Span — definition: Question bank — Maths deep dive

True or false — justify

True or false: A span can be a line that does not pass through the origin.

False. Choosing all scalars zero always yields

0

, so every span contains the origin. Picture it: an off-origin line (like

y = x + 1

) is a legal set of points, but you cannot "stay home" at

0

on it, so it fails the very first membership check — see the red dashed line in the figure below, which is not a span.

True or false: If

v_{2} = 2 v_{1}

, then

span {v_{1}, v_{2}}

is a plane.

False.

v_{2}

lies on the same line as

v_{1}

(just twice as long), so laying them tip-to-tail never leaves that line; the span stays one-dimensional. See the collapse figure below where both arrows sit on one ray.

True or false:

span {v_{1}, v_{2}} = span {v_{2}, v_{1}}

True. Reordering the arrows changes nothing you can build — the same tip-to-tail combinations exist either way, so the reachable territory is identical.

True or false: The span of the empty set is empty.

False. The span of

{}

{0}

(the empty combination gives zero), which is the smallest possible subspace — a single point at the origin, not "nothing."

True or false: If

w \in span {v_{1}, v_{2}}

, then

span {v_{1}, v_{2}, w} = span {v_{1}, v_{2}}

True.

w

already sits inside the current reachable region (the plane or line), so throwing it into the generator list adds no new direction to travel in; the shape is unchanged.

True or false: Three vectors in

R^{3}

always span all of

R^{3}

False. They fill

R^{3}

only if they are linearly independent. If all three lie in one flat plane (coplanar), every combination stays in that plane — you get a 2D sheet, not the whole space.

True or false: Restricting scalars to

c_{i} \geq 0

still gives the span.

False. With only nonnegative scalars you can never go "backwards," so instead of a full line through the origin you get a one-sided cone/wedge. See the cone-vs-line panel below: the span (green) opens both ways; the cone (orange) opens one way.

True or false: A span is closed under subtraction.

True. Subtracting is the same as adding the flipped vector

(- 1) \cdot y

; spans are closed under both flipping and adding, so the tip-to-tail result

x - y

lands back inside.

True or false:

span {v}

for a nonzero

v

R^{2}

is one-dimensional.

True. A single nonzero arrow, stretched and flipped, sweeps out exactly a line through the origin — one free scalar, one dimension.

True or false: If

v_{1}, v_{2}

span

R^{2}

, the matrix with columns

v_{1}, v_{2}

has determinant zero.

False. Spanning

R^{2}

means the two arrows point in different directions, so the parallelogram they make has nonzero area — and area is exactly

∣ det ∣

, which is therefore nonzero (see Determinant).

Spot the error

Claim: "

span {v_{1}, \dots, v_{k}}

is just the set

{v_{1}, \dots, v_{k}}

Wrong. The mental slip is treating the generators as the whole answer. But the span is the infinite fill you get by stretching and adding them — a line or plane, not

k

isolated dots. The

k

vectors are seeds; the span is the whole field they grow (all linear combinations).

Claim: "

(2, 2)

is in

span {(2, 1)}

because both entries are positive."

Wrong. The slip is confusing "looks similar / positive" with "on the line." The span of

(2, 1)

is the single line of slope

\frac{1}{2}

; the point

(2, 2)

has slope

1

, so it sits off that line. Algebra confirms:

c (2, 1) = (2, 2)

needs

c = 1

from the first entry but then gives

1 \neq = 2

in the second. Sign of entries is irrelevant — geometry is which line.

Claim: "Adding any extra vector always makes the span bigger."

Wrong. The slip: "more arrows = more reach." But if the new arrow already lies inside the current line/plane, it points nowhere new, so the shape does not grow at all. In the collapse figure, adding

(2, 2)

{(1, 1)}

keeps a line. Growth needs a genuinely independent direction.

Claim: "

span {(1, 0, 0), (0, 1, 0)}

contains

(0, 0, 5)

because we can scale freely."

Wrong. Those two arrows both live flat in the floor (

x y

-plane), so every combination stays on the floor with third coordinate

a \cdot 0 + b \cdot 0 = 0

. The point

(0, 0, 5)

floats straight up off the floor — unreachable. Free scaling still can't lift you out of the plane the generators lie in.

