Linear Combinations and the Idea of Span

≈ 60 min

The linear combination

Lesson 1 supplied two operations on $\mathbf{R}^n$ : addition of vectors and multiplication of a vector by a real number. Everything in this course is built from those two and nothing else. The object that packages them together is the following.

Definition (linear combination). Let $\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k$ be vectors in $\mathbf{R}^n$ and let $c_1, c_2, \ldots, c_k$ be real numbers. The vector

c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_k\mathbf{v}_k

is called a linear combination of $\mathbf{v}_1, \ldots, \mathbf{v}_k$ . The numbers $c_1, \ldots, c_k$ are the weights (equivalently, the coefficients) of the combination.

Two remarks on the definition, both of which are used constantly.

First, the expression needs no brackets. Addition in $\mathbf{R}^n$ is associative and commutative (Lesson 1), so the terms may be grouped and reordered freely, and the sum of $k$ terms is unambiguous. The result is a single vector in $\mathbf{R}^n$ — the same $\mathbf{R}^n$ the $\mathbf{v}_i$ live in, since neither operation changes the number of components.

Second, and more importantly: the weights range over the whole of $\mathbf{R}$ . They may be negative, they may be zero, they may be fractions or irrational numbers. There is no requirement that they be positive, that they be integers, or that they sum to anything in particular.

Example. Take $\mathbf{v}_1 = (1,2)^{\mathsf{T}}$ and $\mathbf{v}_2 = (3,-1)^{\mathsf{T}}$ in $\mathbf{R}^2$ .

weights $(c_1,c_2)$	$c_1\mathbf{v}_1 + c_2\mathbf{v}_2$	value
$(2,1)$	$(2,4)^{\mathsf{T}} + (3,-1)^{\mathsf{T}}$	$(5,3)^{\mathsf{T}}$
$(-1,0)$	$(-1,-2)^{\mathsf{T}} + (0,0)^{\mathsf{T}}$	$(-1,-2)^{\mathsf{T}}$
$(\tfrac{1}{2},-\tfrac{1}{2})$	$(\tfrac{1}{2},1)^{\mathsf{T}} + (-\tfrac{3}{2},\tfrac{1}{2})^{\mathsf{T}}$	$(-1,\tfrac{3}{2})^{\mathsf{T}}$
$(0,0)$	$(0,0)^{\mathsf{T}} + (0,0)^{\mathsf{T}}$	$(0,0)^{\mathsf{T}}$

The last row is worth a name: the combination with every weight equal to $0$ is the trivial combination, and it always produces $\mathbf{0}$ , whatever the vectors are.

The familiar case is already an instance. With the vectors $\mathbf{e}_1 = (1,0)^{\mathsf{T}}$ and $\mathbf{e}_2 = (0,1)^{\mathsf{T}}$ of Lesson 1,

\begin{bmatrix} 7 \\ -4 \end{bmatrix} = 7\begin{bmatrix} 1 \\ 0 \end{bmatrix} + (-4)\begin{bmatrix} 0 \\ 1 \end{bmatrix} = 7\mathbf{e}_1 - 4\mathbf{e}_2,

so writing a vector in components is writing it as a linear combination. Nothing new has been introduced; what is new is the decision to allow the $\mathbf{v}_i$ to be arbitrary vectors rather than the $\mathbf{e}_i$ .

Span: the set of everything reachable

One linear combination is one vector. Fixing the vectors $\mathbf{v}_1, \ldots, \mathbf{v}_k$ and letting the weights vary over all of $\mathbf{R}$ produces not one vector but a whole collection of them, and that collection is the object of interest.

Definition (span). Let $\mathbf{v}_1, \ldots, \mathbf{v}_k \in \mathbf{R}^n$ . The span of $\mathbf{v}_1, \ldots, \mathbf{v}_k$ is the set of all their linear combinations:

\operatorname{span}\{\mathbf{v}_1, \ldots, \mathbf{v}_k\} = \{\, c_1\mathbf{v}_1 + \cdots + c_k\mathbf{v}_k \; : \; c_1, \ldots, c_k \in \mathbf{R} \,\}.

