Span
A few vectors can be scaled and added. Which vectors can be reached that way at all, and what shape does that collection make?
Where you are 8 in Vectors in n dimensions
Why this exists
Nemanja, Branka and Dragica run the camera at a small observatory. The camera rides on a mount driven by two motors. One turn of the first motor moves the camera by a fixed vector \mathbf{p}, in centimeters from its home position; one turn of the second moves it by \mathbf{q}. Each motor runs forward or in reverse and can stop at any fraction of a turn, so after c_1 turns of the first and c_2 of the second the camera sits at c_1\mathbf{p} + c_2\mathbf{q}.
Linear combination answered whether one given position can be reached. The crew needs every position the mount can reach, the shape they make, and whether a third motor would add any. Everything reachable from a list by scaling and adding is its span.
Definition
Let \mathbf{v}_1, \dots, \mathbf{v}_k be vectors in \mathbb{R}^n. Their span is the set of all their linear combinations:
A vector \mathbf{b} is in the span, written \mathbf{b} \in \operatorname{span}\{\mathbf{v}_1, \dots, \mathbf{v}_k\}, when it is a linear combination of \mathbf{v}_1, \dots, \mathbf{v}_k. The vectors span a set W (for instance all of \mathbb{R}^n) when their span is exactly W.
The list is finite, but the span is usually an infinite set. It always holds \mathbf{0} (all weights zero) and each \mathbf{v}_i (weight 1 on \mathbf{v}_i, 0 on the rest).
Intuition
Start with one nonzero vector \mathbf{u}. Its span is every multiple c\mathbf{u}. As c runs through the real numbers, the tip of c\mathbf{u} runs along the whole line through the origin in the direction of \mathbf{u}: positive c on one side, negative c on the other, c = 0 at the origin. For \mathbf{u} = (3, 1) that line is x = 3y.
A second vector \mathbf{v} off that line adds a second direction. To reach \mathbf{b}, draw the line through \mathbf{b} parallel to \mathbf{v}. It crosses the line of \mathbf{u} at some c_1\mathbf{u}, and the rest of the way runs along \mathbf{v}, so it is some c_2\mathbf{v}. So \mathbf{u} and \mathbf{v} span \mathbb{R}^2 (proved by algebra under Why it is true).
In \mathbb{R}^3 the same two moves sweep out a flat sheet: two vectors not on one line span a plane through the origin. A third vector on the sheet adds nothing; one pointing off it reaches all of space. Past three entries there is no picture, but the words carry over: a span is \{\mathbf{0}\}, a line, a plane, or a larger flat piece of \mathbb{R}^n, always through the origin.
Results
Testing membership. \mathbf{b} \in \operatorname{span}\{\mathbf{v}_1, \dots, \mathbf{v}_k\} exactly when the system with augmented matrix [\,\mathbf{v}_1 \ \cdots \ \mathbf{v}_k \mid \mathbf{b}\,] is consistent, which row reduction settles.
The span of one or two vectors. \operatorname{span}\{\mathbf{v}\} is \{\mathbf{0}\} when \mathbf{v} = \mathbf{0}, and otherwise the line through the origin along \mathbf{v}. For two vectors, if neither is a multiple of the other, \operatorname{span}\{\mathbf{v}_1, \mathbf{v}_2\} is the plane through the origin containing both; if \mathbf{v}_2 = t\mathbf{v}_1, it equals \operatorname{span}\{\mathbf{v}_1\}. In \mathbb{R}^2 such a plane is all of \mathbb{R}^2. In \mathbb{R}^3 it is the solution set of one equation a_1x + a_2y + a_3z = 0 with a_1, a_2, a_3 not all zero.
Spanning \mathbb{R}^m. Let A be the m \times k matrix whose columns are \mathbf{v}_1, \dots, \mathbf{v}_k in \mathbb{R}^m. The vectors span \mathbb{R}^m exactly when an echelon form of A (Echelon forms and Gaussian elimination) has a pivot in every row. Each column holds at most one pivot, so this needs k \geq m: fewer than m vectors never span \mathbb{R}^m.
