Linear combinations and span

Day: 2 of 40
Length: 2:32
Textbook: Lay §1.3
Syllabus: Groundwork before Lesson 10

The span of some vectors is every place you can reach by scaling and adding them.

Yesterday ended with the combinations $a\,\yellow{\vec v} + b\,\blue{\vec w}$ covering the whole plane. That raises today's question: given a few vectors, which targets can their combinations reach? We give that set a name, the span, and learn to test whether one particular vector belongs to it. The test turns into a system of equations, which is why Day 3 teaches a method for solving systems and Day 4 rewrites the same question as $A\mathbf x = \mathbf b$ .

Each section below opens with a problem. Try it before you open the walkthrough, even if you only get partway. The box after the walkthrough names what you just did.

Span

Let $\yellow{\vec v} = \begin{bmatrix} 3 \\ 2 \end{bmatrix}$ and $\blue{\vec w} = \begin{bmatrix} -1 \\ 2 \end{bmatrix}$ . Write down five different vectors of the form $a\,\yellow{\vec v} + b\,\blue{\vec w}$ . Is the zero vector one of them? Is $\yellow{\vec v}$ itself?

Walk through the solution

We are free to pick any weights, so start with the easiest ones. Setting both weights to $0$ gives

0\begin{bmatrix} 3 \\ 2 \end{bmatrix} + 0\begin{bmatrix} -1 \\ 2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix},

so the zero vector is reachable. Setting $a = 1$ and $b = 0$ gives $\yellow{\vec v}$ itself, and $a = 0$ , $b = 1$ gives $\blue{\vec w}$ .

Two more come from using both vectors at once:

\begin{aligned} 1\begin{bmatrix} 3 \\ 2 \end{bmatrix} + 1\begin{bmatrix} -1 \\ 2 \end{bmatrix} &= \begin{bmatrix} 3 - 1 \\ 2 + 2 \end{bmatrix} = \begin{bmatrix} 2 \\ 4 \end{bmatrix} \\ 1\begin{bmatrix} 3 \\ 2 \end{bmatrix} - 1\begin{bmatrix} -1 \\ 2 \end{bmatrix} &= \begin{bmatrix} 3 + 1 \\ 2 - 2 \end{bmatrix} = \begin{bmatrix} 4 \\ 0 \end{bmatrix} \end{aligned}

We could keep going forever, because every new pair of weights gives another vector. The list had two vectors, but the set of their combinations has infinitely many.

Span

If $\vec v_1, \dots, \vec v_p$ are in $\mathbb{R}^n$ , then $\operatorname{Span}\{\vec v_1, \dots, \vec v_p\}$ is the set of all linear combinations

c_1 \vec v_1 + c_2 \vec v_2 + \cdots + c_p \vec v_p

where $c_1, \dots, c_p$ can be any scalars. We say the vectors span (or generate) this set.

Problem 1 found three members that every span has. The zero vector is in every span, because all the weights can be $0$ . Each $\vec v_i$ is in the span too, along with all of its multiples, since $c\,\vec v_1 = c\,\vec v_1 + 0\,\vec v_2 + \cdots + 0\,\vec v_p$ .

Let $\vec a_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $\vec a_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$ . Which statement is true?

Two vectors in the plane

Problem 1's vectors seemed to reach everywhere. Now we ask when two vectors can fail to reach a point.

Let $\yellow{\vec v} = \begin{bmatrix} 2 \\ 1 \end{bmatrix}$ and $\blue{\vec w} = \begin{bmatrix} -4 \\ -2 \end{bmatrix}$ . Can a combination $a\,\yellow{\vec v} + b\,\blue{\vec w}$ reach $\begin{bmatrix} 1 \\ 3 \end{bmatrix}$ ?

Walk through the solution

Before solving anything, compare the two vectors. Each entry of $\blue{\vec w}$ is $-2$ times the matching entry of $\yellow{\vec v}$ , since $-2 \cdot 2 = -4$ and $-2 \cdot 1 = -2$ . So $\blue{\vec w} = -2\yellow{\vec v}$ .