Read the notation carefully. The braces on the left enclose a list of vectors; the braces on the right enclose a set defined by a condition, and the colon is read "such that". The whole expression names a subset of $\mathbf{R}^n$ — typically an infinite one.

The grammar follows from this. If $\mathbf{b} \in \mathbf{R}^n$ , then " $\mathbf{b} \in \operatorname{span}\{\mathbf{v}_1,\ldots,\mathbf{v}_k\}$ " is a statement that is either true or false; it asserts that at least one choice of weights produces $\mathbf{b}$ . By contrast " $\operatorname{span}\{\mathbf{v}\} = \mathbf{v}$ " is not false but meaningless, because it equates a set with one of its elements.

Three facts fall straight out of the definition.

Proposition 1. For any $\mathbf{v}_1, \ldots, \mathbf{v}_k \in \mathbf{R}^n$ :

$\mathbf{0} \in \operatorname{span}\{\mathbf{v}_1, \ldots, \mathbf{v}_k\}$ ;
$\mathbf{v}_j \in \operatorname{span}\{\mathbf{v}_1, \ldots, \mathbf{v}_k\}$ for each $j$ ;
if $\mathbf{x}$ and $\mathbf{y}$ lie in the span and $t \in \mathbf{R}$ , then $\mathbf{x}+\mathbf{y}$ and $t\mathbf{x}$ lie in the span.

Proof. (1) Take the trivial combination: $0\mathbf{v}_1 + \cdots + 0\mathbf{v}_k = \mathbf{0}$ , using $0\mathbf{v} = \mathbf{0}$ from Lesson 1. (2) Take $c_j = 1$ and every other weight $0$ ; the combination collapses to $1\mathbf{v}_j = \mathbf{v}_j$ . (3) Write $\mathbf{x} = \sum_i c_i \mathbf{v}_i$ and $\mathbf{y} = \sum_i d_i \mathbf{v}_i$ . Collecting terms with the distributive laws gives $\mathbf{x}+\mathbf{y} = \sum_i (c_i+d_i)\mathbf{v}_i$ and $t\mathbf{x} = \sum_i (tc_i)\mathbf{v}_i$ , and both right-hand sides are linear combinations of the same vectors. $\blacksquare$

Part (1) has a geometric consequence used repeatedly from here on: every span contains the origin. Part (3) says a span is closed under exactly the two operations that built it; the standard name for a set with this property is introduced in Section 5, and there is no need for it yet.

What spans look like

Call two vectors $\mathbf{u}, \mathbf{v} \in \mathbf{R}^n$ parallel if one is a scalar multiple of the other: that is, if $\mathbf{v} = c\mathbf{u}$ for some $c \in \mathbf{R}$ , or $\mathbf{u} = d\mathbf{v}$ for some $d \in \mathbf{R}$ . Both clauses are needed because of the zero vector: $\mathbf{0} = 0\mathbf{u}$ holds for every $\mathbf{u}$ , but if $\mathbf{u} = \mathbf{0}$ and $\mathbf{v} \neq \mathbf{0}$ then $\mathbf{v}$ is not a multiple of $\mathbf{u}$ .

One vector

Theorem 2. Let $\mathbf{v} \in \mathbf{R}^n$ .

If $\mathbf{v} = \mathbf{0}$ then $\operatorname{span}\{\mathbf{v}\} = \{\mathbf{0}\}$ : a set with exactly one element.
If $\mathbf{v} \neq \mathbf{0}$ then $\operatorname{span}\{\mathbf{v}\} = \{\, c\mathbf{v} : c \in \mathbf{R} \,\}$ is the line through the origin in the direction of $\mathbf{v}$ .

Proof. If $\mathbf{v} = \mathbf{0}$ then $c\mathbf{v} = \mathbf{0}$ for every $c$ , so the set of combinations has the single element $\mathbf{0}$ . If $\mathbf{v} \neq \mathbf{0}$ , then by Lesson 1 the vector $c\mathbf{v}$ has magnitude $|c|\,\|\mathbf{v}\|$ and lies on the line through $\mathbf{0}$ and $\mathbf{v}$ , pointing the same way when $c>0$ and the opposite way when $c<0$ ; as $c$ runs over $\mathbf{R}$ every point of that line is obtained exactly once. $\blacksquare$

The hypothesis $\mathbf{v} \neq \mathbf{0}$ cannot be dropped: $\operatorname{span}\{(0,0)^{\mathsf{T}}\}$ is a single point, not a line.