Claim: "Two vectors always span a plane in

R^{3}

Wrong. The slip counts how many vectors instead of how many directions. Two arrows span a plane only if they point differently; if one is a multiple of the other (or one is

0

), they share a line and span at most that line — as the collapse figure shows.

Claim: "

0

is only in the span if one of the generators is

0

Wrong. The origin is guaranteed no matter what the generators are: set all scalars to zero and you land at

0

. Every span shape in the overview figure passes through the black origin dot for exactly this reason.

Why questions

Why does a span always pass through the origin?

Because

c_{1} = \dots = c_{k} = 0

is always a legal choice of scalars, and it yields

0

— you can always choose to "not move," landing exactly at the origin.

Why is a span always a subspace and never just a random blob?

It contains

0

and is closed under both addition (

\sum a_{i} v_{i} + \sum b_{i} v_{i} = \sum (a_{i} + b_{i}) v_{i}

) and scaling (

λ \sum a_{i} v_{i} = \sum (λ a_{i}) v_{i}

) — exactly the three subspace conditions — so it is a perfectly flat sheet/line through the origin, never a bent or floating blob.

Why do we allow negative and zero scalars, not just positive ones?

Negatives let us reach the opposite direction and zeros let us drop a vector; together they turn a one-sided cone into a full symmetric flat through the origin — this is the exact difference drawn in the cone-vs-line figure.

Why does testing "

w \in span {v_{i}}

" reduce to solving a linear system?

The question "do scalars

c_{i}

exist with

\sum c_{i} v_{i} = w

?" is literally a set of linear equations in the unknowns

c_{i}

; a solution existing means yes, none means no.

Why does a nonzero determinant of the

2 \times 2

matrix

[v_{1} v_{2}]

prove they span

R^{2}

Nonzero determinant means the matrix is invertible, so

\sum c_{i} v_{i} = (x, y)

has a (unique) solution for every target

(x, y)

— every vector is reachable. Geometrically the two arrows enclose a real, non-flattened parallelogram of nonzero area.

Why does

v_{2} = 2 v_{1}

collapse the span to a line?

Every combination becomes

c_{1} v_{1} + c_{2} (2 v_{1}) = (c_{1} + 2 c_{2}) v_{1}

, a single scalar times

v_{1}

, so all outputs lie on the one line through

v_{1}

Why is the number of generators not the same as the dimension of the span?

Generators can be redundant (dependent); dimension counts only genuinely independent directions, which is the size of a basis — often fewer than the number of generators.

Edge cases

What is

span {0}

(the zero vector alone)?

Just

{0}

— every scalar multiple of

0

is still

0

, so the span is the single-point origin subspace.

What is the span of the empty set of vectors?

The zero subspace

{0}

, since the "empty sum" of scaled vectors is defined to be

0

What does

span {v}

look like if

v \neq = 0

A line through the origin in the direction of

v

, extending infinitely both ways because scalars can be positive or negative.

What is

span {v_{1}, \dots, v_{k}}

if every

v_{i} = 0

Still

{0}

; no matter how many zero vectors you list, every combination is zero.

v_{1}, v_{2}, v_{3} \in R^{2}

, can their span be all of

R^{2}

Yes — three vectors in

R^{2}

must be dependent, but two of them can already be independent and span the whole plane; extra vectors just don't shrink it.

Can a span in

R^{3}

be exactly a single point?

Yes — the span of only zero vectors (or the empty set) is

{0}

, a zero-dimensional point at the origin.

If one generator is a linear combination of the others, does the span change when you delete it?

No. That generator was redundant, so removing it leaves the same reachable set — this is exactly how you prune down toward a basis.

The trap-spotter's strategy, as a flowchart

Run every span question through these three gates in order — the diagram mirrors the mnemonic below.

Connections

Linear combination — every membership test is built from these.
Linear independence — decides whether extra generators enlarge the span.
Subspace — what every span is.
Basis — a span's minimal, non-redundant generating set.
Dimension — counts independent directions, not raw generators.
Determinant — the full-span test in square cases.
Column space — the span of a matrix's columns.