That lets us rewrite every combination using $\yellow{\vec v}$ alone:

a\,\yellow{\vec v} + b\,\blue{\vec w} = a\,\yellow{\vec v} + b(-2\yellow{\vec v}) = (a - 2b)\,\yellow{\vec v}.

Every combination is a multiple of $\yellow{\vec v} = (2, 1)$ , so its second entry is half of its first. The target $(1, 3)$ has second entry $3$ , which is not half of $1$ . No weights reach it.

The span here is the line through the origin in the direction of $\yellow{\vec v}$ . Adding $\blue{\vec w}$ gave us nothing new, because it already lay on that line.

Let $\yellow{\vec v} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$ and $\blue{\vec w} = \begin{bmatrix} 1 \\ -1 \end{bmatrix}$ . Find weights that reach $\begin{bmatrix} 5 \\ 1 \end{bmatrix}$ . Then find weights that reach any point $\begin{bmatrix} h \\ k \end{bmatrix}$ .

Walk through the solution

We want $a\begin{bmatrix} 1 \\ 1 \end{bmatrix} + b\begin{bmatrix} 1 \\ -1 \end{bmatrix} = \begin{bmatrix} 5 \\ 1 \end{bmatrix}$ . The left side is the single vector $\begin{bmatrix} a + b \\ a - b \end{bmatrix}$ , and two vectors are equal when their entries match, so we need

\begin{aligned} a + b &= 5 \\ a - b &= 1. \end{aligned}

Adding the two equations cancels $b$ and leaves $2a = 6$ , so $a = 3$ . Then the first equation gives $b = 5 - 3 = 2$ .

Nothing about the numbers $5$ and $1$ mattered. With a general target, the same two moves give $2a = h + k$ and $2b = h - k$ , so

a = \frac{h + k}{2}, \qquad b = \frac{h - k}{2}.

These weights exist for every $h$ and $k$ , so every point of the plane is in $\operatorname{Span}\{\yellow{\vec v}, \blue{\vec w}\}$ .

Two panels. On the left, v = (3, 2) and w = (−1, 2) with a slanted teal grid of combination lines that covers the plane. On the right, v = (3, 2) and w = (−1.5, −1), which points opposite v, and a single teal line through the origin along v.

Two vectors span a line or the whole plane

In $\mathbb{R}^2$ the span of two vectors is one of three sets. It is the whole plane when the vectors point in different directions, as in Problem 3. It is a line through the origin when one vector is a multiple of the other, as in Problem 2. It is just $\{\vec 0\}$ when both vectors are zero.

The same elimination as Problem 3 works for any $\yellow{\vec v} = (v_1, v_2)$ and $\blue{\vec w} = (w_1, w_2)$ . It gives

x_1 = \frac{w_2 h - w_1 k}{v_1 w_2 - v_2 w_1}, \qquad x_2 = \frac{v_1 k - v_2 h}{v_1 w_2 - v_2 w_1}.

In Problem 3 the bottom number is $1 \cdot (-1) - 1 \cdot 1 = -2$ , and in Problem 2 it is $2 \cdot (-2) - 1 \cdot (-4) = 0$ . Whenever $v_1 w_2 - v_2 w_1 \neq 0$ , the weights exist for every target and the span is $\mathbb{R}^2$ . For nonzero vectors, the number is zero exactly when $\blue{\vec w}$ is a multiple of $\yellow{\vec v}$ . Day 7 names it the determinant, and Day 25 shows that it measures area.

Drag either arrow and watch the teal grid of combinations. The number $v_1 w_2 - v_2 w_1$ in the readout drops to $0$ at the moment the grid folds onto one line.

The teal lines show where

a\vec v + b\vec w

can land. Move

\vec w

until the span shrinks to a single line.

\begin{aligned} &\operatorname{Span}\{\textcolor{#FFD447}{\begin{bmatrix} 3 \\ 2 \end{bmatrix}}, \textcolor{#5BC0EB}{\begin{bmatrix} -1 \\ 2 \end{bmatrix}}\} \\ &= \mathbb{R}^2 \end{aligned}

v_1 w_2 - v_2 w_1 = 8

When the span is a line, you can tell whether a point is in it by asking whether the point is a multiple of $\yellow{\vec v}$ . Try it on these dots.