Two vectors

Theorem 3. Let $\mathbf{u}, \mathbf{v} \in \mathbf{R}^3$ .

If $\mathbf{u}$ and $\mathbf{v}$ are parallel, then $\operatorname{span}\{\mathbf{u},\mathbf{v}\} = \operatorname{span}\{\mathbf{u}\}$ or $\operatorname{span}\{\mathbf{v}\}$ , hence is a line through the origin (or $\{\mathbf{0}\}$ if both vectors are $\mathbf{0}$ ).
If neither is a scalar multiple of the other — informally, if they point in genuinely different directions — then $\operatorname{span}\{\mathbf{u},\mathbf{v}\}$ is a plane through the origin containing both $\mathbf{u}$ and $\mathbf{v}$ .

Proof of the first part. Suppose $\mathbf{v} = c\mathbf{u}$ . Any combination becomes

c_1\mathbf{u} + c_2\mathbf{v} = c_1\mathbf{u} + c_2(c\mathbf{u}) = (c_1 + c_2 c)\mathbf{u},

which is a multiple of $\mathbf{u}$ ; conversely every multiple $t\mathbf{u}$ arises by taking $c_1 = t$ , $c_2 = 0$ . So the two sets coincide. The case $\mathbf{u} = d\mathbf{v}$ is identical with the roles exchanged. $\blacksquare$

For the second part, note that the weights $c_1$ and $c_2$ act like two coordinates that may be set separately: sliding $c_1$ moves the result along the line of $\mathbf{u}$ , sliding $c_2$ moves it along the line of $\mathbf{v}$ , and because the two directions genuinely differ, the two sliders sweep out a flat two-directional sheet rather than a single line. Lesson 3 gives this sheet its parametric equation and pins down the argument.

The condition cannot be dropped. With $\mathbf{u} = (1,2,3)^{\mathsf{T}}$ and $\mathbf{v} = (2,4,6)^{\mathsf{T}}$ we have $\mathbf{v} = 2\mathbf{u}$ , so

c_1\mathbf{u} + c_2\mathbf{v} = (c_1 + 2c_2)\mathbf{u},

and $\operatorname{span}\{\mathbf{u},\mathbf{v}\}$ is the line through the origin in the direction $(1,2,3)^{\mathsf{T}}$ , not a plane. Two vectors are supplied but only one direction is; the second vector adds nothing. Recognising when this collapse happens is important enough that Section 5 devotes a definition to it.

Consequence: spans never miss the origin

By Proposition 1(1), $\mathbf{0}$ belongs to every span. So the line $\ell = \{\,(x,y)^{\mathsf{T}} : y = x+1\,\}$ in $\mathbf{R}^2$ is not the span of any list of vectors whatsoever, since $(0,0)^{\mathsf{T}} \notin \ell$ . A line or plane that misses the origin can be described — Lesson 3 does exactly that — but it is never a span.

Membership is a system of linear equations

The question "is $\mathbf{b} \in \operatorname{span}\{\mathbf{a}_1, \ldots, \mathbf{a}_k\}$ ?" asks whether weights exist with $c_1\mathbf{a}_1 + \cdots + c_k\mathbf{a}_k = \mathbf{b}$ . Two vectors in $\mathbf{R}^n$ are equal precisely when they agree in every one of the $n$ components, so this single vector equation is $n$ scalar equations in the $k$ unknowns $c_1, \ldots, c_k$ . Membership holds if and only if that system has at least one solution.

Example A (membership holds). Is $\mathbf{b} = (4,5)^{\mathsf{T}}$ in $\operatorname{span}\{\mathbf{a}_1, \mathbf{a}_2\}$ , where $\mathbf{a}_1 = (1,2)^{\mathsf{T}}$ and $\mathbf{a}_2 = (2,1)^{\mathsf{T}}$ ?