A vector already in the span adds nothing. If \mathbf{w} \in \operatorname{span}\{\mathbf{v}_1, \dots, \mathbf{v}_k\}, then
If \mathbf{w} is not in the span, the span grows, since the new one contains \mathbf{w}.
A span is closed under combinations. A linear combination of vectors in a span is again in that span (a combination of combinations is a combination, from Linear combination). This makes every span a subspace.
Why it is true
One or two vectors. \operatorname{span}\{\mathbf{v}\} = \{c\mathbf{v}\}, which is \{\mathbf{0}\} when \mathbf{v} = \mathbf{0} and the line along \mathbf{v} otherwise. If \mathbf{v}_2 = t\mathbf{v}_1, then c_1\mathbf{v}_1 + c_2\mathbf{v}_2 = (c_1 + tc_2)\mathbf{v}_1, and c_1 + tc_2 takes every real value, so the two spans agree. If neither is a multiple of the other, the set of all c_1\mathbf{v}_1 + c_2\mathbf{v}_2 is, by definition, the plane through the origin with directions \mathbf{v}_1 and \mathbf{v}_2.
In \mathbb{R}^2 that plane is all of \mathbb{R}^2. Write \mathbf{v}_1 = (a, b), \mathbf{v}_2 = (c, d) and D = ad - bc. When D \neq 0, the weights \displaystyle c_1 = \frac{dx - cy}{D} and \displaystyle c_2 = \frac{ay - bx}{D} give c_1\mathbf{v}_1 + c_2\mathbf{v}_2 = (x, y), as substituting shows. And D \neq 0 here: \mathbf{v}_1 \neq \mathbf{0}, since \mathbf{0} = 0\mathbf{v}_2; if ad = bc with a \neq 0, then \displaystyle \mathbf{v}_2 = \frac{c}{a}\mathbf{v}_1, and with a = 0, b \neq 0 forces c = 0 and \displaystyle \mathbf{v}_2 = \frac{d}{b}\mathbf{v}_1. In \mathbb{R}^3, an echelon form of the 3 \times 2 matrix [\,\mathbf{v}_1 \ \mathbf{v}_2\,] has at most two pivots, so its third row is zero. The same row operations on [\,\mathbf{v}_1 \ \mathbf{v}_2 \mid \mathbf{b}\,] with \mathbf{b} = (x, y, z) add and scale the right sides, leaving a third row [\,0 \ \ 0 \mid a_1x + a_2y + a_3z\,], and the system is consistent exactly when a_1x + a_2y + a_3z = 0. The a_i are not all zero, since otherwise every \mathbf{b} would be in the span, and two vectors do not span \mathbb{R}^3.
Spanning \mathbb{R}^m. Row operations turn [\,A \mid \mathbf{b}\,] into [\,U \mid \mathbf{d}\,], with U an echelon form of A, without changing the solutions. If U has a pivot in every row, no row of [\,U \mid \mathbf{d}\,] reads [\,0 \ \cdots \ 0 \mid \text{nonzero}\,], so the system is consistent for every \mathbf{b}. If some row of U has no pivot, it is a zero row, and zero rows sit at the bottom, so the last row is all zeros. Then [\,U \mid \mathbf{e}_m\,] has last row [\,0 \ \cdots \ 0 \mid 1\,] and no solution. Row operations can be undone, so running them backward turns [\,U \mid \mathbf{e}_m\,] into [\,A \mid \mathbf{b}\,] for some \mathbf{b}, with the same empty solution set: that \mathbf{b} is not in the span. Both directions hold for any echelon form U, so all echelon forms of A agree on the test.
Adding a vector already in the span. A combination of \mathbf{v}_1, \dots, \mathbf{v}_k is a combination of the longer list with weight 0 on \mathbf{w}. Conversely, write \mathbf{w} = a_1\mathbf{v}_1 + \dots + a_k\mathbf{v}_k. Distributing d and collecting the terms in each \mathbf{v}_j,
Each span contains the other, so they are equal.