Tap every dot that is in

\operatorname{Span}\{\vec v, \vec w\}

and leave the others alone.

\vec w = -2\,\vec v

a\vec v + b\vec w = (a - 2b)\,\vec v

Membership as an equation

Problems 2 and 3 both came down to matching entries and solving equations. This section turns that into a routine that works for any target.

Is $\pink{\vec b} = \begin{bmatrix} 5 \\ -2 \end{bmatrix}$ in $\operatorname{Span}\left\{\yellow{\begin{bmatrix} 3 \\ 2 \end{bmatrix}}, \blue{\begin{bmatrix} -1 \\ 2 \end{bmatrix}}\right\}$ ? If it is, find the weights.

Walk through the solution

The question asks whether some weights $x_1, x_2$ make a combination equal $\pink{\vec b}$ . Write that down as one equation between vectors and solve it.

1
Write the vector equation, with the unknown weights in front of the vectors:
$x_1 \yellow{\begin{bmatrix} 3 \\ 2 \end{bmatrix}} + x_2 \blue{\begin{bmatrix} -1 \\ 2 \end{bmatrix}} = \pink{\begin{bmatrix} 5 \\ -2 \end{bmatrix}}$
2
3
4

So $\pink{\vec b} = \yellow{\vec v} - 2\blue{\vec w}$ , and $\pink{\vec b}$ is in the span.

The vectors v = (3, 2) and w = (−1, 2). From the tip of v, the arrow −2w runs down to the point (5, −2), which is the tip of the pink vector b = v − 2w, marked with an orange dot.

\yellow{\vec v}

Membership is a system of equations

Asking whether $\pink{\vec b}$ is in $\operatorname{Span}\{\vec a_1, \dots, \vec a_n\}$ is the same as asking whether the vector equation

x_1 \vec a_1 + x_2 \vec a_2 + \cdots + x_n \vec a_n = \pink{\vec b}

has a solution. Matching entries turns it into a linear system with one equation per entry. If the system has a solution, the solution is the list of weights. If it has none, $\pink{\vec b}$ is not in the span.

The augmented matrix of that system keeps only the numbers. Its columns are the vectors themselves, in order, so we write it $[\,\vec a_1\ \cdots\ \vec a_n\ \ \pink{\vec b}\,]$ . For Problem 4 it is

\left[\begin{array}{cc|c} \yellow{3} & \blue{-1} & \pink{5} \\ \yellow{2} & \blue{2} & \pink{-2} \end{array}\right].

On Day 3 this grid of numbers is exactly what we row reduce to solve the system.

The translation also runs backward. The system $2x_1 - x_2 = 4$ , $\ x_1 + 3x_2 = -5$ is the vector equation $x_1\begin{bmatrix} 2 \\ 1 \end{bmatrix} + x_2\begin{bmatrix} -1 \\ 3 \end{bmatrix} = \begin{bmatrix} 4 \\ -5 \end{bmatrix}$ , because the coefficients of $x_1$ read down the rows form the first vector.

Which vector equation has the same solutions as the system $x_1 + 4x_2 = 7$ , $\;3x_1 - 2x_2 = 0$ ?

Solve one more membership question with the sliders. Set up the two equations first, and then check your weights on the picture.

Choose weights

a

and

b

so that

a\,\vec v + b\,\vec w

lands on the ringed point

\begin{bmatrix} 4 \\ -4 \end{bmatrix}

a = 1

b = 0

1\textcolor{#FFD447}{\begin{bmatrix} 2 \\ -1 \end{bmatrix}} + 0\textcolor{#5BC0EB}{\begin{bmatrix} 2 \\ 1 \end{bmatrix}} = \textcolor{#3DD6B5}{\begin{bmatrix} 2 \\ -1 \end{bmatrix}}

A plane in space

The same routine works in $\mathbb{R}^3$ , with one new twist. Now there are three equations but still only two unknowns.

Let $\yellow{\vec u} = \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix}$ and $\blue{\vec v} = \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}$ . Is $\pink{\vec b} = \begin{bmatrix} 4 \\ 3 \\ -1 \end{bmatrix}$ in $\operatorname{Span}\{\yellow{\vec u}, \blue{\vec v}\}$ ?