Write the requirement out:

c_1\begin{bmatrix} 1 \\ 2 \end{bmatrix} + c_2\begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 4 \\ 5 \end{bmatrix} \quad\Longleftrightarrow\quad \begin{aligned} c_1 + 2c_2 &= 4 \\ 2c_1 + \phantom{2}c_2 &= 5 \end{aligned}

Double the first equation to get $2c_1 + 4c_2 = 8$ and subtract the second: $3c_2 = 3$ , so $c_2 = 1$ and then $c_1 = 4 - 2 = 2$ . Check by substitution, always:

2(1,2)^{\mathsf{T}} + 1(2,1)^{\mathsf{T}} = (2,4)^{\mathsf{T}} + (2,1)^{\mathsf{T}} = (4,5)^{\mathsf{T}} = \mathbf{b}.

The answer is yes, and the weights $(c_1,c_2) = (2,1)$ are the certificate.

Example B (membership fails). Is $\mathbf{b} = (1,1,1)^{\mathsf{T}}$ in $\operatorname{span}\{\mathbf{a}_1, \mathbf{a}_2\}$ , where $\mathbf{a}_1 = (1,0,1)^{\mathsf{T}}$ and $\mathbf{a}_2 = (2,1,3)^{\mathsf{T}}$ ?

Here three components give three equations in two unknowns:

\begin{aligned} c_1 + 2c_2 &= 1 \\ \phantom{c_1 + 2}c_2 &= 1 \\ c_1 + 3c_2 &= 1 \end{aligned}

The second equation forces $c_2 = 1$ ; the first then forces $c_1 = 1 - 2 = -1$ . Every unknown is now determined, so the third equation is a consistency test, and it fails: $c_1 + 3c_2 = -1 + 3 = 2 \neq 1$ . No weights exist, so $\mathbf{b} \notin \operatorname{span}\{\mathbf{a}_1,\mathbf{a}_2\}$ .

Geometrically, $\mathbf{a}_1$ and $\mathbf{a}_2$ are not parallel (no scalar $c$ has $c(1,0,1)^{\mathsf{T}} = (2,1,3)^{\mathsf{T}}$ , since the middle component would need $0 = 1$ ), so by Theorem 3 their span is a plane through the origin in $\mathbf{R}^3$ — and $\mathbf{b}$ is a point of $\mathbf{R}^3$ off that plane.

Notice what has happened. A geometric question about membership of an infinite set has become a finite piece of arithmetic: eliminate, solve, check. It also runs in reverse — the exercises below ask both "find the weights" and "show no weights exist". Deciding solvability of such systems efficiently and in general, for any number of equations and unknowns, is the central computational question of Section 2.

Three failures to guard against

1. Treating a span as one vector, or as a number. $\operatorname{span}\{(1,2)^{\mathsf{T}}\}$ is not $(1,2)^{\mathsf{T}}$ , and it is not $1$ or $2$ . It is the infinite set $\{(c,2c)^{\mathsf{T}} : c \in \mathbf{R}\}$ , which contains $(1,2)^{\mathsf{T}}$ , $(0,0)^{\mathsf{T}}$ , $(-3,-6)^{\mathsf{T}}$ , $(\pi, 2\pi)^{\mathsf{T}}$ and infinitely many others. Any answer to "what is the span?" must describe a set.

2. Assuming two vectors span a plane. The phrase "two vectors" refers to the length of the list, not to the number of directions supplied. If $\mathbf{u} = (3,-1,2)^{\mathsf{T}}$ and $\mathbf{v} = (-6,2,-4)^{\mathsf{T}}$ then $\mathbf{v} = -2\mathbf{u}$ and the span is a line; if $\mathbf{u} = \mathbf{v} = \mathbf{0}$ the span is $\{\mathbf{0}\}$ . Before asserting "plane", verify that neither vector is a scalar multiple of the other.