Worked examples
Part 1, Examples 1 to 3: what a span looks like and whether a vector is in it. Part 2, Examples 4 to 6: spanning \mathbb{R}^m, vectors that add nothing, and equations of spans.
One vector, then two, in the plane
Let \mathbf{u} = (3, 1) and \mathbf{v} = (-2, 5). (a) Describe \operatorname{span}\{\mathbf{u}\} and decide whether (-6, -2) and (6, 1) are in it. (b) Show that \mathbf{u} and \mathbf{v} span \mathbb{R}^2, and write \mathbf{b} = (4, 7) as a combination of them.
Solution(a) The span is every c\mathbf{u} = (3c, c): the line through the origin with x = 3y. Since (-6, -2) = -2\mathbf{u}, it is in the span. For (6, 1) the first entry forces c = 2, and then the second entry would be 2, not 1, so (6, 1) is not in the span.
(b) For a general (x, y) we need 3c_1 - 2c_2 = x and c_1 + 5c_2 = y. The second equation gives c_1 = y - 5c_2; substituting into the first, 3y - 17c_2 = x, so
Weights exist for every (x, y), so every vector of \mathbb{R}^2 is in \operatorname{span}\{\mathbf{u}, \mathbf{v}\}. For \mathbf{b}, \displaystyle c_1 = \frac{20 + 14}{17} = 2 and \displaystyle c_2 = \frac{21 - 4}{17} = 1, so \mathbf{b} = 2\mathbf{u} + \mathbf{v}, the walk in the second figure. Check: (6, 2) + (-2, 5) = (4, 7).
In the span or not, in three dimensions
Let \mathbf{v}_1 = (2, -3, 5) and \mathbf{v}_2 = (1, 3, -4). Decide whether \mathbf{b} = (7, 3, -2) and \mathbf{d} = (1, 2, 3) are in \operatorname{span}\{\mathbf{v}_1, \mathbf{v}_2\}.
Solution\mathbf{b} is in the span exactly when this system in c_1, c_2 is consistent:
Replace row 2 by 2(\text{row } 2) + 3(\text{row } 1) and row 3 by 2(\text{row } 3) - 5(\text{row } 1); scaling by 2 \neq 0 avoids fractions:
Row 2 gives c_2 = 3 and row 3 gives c_2 = 3 too, so the rows agree. Row 1 then gives 2c_1 = 7 - 3, so c_1 = 2, and
Check: (4, -6, 10) + (3, 9, -12) = (7, 3, -2).
For \mathbf{d} the same operations act on the right sides 1, 2, 3: row 2 gets 2(2) + 3(1) = 7 and row 3 gets 2(3) - 5(1) = 1. Now row 2 asks 9c_2 = 7 and row 3 asks -13c_2 = 1, which disagree. Replacing row 3 by 9(\text{row } 3) + 13(\text{row } 2) shows the conflict as a row [\,0 \ \ 0 \mid 100\,], the equation 0 = 100. The system is inconsistent, so \mathbf{d} is not in the span.
Two vectors, one line
Describe \operatorname{span}\{(2, -6, 4), (-3, 9, -6)\} in \mathbb{R}^3 by equations, and decide whether (5, -15, 10) and (1, -3, 3) are in it.
SolutionComparing entries, \displaystyle (-3, 9, -6) = -\frac{3}{2}(2, -6, 4). Writing \mathbf{w} = (2, -6, 4),
and the weight \displaystyle c_1 - \frac{3}{2}c_2 takes every real value. So the span is the line through the origin along \mathbf{w}.
For equations, carry a general right side (x, y, z). Row 2 plus 3 times row 1, and row 3 minus 2 times row 1, clear both columns at once:
The system is consistent exactly when both zero rows have right side 0, so the span is the line
For (5, -15, 10), 15 - 15 = 0 and 10 - 10 = 0: in the span, and row 1 leaves c_2 free, so it is reached in many ways, for example (2, -6, 4) - (-3, 9, -6). For (1, -3, 3), 3 - 3 = 0 but 2 - 3 = -1 \neq 0: not in the span. Two vectors, but only a line: the number of vectors alone does not decide the shape.
Do they span all of space?