Walk through the solution

Write the vector equation $x_1\yellow{\vec u} + x_2\blue{\vec v} = \pink{\vec b}$ and match the three entries:

\begin{aligned} 2x_1 + x_2 &= 4 \\ -x_1 + 2x_2 &= 3 \\ x_1 + x_2 &= -1 \end{aligned}

Two unknowns are usually pinned down by two equations, so solve the first two and then test the third. Adding twice row 2 to row 1 cancels $x_1$ , because $2x_1 + 2(-x_1) = 0$ . That leaves $x_2 + 4x_2 = 4 + 6$ , so $5x_2 = 10$ and $x_2 = 2$ . Row 1 then gives $2x_1 + 2 = 4$ , so $x_1 = 1$ .

These are the only weights that satisfy rows 1 and 2. Row 3 asks for $x_1 + x_2 = -1$ , but $1 + 2 = 3$ . Row 3 fails, so no weights satisfy all three rows, and $\pink{\vec b}$ is not in the span.

Three-dimensional axes with a teal plane through the origin spanned by u = (2, −1, 1) and v = (1, 2, 1). The pink vector to (4, 3, 3) ends on the plane at an orange dot. A second pink vector to (4, 3, −1) ends below the plane, joined to the dot by a dashed orange segment.

The figure shows why. The combinations of $\yellow{\vec u}$ and $\blue{\vec v}$ fill a flat plane through the origin. The point $(4, 3, -1)$ sits below that plane, while $(4, 3, 3)$ lies on it.

Keep $\yellow{\vec u}$ and $\blue{\vec v}$ from Problem 5. For which value of $h$ is $\begin{bmatrix} 4 \\ 3 \\ h \end{bmatrix}$ in $\operatorname{Span}\{\yellow{\vec u}, \blue{\vec v}\}$ ?

Walk through the solution

The first two rows do not involve $h$ , so they are the same as in Problem 5, and they still force $x_1 = 1$ and $x_2 = 2$ . The third row now reads $x_1 + x_2 = h$ , which says $1 + 2 = h$ . It holds only when $h = 3$ .

Check the whole vector: $1\yellow{\vec u} + 2\blue{\vec v} = (2 + 2,\ -1 + 4,\ 1 + 2) = (4, 3, 3)$ .

Two vectors in space span a plane

In $\mathbb{R}^3$ , two vectors that are not multiples of each other span a plane through the origin. Most points of space are off that plane. The system for $\pink{\vec b}$ has three equations and two unknowns, and a point is in the span only when one pair of weights satisfies all three at once.

Here is Problem 6 as a picture of the weights. Each row of the system is a line in the $(x_1, x_2)$ plane, and a pair of weights satisfies a row when it lies on that row's line. Rows 1 and 2 cross at $(1, 2)$ . Slide $h$ and watch the pink third line.

Each row of the system is a line of weights. Slide

h

until all three lines meet, then drag the teal weight point onto the meeting point.

h = -2

\begin{aligned} \textcolor{#FFD447}{2x_1 + x_2} &= 4 && \phantom{\checkmark} \\ \textcolor{#5BC0EB}{-x_1 + 2x_2} &= 3 && \phantom{\checkmark} \\ \textcolor{#F28AB2}{x_1 + x_2} &= -2 && \phantom{\checkmark} \end{aligned}

(x_1, x_2) = (0, 0)

The same idea works in the plane with parallel vectors. Below, $\blue{\vec w} = -2\yellow{\vec v}$ , so the span is one line, and only one height $h$ puts $\pink{\vec b}$ on it.

Every

\vec b = \begin{bmatrix} 4 \\ h \end{bmatrix}

sits on the dashed line. Find the

h

that puts

\vec b

\operatorname{Span}\{\vec v, \vec w\}

h = -3

x_1\textcolor{#FFD447}{\begin{bmatrix} 2 \\ 1 \end{bmatrix}} + x_2\textcolor{#5BC0EB}{\begin{bmatrix} -4 \\ -2 \end{bmatrix}} = \textcolor{#F28AB2}{\begin{bmatrix} 4 \\ -3 \end{bmatrix}}

Closed under addition

A span has one more useful property. Adding two vectors from a span never takes you out of it.