3. Assuming that more vectors means a bigger span. Adding a vector to the list can leave the span completely unchanged. Take $\mathbf{u} = (1,0)^{\mathsf{T}}$ , $\mathbf{v} = (0,1)^{\mathsf{T}}$ and $\mathbf{w} = (1,1)^{\mathsf{T}}$ in $\mathbf{R}^2$ . Then $\mathbf{w} = \mathbf{u} + \mathbf{v}$ , so

c_1\mathbf{u} + c_2\mathbf{v} + c_3\mathbf{w} = (c_1+c_3)\mathbf{u} + (c_2+c_3)\mathbf{v},

which already lies in $\operatorname{span}\{\mathbf{u},\mathbf{v}\}$ ; and the reverse inclusion holds by taking $c_3 = 0$ . The two spans are equal. In general $\operatorname{span}\{\mathbf{v}_1,\ldots,\mathbf{v}_k\} \subseteq \operatorname{span}\{\mathbf{v}_1,\ldots,\mathbf{v}_k,\mathbf{w}\}$ always holds, so a span can never shrink when a vector is appended — but it need not grow, and it grows precisely when $\mathbf{w}$ was not already reachable.

Self-check

Before starting the exercises, answer these in writing and compare with the worked solutions that follow.

Give three distinct elements of $\operatorname{span}\{(2,-1)^{\mathsf{T}}\}$ , one of which has a negative first component.
Explain in one sentence why $\{(x,y)^{\mathsf{T}} : y = 2x + 3\}$ cannot equal $\operatorname{span}\{\mathbf{v}\}$ for any $\mathbf{v} \in \mathbf{R}^2$ .
State the exact hypothesis under which $\operatorname{span}\{\mathbf{u},\mathbf{v}\}$ is a plane through the origin in $\mathbf{R}^3$ , and give a pair of vectors showing the hypothesis cannot be weakened to " $\mathbf{u}$ and $\mathbf{v}$ are both non-zero".

Let $\mathbf{u} = (1,-2,3)^{\mathsf{T}}$ and $\mathbf{v} = (4,0,-1)^{\mathsf{T}}$ in $\mathbf{R}^3$ , and let $\mathbf{w} = 3\mathbf{u} - 2\mathbf{v}$ . Enter the third component $w_3$ .

In $\mathbf{R}^2$ let $\mathbf{a} = (1,1)^{\mathsf{T}}$ , $\mathbf{b} = (3,-2)^{\mathsf{T}}$ and $\mathbf{c} = (0,5)^{\mathsf{T}}$ . Compute the linear combination $\mathbf{x} = 2\mathbf{a} + 0\mathbf{b} + (-1)\mathbf{c}$ and enter its second component $x_2$ .

Let $\mathbf{a} = (-2,3)^{\mathsf{T}}$ and $\mathbf{b} = (6,-9)^{\mathsf{T}}$ in $\mathbf{R}^2$ . There is exactly one scalar $c$ with $\mathbf{b} = c\,\mathbf{a}$ . Enter $c$ .

Is $\mathbf{b} = (4,6,-3)^{\mathsf{T}}$ a scalar multiple of $\mathbf{a} = (2,3,1)^{\mathsf{T}}$ ?

Yes, with $c = 2$ , since $4 = 2 \cdot 2$ and $6 = 2 \cdot 3$ .

No. The first component forces $c = 2$ and the second agrees, but the third would then need $2 \cdot 1 = -3$ , which is false; no single scalar works for all three components.

Yes, with $c = -3$ , since the third component is $-3$ .

No, because a scalar multiple must have all of its components with the same sign as the original.

In $\mathbf{R}^3$ write $\mathbf{b} = (5,0,-2)^{\mathsf{T}}$ as a linear combination $c_1\mathbf{e}_1 + c_2\mathbf{e}_2 + c_3\mathbf{e}_3$ , where $\mathbf{e}_1 = (1,0,0)^{\mathsf{T}}$ , $\mathbf{e}_2 = (0,1,0)^{\mathsf{T}}$ and $\mathbf{e}_3 = (0,0,1)^{\mathsf{T}}$ . Enter the weight $c_3$ .

Describe $\operatorname{span}\{(3,-1)^{\mathsf{T}}\}$ as a subset of $\mathbf{R}^2$ .

The line through the origin containing $(3,-1)^{\mathsf{T}}$ , that is, the set $\{\,(3c,-c)^{\mathsf{T}} : c \in \mathbf{R}\,\}$ .