(a) Do \mathbf{u}_1 = (1, -3, 2), \mathbf{u}_2 = (2, -5, 3) and \mathbf{u}_3 = (-1, 4, 1) span \mathbb{R}^3? (b) Do \mathbf{u}_1, \mathbf{u}_2 and \mathbf{w} = (4, -11, 7)? If not, find a vector outside their span.
Solution(a) Put the vectors in as columns and reduce. Row 2 plus 3 times row 1 is (0, 1, 1), row 3 minus 2 times row 1 is (0, -1, 3), and then row 3 plus row 2 is (0, 0, 4):
There is a pivot in every row, so the vectors span \mathbb{R}^3, without solving for any particular \mathbf{b}.
(b) The same operations give rows (1, 2, 4), then (0, 1, 1), then (0, -1, -1) and finally (0, 0, 0). Row 3 has no pivot, so these vectors do not span \mathbb{R}^3. To find a vector outside, carry \mathbf{b} = (x, y, z) along: row 2’s right side becomes y + 3x, row 3’s becomes z - 2x and then (z - 2x) + (y + 3x) = x + y + z. The system is consistent exactly when
so (1, 0, 0) is not in the span. The cause: \mathbf{w} = 2\mathbf{u}_1 + \mathbf{u}_2, so by the adding result the span of all three is \operatorname{span}\{\mathbf{u}_1, \mathbf{u}_2\}, a plane.
The equation of a plane through the origin
With \mathbf{v}_1 = (2, -3, 5) and \mathbf{v}_2 = (1, 3, -4) from Example 2, find one equation satisfied by exactly the vectors of \operatorname{span}\{\mathbf{v}_1, \mathbf{v}_2\}, and check it against Example 2.
SolutionRun Example 2’s operations with a general right side \mathbf{b} = (x, y, z):
Replacing row 3 by 9(\text{row } 3) + 13(\text{row } 2) leaves [\,0 \ \ 0 \mid -6x + 26y + 18z\,], since 9(2z - 5x) + 13(3x + 2y) = -6x + 26y + 18z. When that entry is 0, rows 1 and 2 supply the weights; when it is not, there are none. Dividing by -2, the span is the plane
Check: \mathbf{v}_1 gives 6 + 39 - 45 = 0 and \mathbf{v}_2 gives 3 - 39 + 36 = 0. For \mathbf{b} = (7, 3, -2), 21 - 39 + 18 = 0, in the span; for \mathbf{d} = (1, 2, 3), 3 - 26 - 27 = -50 \neq 0, not in the span; before dividing by -2 that entry was 100, the one Example 2 found. In dot product terms, the span is every vector orthogonal to (3, -13, -9).
Which positions can the camera reach?
At the observatory, one turn of the first motor moves the camera by \mathbf{p} = (2, 0, 1) centimeters and one turn of the second by \mathbf{q} = (1, 3, -1). (a) Find the equation of the reachable positions, and decide whether (5, 3, 1) and (4, 1, 2) can be reached. (b) Dragica suggests a third motor moving \mathbf{r} = (3, -3, 3) per turn. Does it reach anything new? (c) Branka suggests instead a lift moving \mathbf{s} = (0, 0, 1) per turn. Show that \mathbf{p}, \mathbf{q}, \mathbf{s} reach every position, and find the turns that reach (4, 1, 2).
Solution(a) The reachable positions are \operatorname{span}\{\mathbf{p}, \mathbf{q}\}. Swap rows 1 and 3 to put a 1 in the corner, then subtract twice the new row 1 from row 3:
Row 3 minus row 2 is [\,0 \ \ 0 \mid x - y - 2z\,], so the reachable positions form the plane x - y - 2z = 0 through home. For (5, 3, 1), 5 - 3 - 2 = 0: reachable. Row 2 gives 3c_2 = 3, so c_2 = 1, and row 1 gives c_1 = 1 + 1 = 2: two turns of the first motor and one of the second. Check: (4, 0, 2) + (1, 3, -1) = (5, 3, 1). For (4, 1, 2), 4 - 1 - 4 = -1 \neq 0: unreachable.