Let $\vec v_1 = \begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix}$ and $\vec v_2 = \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix}$ . Take $\vec x = 2\vec v_1 + \vec v_2$ and $\vec y = -\vec v_1 + 3\vec v_2$ , which are both in $\operatorname{Span}\{\vec v_1, \vec v_2\}$ . Find weights that write $\vec x + \vec y$ as a combination of $\vec v_1$ and $\vec v_2$ .

Walk through the solution

One way is to compute both vectors and add them. $\vec x = (2, 0, 4) + (0, 1, -1) = (2, 1, 3)$ and $\vec y = (-1, 0, -2) + (0, 3, -3) = (-1, 3, -5)$ , so $\vec x + \vec y = (1, 4, -2)$ . Then we would still have to find the weights.

A faster way is to add the combinations themselves and group the terms that share a vector:

\begin{aligned} \vec x + \vec y &= (2\vec v_1 + \vec v_2) + (-\vec v_1 + 3\vec v_2) \\ &= (2 - 1)\,\vec v_1 + (1 + 3)\,\vec v_2 \\ &= 1\,\vec v_1 + 4\,\vec v_2. \end{aligned}

Check: $(1, 0, 2) + (0, 4, -4) = (1, 4, -2)$ , which matches. The weights of a sum are the sums of the weights.

Sums stay in the span

If $\vec x = c_1 \vec v_1 + c_2 \vec v_2$ and $\vec y = d_1 \vec v_1 + d_2 \vec v_2$ , then

\vec x + \vec y = (c_1 + d_1)\,\vec v_1 + (c_2 + d_2)\,\vec v_2.

The sums $c_i + d_i$ are ordinary numbers, so $\vec x + \vec y$ is another combination of $\vec v_1$ and $\vec v_2$ . The same grouping works for any number of vectors, and scaling works the same way, since $k\vec x = (kc_1)\vec v_1 + (kc_2)\vec v_2$ . On Day 11 these two facts make every span a subspace.

Each membership test today ended in a small system solved by hand. Larger systems need a method that always works, and tomorrow's lesson teaches it: row reduction on the augmented matrix.

Practice

Warm-up

Find the vector in $\operatorname{Span}\left\{\begin{bmatrix} 1 \\ -2 \\ 0 \end{bmatrix}, \begin{bmatrix} 3 \\ 1 \\ 4 \end{bmatrix}\right\}$ that uses the weights $2$ and $-1$ . (Modeled on Lay §1.3 #15.)

Show the solution

2\begin{bmatrix} 1 \\ -2 \\ 0 \end{bmatrix} - \begin{bmatrix} 3 \\ 1 \\ 4 \end{bmatrix} = \begin{bmatrix} 2 - 3 \\ -4 - 1 \\ 0 - 4 \end{bmatrix} = \begin{bmatrix} -1 \\ -5 \\ -4 \end{bmatrix}

Write the augmented matrix of the vector equation $x_1\begin{bmatrix} 1 \\ 4 \end{bmatrix} + x_2\begin{bmatrix} -2 \\ 3 \end{bmatrix} = \begin{bmatrix} 5 \\ -1 \end{bmatrix}$ . Enter it row by row.

Show the solution

The three vectors become the three columns, in order:

\left[\begin{array}{cc|c} 1 & -2 & 5 \\ 4 & 3 & -1 \end{array}\right]

Row 1 is the equation $x_1 - 2x_2 = 5$ and row 2 is $4x_1 + 3x_2 = -1$ .

What is $\operatorname{Span}\left\{\begin{bmatrix} 2 \\ -4 \\ 6 \end{bmatrix}, \begin{bmatrix} -3 \\ 6 \\ -9 \end{bmatrix}\right\}$ ? (Modeled on Lay §1.3 #19.)