The single vector $(3,-1)^{\mathsf{T}}$ .

All of $\mathbf{R}^2$ .

The line through the point $(3,-1)$ in the direction $(1,0)^{\mathsf{T}}$ .

Find the real numbers $c_1, c_2$ with $c_1(1,2)^{\mathsf{T}} + c_2(3,-1)^{\mathsf{T}} = (9,4)^{\mathsf{T}}$ , and enter $c_1$ .

In $\mathbf{R}^3$ let $\mathbf{a}_1 = (1,1,1)^{\mathsf{T}}$ , $\mathbf{a}_2 = (2,-1,3)^{\mathsf{T}}$ and $\mathbf{b} = (5,2,6)^{\mathsf{T}}$ . It is a fact that $\mathbf{b} = c_1\mathbf{a}_1 + c_2\mathbf{a}_2$ for suitable real $c_1, c_2$ . Enter $c_1$ .

Is $\mathbf{b} = (2,5)^{\mathsf{T}}$ an element of $\operatorname{span}\{(1,2)^{\mathsf{T}}, (2,4)^{\mathsf{T}}\}$ ?

Yes: any two vectors in $\mathbf{R}^2$ span the whole of $\mathbf{R}^2$ .

No. Since $(2,4)^{\mathsf{T}} = 2(1,2)^{\mathsf{T}}$ , the span is only the line $\{(c,2c)^{\mathsf{T}} : c \in \mathbf{R}\}$ , and $(2,5)^{\mathsf{T}}$ is not on it because $2 \cdot 2 = 4 \neq 5$ .

Yes, with weights $c_1 = 2$ and $c_2 = \tfrac{1}{2}$ .

The question is not well posed: a span is a set, so membership cannot be asked of it.

Let $\mathbf{a}_1 = (1,0,1)^{\mathsf{T}}$ and $\mathbf{a}_2 = (0,1,1)^{\mathsf{T}}$ in $\mathbf{R}^3$ . Is $\mathbf{b} = (1,2,3)^{\mathsf{T}}$ in $\operatorname{span}\{\mathbf{a}_1,\mathbf{a}_2\}$ ?

Yes: $\mathbf{b} = 1\mathbf{a}_1 + 2\mathbf{a}_2$ .

No: $\mathbf{b}$ has three components but only two vectors are given, so it cannot be reached.

Yes: $\mathbf{b} = 2\mathbf{a}_1 + 1\mathbf{a}_2$ .

No: the third components of $\mathbf{a}_1$ and $\mathbf{a}_2$ are both $1$ , and two ones cannot combine to give $3$ .

Let $\mathbf{a}_1 = (1,1,0)^{\mathsf{T}}$ and $\mathbf{a}_2 = (0,1,1)^{\mathsf{T}}$ in $\mathbf{R}^3$ . Is $\mathbf{b} = (3,-1,2)^{\mathsf{T}}$ in $\operatorname{span}\{\mathbf{a}_1,\mathbf{a}_2\}$ ?

Yes, with $c_1 = 3$ and $c_2 = 2$ .

No. The first component forces $c_1 = 3$ and the third forces $c_2 = 2$ , but the second component would then be $c_1 + c_2 = 5$ , not $-1$ ; the system is inconsistent.

Yes, with $c_1 = -1$ and $c_2 = 2$ .

No, because $\mathbf{b}$ is not a scalar multiple of $\mathbf{a}_1$ and not a scalar multiple of $\mathbf{a}_2$ .

Describe $\operatorname{span}\{(2,-4,6)^{\mathsf{T}}, (-1,2,-3)^{\mathsf{T}}\}$ as a geometric object in $\mathbf{R}^3$ . Answer with a short phrase, for example "a plane through the origin" or "a line through the origin".

Let $\mathbf{u}, \mathbf{v}$ be arbitrary vectors in $\mathbf{R}^n$ . Which of the following is a correct and complete justification that $\mathbf{0} \in \operatorname{span}\{\mathbf{u},\mathbf{v}\}$ ?