(b) For \mathbf{r}, 3 + 3 - 6 = 0, so \mathbf{r} is already in the span; in fact \mathbf{r} = 2\mathbf{p} - \mathbf{q}. By the adding result, \operatorname{span}\{\mathbf{p}, \mathbf{q}, \mathbf{r}\} = \operatorname{span}\{\mathbf{p}, \mathbf{q}\}: the third motor reaches nothing new.
(c) For \mathbf{s}, 0 - 0 - 2 \neq 0, so the lift points off the plane. Swapping rows 1 and 3 of [\,\mathbf{p} \ \mathbf{q} \ \mathbf{s}\,], subtracting twice row 1 from row 3, then row 2 from row 3:
A pivot in every row, so \mathbf{p}, \mathbf{q}, \mathbf{s} span \mathbb{R}^3. For (4, 1, 2) the entries give 3c_2 = 1, 2c_1 + c_2 = 4 and c_1 - c_2 + c_3 = 2, so
Check: \displaystyle \left(\frac{11}{3}, 0, \frac{11}{6}\right) + \left(\frac{1}{3}, 1, -\frac{1}{3}\right) + \left(0, 0, \frac{1}{2}\right) = (4, 1, 2).
Boundaries
- Every real weight. A span allows every real weight. If the motors could only run forward (c_i \geq 0), the reachable set would be a wedge, not a plane; whole turns only would give a grid of separate points. Neither is a span.
- Always through the origin. \mathbf{0} is in every span, so a line or plane that misses the origin, such as y = x + 1 in \mathbb{R}^2, is never a span. By convention the span of the empty list is \{\mathbf{0}\}.
- Counting vectors is not enough. Fewer than m vectors never span \mathbb{R}^m, but m or more may still fail (Example 4(b)). The test counts pivots in rows; a pivot in every column is a different property, the subject of Linear independence.
- One equation only for a plane in \mathbb{R}^3. A line in \mathbb{R}^3 takes two (Example 3). In \mathbb{R}^4 two vectors, neither a multiple of the other, still span a plane, but an echelon form of the 4 \times 2 matrix leaves two zero rows, so it takes two equations.
Where this goes next
A span is the first example of a subspace. The adding result asks which vectors in a list can be dropped without shrinking the span. A list with none to drop is linearly independent, and an independent list that spans a space is a basis, whose size is the dimension. Since A\mathbf{x} is the combination of the columns of A with weights from \mathbf{x} (Multiplying a matrix by a vector), A\mathbf{x} = \mathbf{b} is solvable exactly when \mathbf{b} is in the span of the columns, the image of A.
Exercises
Exercises 1 to 9 go with Part 1, 10 to 17 with Part 2; the reasoning group draws on both.
Part 1: describing a span and testing membership
- Describe \operatorname{span}\{(4, -2)\} in \mathbb{R}^2 and decide whether (-10, 5) and (6, -2) are in it.
- Describe \operatorname{span}\{\mathbf{0}\} in \mathbb{R}^3, and \operatorname{span}\{\mathbf{0}, (1, -1, 2)\}.
- Find weights that write a general (x, y) as a combination of (1, 4) and (2, -3). What does this show about their span? Write (5, 9) as a combination of them.
- Describe \operatorname{span}\{(6, -9), (-4, 6)\} in \mathbb{R}^2 by an equation in x and y.
In Exercises 5 to 7, \mathbf{w}_1 = (1, 4, -1) and \mathbf{w}_2 = (2, 3, 1).
- Is (4, 1, 5) in \operatorname{span}\{\mathbf{w}_1, \mathbf{w}_2\}? If so, give the weights.
- Is (3, 2, 1) in \operatorname{span}\{\mathbf{w}_1, \mathbf{w}_2\}?
- Find h so that (1, h, 5) is in \operatorname{span}\{\mathbf{w}_1, \mathbf{w}_2\}.
- Decide whether (2, 3, 1, 3) and (2, 0, 1, -1) are in \operatorname{span}\{(1, 1, 0, 2), (0, 1, 1, -1)\}.