Core

Let $\vec a_1 = \begin{bmatrix} 1 \\ 3 \\ -1 \end{bmatrix}$ , $\vec a_2 = \begin{bmatrix} 2 \\ -1 \\ 4 \end{bmatrix}$ and $\vec b = \begin{bmatrix} -1 \\ 11 \\ -11 \end{bmatrix}$ . Find weights with $x_1 \vec a_1 + x_2 \vec a_2 = \vec b$ . (Modeled on Lay §1.3, Example 5.)

Show a hint

Write the three equations. Use the first to express $x_1$ in terms of $x_2$ , substitute into the second, and then test the third.

Show the solution

The system is $x_1 + 2x_2 = -1$ , $\ 3x_1 - x_2 = 11$ and $-x_1 + 4x_2 = -11$ . The first row gives $x_1 = -1 - 2x_2$ . Substituting into the second gives $-3 - 6x_2 - x_2 = 11$ , so $-7x_2 = 14$ and $x_2 = -2$ . Then $x_1 = -1 + 4 = 3$ .

The third row must agree too, and it does: $-3 + 4(-2) = -11$ . So $\vec b = 3\vec a_1 - 2\vec a_2$ is in $\operatorname{Span}\{\vec a_1, \vec a_2\}$ .

Let $\vec v_1 = \begin{bmatrix} 1 \\ -2 \\ 3 \end{bmatrix}$ , $\vec v_2 = \begin{bmatrix} 2 \\ 1 \\ -1 \end{bmatrix}$ and $\vec y = \begin{bmatrix} 4 \\ h \\ 5 \end{bmatrix}$ . For what value of $h$ is $\vec y$ in $\operatorname{Span}\{\vec v_1, \vec v_2\}$ ? (Modeled on Lay §1.3 #17.)

Show a hint

Two of the three rows do not contain $h$ . Solve those two for the weights first.

Show the solution

The first and third rows do not involve $h$ :

\begin{aligned} x_1 + 2x_2 &= 4 \\ 3x_1 - x_2 &= 5 \end{aligned}

Adding twice the second row to the first gives $7x_1 = 14$ , so $x_1 = 2$ , and then $x_2 = 1$ . The second row says $-2x_1 + x_2 = h$ , so $h = -4 + 1 = -3$ . For any other $h$ that row is false, so $\vec y$ is in the span only when $h = -3$ .

Is $\begin{bmatrix} 1 \\ 5 \\ 2 \end{bmatrix}$ in $\operatorname{Span}\left\{\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ 3 \\ 2 \end{bmatrix}\right\}$ ? (Modeled on Lay §1.3, Example 6.)

A workshop has two production lines. One hour on line A makes $4$ chairs and $1$ table. One hour on line B makes $2$ chairs and $3$ tables. Write a vector equation for running A for $x_1$ hours and B for $x_2$ hours to make $16$ chairs and $14$ tables, then solve it. (Modeled on Lay §1.3 #27.)

Show a hint

Each line's hourly output is a vector (chairs, tables). The total output is a linear combination of those two vectors.

Show the solution

x_1\begin{bmatrix} 4 \\ 1 \end{bmatrix} + x_2\begin{bmatrix} 2 \\ 3 \end{bmatrix} = \begin{bmatrix} 16 \\ 14 \end{bmatrix}

The rows are $4x_1 + 2x_2 = 16$ and $x_1 + 3x_2 = 14$ . The first gives $x_2 = 8 - 2x_1$ . Substituting, $x_1 + 24 - 6x_1 = 14$ , so $-5x_1 = -10$ and $x_1 = 2$ . Then $x_2 = 4$ . Run line A for $2$ hours and line B for $4$ hours.

Let $\vec u = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$ and $\vec v = \begin{bmatrix} 3 \\ -1 \end{bmatrix}$ . Find the weights that reach $\begin{bmatrix} 7 \\ 0 \end{bmatrix}$ . Then explain why $\operatorname{Span}\{\vec u, \vec v\} = \mathbb{R}^2$ . (Modeled on Lay §1.3 #21.)

Show a hint

The second row, $2x_1 - x_2 = 0$ , tells you $x_2$ in terms of $x_1$ . For the second part, redo the elimination with a general target $(h, k)$ .