Take $c_1 = c_2 = 0$ . Then $c_1\mathbf{u} + c_2\mathbf{v} = 0\mathbf{u} + 0\mathbf{v} = \mathbf{0} + \mathbf{0} = \mathbf{0}$ , and this works whatever $\mathbf{u}$ and $\mathbf{v}$ are.

Take $c_1 = 1$ and $c_2 = -1$ . Then $c_1\mathbf{u} + c_2\mathbf{v} = \mathbf{u} - \mathbf{v} = \mathbf{0}$ .

It holds only in the case where $\mathbf{u}$ or $\mathbf{v}$ is itself the zero vector.

It holds only when $\mathbf{u}$ and $\mathbf{v}$ point in genuinely different directions.

For which value of $t$ does the vector $(t,6)^{\mathsf{T}}$ lie in $\operatorname{span}\{(2,-3)^{\mathsf{T}}\}$ ? Enter $t$ .

For which value of $k$ does $\mathbf{b} = (1,k,5)^{\mathsf{T}}$ lie in $\operatorname{span}\{(1,2,-1)^{\mathsf{T}}, (0,1,3)^{\mathsf{T}}\}$ ? Enter $k$ .

Consider $\mathbf{u} = (1,-2,t)^{\mathsf{T}}$ and $\mathbf{v} = (-3,6,9)^{\mathsf{T}}$ in $\mathbf{R}^3$ . For exactly one value of $t$ the set $\operatorname{span}\{\mathbf{u},\mathbf{v}\}$ is a line through the origin rather than a plane. Enter that value of $t$ .

Consider the claim: if $\mathbf{u}, \mathbf{v}, \mathbf{w} \in \mathbf{R}^n$ and $\mathbf{w}$ is different from both $\mathbf{u}$ and $\mathbf{v}$ , then $\operatorname{span}\{\mathbf{u},\mathbf{v},\mathbf{w}\}$ contains strictly more vectors than $\operatorname{span}\{\mathbf{u},\mathbf{v}\}$ . Must this be true?

True: appending a new vector to the list always creates combinations that were not available before.

False. Take $\mathbf{u} = (1,0)^{\mathsf{T}}$ , $\mathbf{v} = (0,1)^{\mathsf{T}}$ and $\mathbf{w} = (1,1)^{\mathsf{T}}$ in $\mathbf{R}^2$ . Then $\mathbf{w}$ differs from both, yet $\mathbf{w} = \mathbf{u}+\mathbf{v}$ , so every combination of the three is already a combination of $\mathbf{u}$ and $\mathbf{v}$ and the two spans are equal.

True provided $\mathbf{u}$ and $\mathbf{v}$ are not parallel.

False, but only because appending a vector can make the span smaller.

The following argument is offered.

Claim. For $\mathbf{u} = (1,-2,3)^{\mathsf{T}}$ and $\mathbf{v} = (-2,4,-6)^{\mathsf{T}}$ , the set $\operatorname{span}\{\mathbf{u},\mathbf{v}\}$ is a plane through the origin in $\mathbf{R}^3$ . Step 1. $\mathbf{u}$ and $\mathbf{v}$ are both non-zero vectors in $\mathbf{R}^3$ . Step 2. The span of two non-zero vectors in $\mathbf{R}^3$ is a plane through the origin. Step 3. Hence $\operatorname{span}\{\mathbf{u},\mathbf{v}\}$ is a plane through the origin. Step 4. In particular $(0,0,0)^{\mathsf{T}}$ belongs to $\operatorname{span}\{\mathbf{u},\mathbf{v}\}$ .

At which step does the argument first fail, and why?

Step 1 — the vector $\mathbf{v}$ has been written down incorrectly and is not in $\mathbf{R}^3$ .

Step 2 — being non-zero is not sufficient. The span is a plane only when neither vector is a scalar multiple of the other; here $\mathbf{v} = -2\mathbf{u}$ , so $\operatorname{span}\{\mathbf{u},\mathbf{v}\} = \operatorname{span}\{\mathbf{u}\}$ , a line through the origin.

Step 3 — the inference is invalid even though Steps 1 and 2 are both correct.

Step 4 — a span need not contain the zero vector, so this conclusion is unwarranted.