- Describe \operatorname{span}\{(0, 2, 0), (0, 0, 3)\} in \mathbb{R}^3 by an equation.
Part 2: spanning and equations of spans
- Do (1, 3, -1), (2, 7, 1) and (0, 1, 4) span \mathbb{R}^3?
- Do (1, -1, 2), (3, -2, 5) and (1, 0, 1) span \mathbb{R}^3? If not, find an equation for their span and a vector outside it.
- Do (1, 2, 3, 4), (0, 1, 2, 3) and (0, 0, 1, 2) span \mathbb{R}^4?
- Do (2, 1), (-4, -2) and (6, 3) span \mathbb{R}^2?
- Find an equation for \operatorname{span}\{(1, 2, 1), (0, 1, -3)\}.
- Show that (0, 1, 2) is in \operatorname{span}\{(1, 0, -1), (2, 1, 0)\}, and give one equation for the span of all three vectors.
- A second mount at the observatory has motors that move (1, 1, 0) and (0, 1, 1) centimeters per turn. Find the equation of the reachable positions, and decide whether (3, 5, 2) and (2, 1, 1) can be reached; where one can, give the turns.
- For which k do (1, 3, 0), (0, 1, k) and (2, 0, 1) span \mathbb{R}^3?
Reasoning
- Explain why two vectors never span \mathbb{R}^3.
- Prove that if \mathbf{x} and \mathbf{y} are in \operatorname{span}\{\mathbf{v}_1, \dots, \mathbf{v}_k\}, then so are \mathbf{x} + \mathbf{y} and c\mathbf{x} for every scalar c.
- Prove that \operatorname{span}\{\mathbf{u}, \mathbf{v}\} = \operatorname{span}\{\mathbf{u} + \mathbf{v}, \mathbf{u} - \mathbf{v}\}.
- Give a counterexample to the claim “three nonzero vectors in \mathbb{R}^3, no two of them multiples of each other, always span \mathbb{R}^3.”
- Explain why the plane x + 2y - z = 4 in \mathbb{R}^3 is not the span of any list of vectors.
Answers
- The line through the origin along (4, -2), which is x + 2y = 0. \displaystyle (-10, 5) = -\frac{5}{2}(4, -2) is in it. For (6, -2) the first entry forces \displaystyle c = \frac{3}{2}, and then the second is -3 \neq -2, so it is not.
- \operatorname{span}\{\mathbf{0}\} = \{\mathbf{0}\}, since every multiple of \mathbf{0} is \mathbf{0}. Adding \mathbf{0}, which is in every span, changes nothing: the second span is the line through the origin along (1, -1, 2).
- \displaystyle c_1 = \frac{3x + 2y}{11} and \displaystyle c_2 = \frac{4x - y}{11}, from c_1 + 2c_2 = x and 4c_1 - 3c_2 = y. Weights exist for every (x, y), so the span is \mathbb{R}^2. (5, 9) = 3(1, 4) + (2, -3).
- The line 3x + 2y = 0. Since \displaystyle (-4, 6) = -\frac{2}{3}(6, -9), the span is all (6c, -9c), and 3(6c) + 2(-9c) = 0.
- Yes: (4, 1, 5) = -2\mathbf{w}_1 + 3\mathbf{w}_2. The equations c_1 + 2c_2 = 4 and 4c_1 + 3c_2 = 1 force c_2 = 3, c_1 = -2, and the third entry, -c_1 + c_2 = 5, matches.
- No. The first two equations force c_1 = -1, c_2 = 2, and then the third entry is -c_1 + c_2 = 3 \neq 1.
- h = -6. Entries 1 and 3 give c_1 + 2c_2 = 1 and -c_1 + c_2 = 5, so c_2 = 2, c_1 = -3, and h = 4(-3) + 3(2) = -6.
- (2, 3, 1, 3) = 2(1, 1, 0, 2) + (0, 1, 1, -1) is in the span. (2, 0, 1, -1) is not: entries 1 and 3 force c_1 = 2 and c_2 = 1, and then the second entry is 3 \neq 0.