Show the solution

The rows are $x_1 + 3x_2 = 7$ and $2x_1 - x_2 = 0$ . The second gives $x_2 = 2x_1$ , so the first becomes $7x_1 = 7$ . That gives $x_1 = 1$ and $x_2 = 2$ .

For a general target, the rows are $x_1 + 3x_2 = h$ and $2x_1 - x_2 = k$ . Adding $3$ times row 2 to row 1 gives $7x_1 = h + 3k$ , and subtracting $2$ times row 1 from row 2 gives $-7x_2 = k - 2h$ . So

x_1 = \frac{h + 3k}{7}, \qquad x_2 = \frac{2h - k}{7},

and these exist for every $h$ and $k$ . Every point is reachable, so the span is $\mathbb{R}^2$ . The number $v_1 w_2 - v_2 w_1 = 1(-1) - 2(3) = -7$ is not zero, which says the same thing.

Challenge

Let $\vec v_1 = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}$ , $\vec v_2 = \begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix}$ , $\vec v_3 = \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}$ and $\vec y = \begin{bmatrix} 3 \\ 2 \\ h \end{bmatrix}$ . For which $h$ is $\vec y$ in $\operatorname{Span}\{\vec v_1, \vec v_2, \vec v_3\}$ ? (Modeled on Lay §1.3, Practice Problem 2.)

Show a hint

Three vectors in $\mathbb{R}^3$ do not always reach everything. Subtract row 1 from row 3 and compare the result with row 2.

Show the solution

The rows are

\begin{aligned} x_1 + 2x_2 &= 3 \\ x_2 + x_3 &= 2 \\ x_1 + 3x_2 + x_3 &= h. \end{aligned}

Subtracting row 1 from row 3 gives $x_2 + x_3 = h - 3$ . Row 2 says $x_2 + x_3 = 2$ , so both can hold only when $h - 3 = 2$ , which means $h = 5$ .

The reason is that $\vec v_3 = \vec v_2 - 2\vec v_1$ , so the three vectors all lie in one plane. Having three vectors did not guarantee a third direction.

Suppose $\vec u$ and $\vec v$ are both in $\operatorname{Span}\{\vec w_1, \vec w_2, \vec w_3\}$ . Show that $\vec u + \vec v$ is in that span too. (Modeled on Lay §1.3, Practice Problem 3.)

Show a hint

Being in the span means having weights. Write $\vec u$ and $\vec v$ with weights $c_1, c_2, c_3$ and $d_1, d_2, d_3$ , add, and group.

Show the solution

Since $\vec u$ and $\vec v$ are in the span, there are scalars with $\vec u = c_1\vec w_1 + c_2\vec w_2 + c_3\vec w_3$ and $\vec v = d_1\vec w_1 + d_2\vec w_2 + d_3\vec w_3$ . Then

\begin{aligned} \vec u + \vec v &= c_1\vec w_1 + c_2\vec w_2 + c_3\vec w_3 + d_1\vec w_1 + d_2\vec w_2 + d_3\vec w_3 \\ &= (c_1 + d_1)\vec w_1 + (c_2 + d_2)\vec w_2 + (c_3 + d_3)\vec w_3. \end{aligned}

Each $c_i + d_i$ is a scalar, so $\vec u + \vec v$ is a linear combination of $\vec w_1, \vec w_2, \vec w_3$ , which is exactly what it means to be in their span.

For which $k$ is $\operatorname{Span}\left\{\begin{bmatrix} 2 \\ k \end{bmatrix}, \begin{bmatrix} 3 \\ 6 \end{bmatrix}\right\}$ only a line instead of all of $\mathbb{R}^2$ ?

Show a hint

The span collapses to a line exactly when $v_1 w_2 - v_2 w_1 = 0$ , which is when one vector is a multiple of the other.

Show the solution

With $\vec v = (2, k)$ and $\vec w = (3, 6)$ , the test number is $v_1 w_2 - v_2 w_1 = 2 \cdot 6 - k \cdot 3 = 12 - 3k$ . It is zero when $k = 4$ .

Check: $(2, 4) = \tfrac23 (3, 6)$ , so the two vectors lie on one line through the origin. For every other $k$ the test number is nonzero, and the span is all of $\mathbb{R}^2$ .