $x$-axis (second coord nonzero) $\Rightarrow$ it adds the $y$-direction, and the span jumps to all of ${\mathbb{R}}^2$. Only $(0,1)$ enlarges the span.

They fail exactly when the columns are dependent, i.e. $\det \left[\begin{array}{cc}1& 2\\ 2& k\end{array}\right]=0$. $\det =1\cdot k-2\cdot 2=k-4$. Setting $k-4=0$ gives $k=4$. At $k=4$: $v_2=(2,4)=2(1,2)=2v_1$, so they span only a line — the span fails to be all of ${\mathbb{R}}^2$. For every other $k$, $\det \ne 0$ and they span ${\mathbb{R}}^2$.

Set up: $a(1,0,1)+b(0,1,1)=(a,b,a+b)$. Match to $(1,1,1)$: $a=1,b=1,a+b=1.$ First two give $a=1,b=1$, but then $a+b=2\ne 1$. The system is inconsistent. Answer: No — $(1,1,1)$ is not in the span. (The span is the plane $\{(a,b,a+b)\}$; the point $(1,1,1)$ lies off it.)

Third vector is already inside the smaller span: $(1,1,0)=1\cdot (1,0,0)+1\cdot (0,1,0)$. So it adds no new direction, and by the "enlarge only if outside" rule the span is unchanged. Formally, both spans are the $xy$-plane $\{(x,y,0)\}$:

• Any combo of the three has third coord $0$ (all three vectors have third coord $0$), so it lands in the $xy$-plane.
• Conversely any $(x,y,0)=x(1,0,0)+y(0,1,0)$ is reachable using just the first two. Hence the two sets have exactly the same span — the third generator is redundant. This is the seed of a minimal spanning set, a Basis.

Let $S=\mathrm{span}\{v_1,\dots ,v_k\}$ and $S^{\prime }=\mathrm{span}\{v_1,\dots ,v_k,w\}$. We show $S=S^{\prime }$ by proving each contains the other.

$S\subseteq S^{\prime }$: any $x=\sum c_i{v}_i$ can be written as $\sum c_i{v}_i+0\cdot w$, a combination of the bigger list. So $x\in S^{\prime }$. ✔

$S^{\prime }\subseteq S$: take any $y=\sum c_i{v}_i+dw\in S^{\prime }$. Since $w\in S$, write $w=\sum a_i{v}_i$. Substitute: $y=\sum c_i{v}_i+d\sum a_i{v}_i=\sum (c_i+d{a}_i)v_i,$ a linear combination of the $v_i$ alone $\Rightarrow y\in S$. ✔

Both inclusions hold, so $S=S^{\prime }$. The redundant vector never enlarges the span — this is exactly why a Basis can be trimmed down to independent vectors.

Span is a line: $\mathrm{span}\{v\}=\{cv:c\in \mathbb{R}\}$. As $c$ sweeps $\mathbb{R}$, the point $cv$ traces the straight line through the origin in direction $v$ — a 1-dimensional flat. It is not all of ${\mathbb{R}}^2$: pick any vector $u$ not parallel to $v$ (one exists in 2D — e.g. rotate $v$ by $90^{\circ }$). If $u=cv$ for some $c$, then $u$ is parallel to $v$, contradiction. So $u\notin \mathrm{span}\{v\}$, meaning the span misses at least one point of ${\mathbb{R}}^2$. Hence it is a proper subset — a line, of Dimension $1$. $■$

The span is the Column space of $A$, and "$w$ reachable" means $Ac=w$ is solvable. ($\Leftarrow$) If $\det A\ne 0$, then $A$ is invertible, so $c=A^{-1}w$ solves $Ac=w$ for every $w\in {\mathbb{R}}^2$. Every vector is reachable $\Rightarrow$ span $={\mathbb{R}}^2$. ($\Rightarrow$) Contrapositive: if $\det A=0$, the columns are dependent, so one is a multiple of the other; their span is a line (or a point), which is a proper subset of ${\mathbb{R}}^2$ $\Rightarrow$ span $\ne {\mathbb{R}}^2$. Combining both directions gives the equivalence. This is precisely the determinant test used throughout the parent note. $■$