- The plane x = 0. Every combination is (0, 2c_1, 3c_2), and any (0, y, z) is reached with \displaystyle c_1 = \frac{y}{2}, \displaystyle c_2 = \frac{z}{3}.
- Yes. With the vectors as columns, row 2 minus 3 times row 1 is (0, 1, 1), row 3 plus row 1 is (0, 3, 4), and row 3 minus 3 times row 2 is (0, 0, 1): a pivot in every row.
- No. Row 2 plus row 1 is (0, 1, 1), row 3 minus 2 times row 1 is (0, -1, -1), and their sum is (0, 0, 0). Carrying (x, y, z), the last right side is (z - 2x) + (y + x), so the span is the plane x - y - z = 0, and (1, 0, 0) is outside it.
- No. Three columns hold at most three pivots, and there are four rows.
- No. (-4, -2) = -2(2, 1) and (6, 3) = 3(2, 1), so the span is the line along (2, 1); (1, 0), for one, is off it.
- 7x - 3y - z = 0. Carrying (x, y, z): row 2 minus 2 times row 1 leaves [\,0 \ \ 1 \mid y - 2x\,], row 3 minus row 1 leaves [\,0 \ \ {-3} \mid z - x\,], and row 3 plus 3 times row 2 leaves [\,0 \ \ 0 \mid -7x + 3y + z\,].
- (0, 1, 2) = -2(1, 0, -1) + (2, 1, 0). By the adding result the span of all three is the span of the first two, the plane x - 2y + z = 0: carrying (x, y, z), row 3 plus row 1 gives [\,0 \ \ 2 \mid x + z\,], then subtracting 2 times row 2 gives [\,0 \ \ 0 \mid x - 2y + z\,].
- The plane x - y + z = 0: row 2 minus row 1 gives [\,0 \ \ 1 \mid y - x\,], then row 3 minus row 2 gives [\,0 \ \ 0 \mid x - y + z\,]. (3, 5, 2) is reachable, 3 - 5 + 2 = 0, with 3 turns of the first motor and 2 of the second. (2, 1, 1) gives 2 - 1 + 1 = 2 \neq 0, so it is not.
- Every k except \displaystyle -\frac{1}{6}. Row 2 minus 3 times row 1 is (0, 1, -6), and row 3 minus k times row 2 is (0, 0, 1 + 6k), a pivot exactly when 1 + 6k \neq 0.
- The 3 \times 2 matrix with the two vectors as columns has at most two pivots, one per column, but three rows. Some row has no pivot, so by the spanning result the vectors do not span \mathbb{R}^3.
- Write \mathbf{x} = a_1\mathbf{v}_1 + \dots + a_k\mathbf{v}_k and \mathbf{y} = b_1\mathbf{v}_1 + \dots + b_k\mathbf{v}_k. Regrouping and collecting, \mathbf{x} + \mathbf{y} = (a_1 + b_1)\mathbf{v}_1 + \dots + (a_k + b_k)\mathbf{v}_k, and distributing, c\mathbf{x} = (ca_1)\mathbf{v}_1 + \dots + (ca_k)\mathbf{v}_k. Both are combinations of the \mathbf{v}_i.
- \mathbf{u} + \mathbf{v} and \mathbf{u} - \mathbf{v} lie in \operatorname{span}\{\mathbf{u}, \mathbf{v}\}, so by closure so does every combination of them. Conversely \displaystyle \mathbf{u} = \frac{1}{2}(\mathbf{u} + \mathbf{v}) + \frac{1}{2}(\mathbf{u} - \mathbf{v}) and \displaystyle \mathbf{v} = \frac{1}{2}(\mathbf{u} + \mathbf{v}) - \frac{1}{2}(\mathbf{u} - \mathbf{v}), so by closure the second span holds the first.
- (1, 0, 0), (0, 1, 0) and (1, 1, 0): nonzero, no two multiples of each other, but every combination has third entry 0, so (0, 0, 1) is not in their span.
- Every span contains \mathbf{0}, from all weights zero, and (0, 0, 0) is not on the plane, since 0 + 0 - 0 \neq 4.