Dimension — basis cardinality: Question bank

Before the traps, let us pin down every symbol they use, and draw the one picture behind the whole story.

Definition Notation used on this page (built from scratch)

$V$ — the vector space we work in. A vector is just an arrow/list of numbers you can add and scale.
A spanning set $S = {w_{1}, w_{2}, \dots, w_{n}}$ — the $w_{i}$ (read "double-u sub $i$ ") are vectors whose combinations reach every vector of $V$ . Think "raw material pile."
An independent set $L = {v_{1}, v_{2}, \dots, v_{m}}$ — the $v_{j}$ ("vee sub $j$ ") are vectors where none is built from the others. Think "vectors that each earn their keep."
Coefficient $c_{j}$ — a plain number multiplying the vector $w_{j}$ when we write one vector as a combination like $c_{1} w_{1} + c_{2} w_{2} + \dots$ . " $c_{j} \neq = 0$ " just means " $w_{j}$ actually shows up in that combination."
$v_{k + 1}$ — the next independent vector we try to swap in during the exchange-lemma proof (the " $(k + 1)$ -th" one).
$∣ B ∣$ — the cardinality (number of vectors) of a set $B$ .

Recall the one weapon behind almost every answer, now stated carefully about counts:

True or false — justify

A basis of

V

must be both independent and spanning.

True — one property alone is not enough; an independent set can be too small and a spanning set too big, so only the both-at-once object has the exact dimension count.

Every spanning set of

V

is a basis.

False — a spanning set may be too big and contain redundant (dependent) vectors; you must first thin it down to an independent subset before it becomes a basis.

Every linearly independent set in

V

is a basis.

False — it may be too small to span; e.g.

{(1, 0)}

is independent in

R^{2}

but reaches only the

x

-axis, so it is not a basis.

Two bases of the same space always contain the same actual vectors.

False — they only share the same count;

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

and

{(1, 1), (1, - 1)}

are both bases of

R^{2}

yet share no vector.

If a space is spanned by

5

vectors, it can still have a basis of size

6

False — a basis is independent, and independent

\leq

spanning

= 5

, so no basis (indeed no independent set) can exceed

5

The Steinitz Exchange Lemma requires the spanning set to be independent.

False — it only needs the set to span; the spanning set may be dependent, and this generality is exactly what makes the lemma powerful.

If every basis of

V

has

3

vectors, then some independent set in

V

has

4

vectors.

False — the max independent size equals the dimension

3

; any

4

vectors must be dependent by the exchange lemma.

Adding a vector to a basis keeps it a basis.

False — it stays spanning but breaks independence (the new vector is a combination of the old ones), so it is no longer a basis.

Removing a vector from a basis of

R^{3}

leaves a basis of a

2

-dimensional subspace.

True — the remaining two vectors are still independent (a subset of an independent set), so they form a basis of their

2

-dimensional span.

The zero vector can belong to a basis.

False — any set containing

0

is dependent, since

1 \cdot 0 = 0

is a nontrivial combination equal to zero.

Spot the error

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

has

4

vectors, so

dim R^{3} = 4

The error is calling these a basis;

R^{3}

is spanned by

3

vectors, so any independent set has

\leq 3

— these

4

are dependent since

(1, 1, 1) = (1, 0, 0) + (0, 1, 0) + (0, 0, 1)

, so they don't count.

"The dimension is the size of the largest set of vectors you can fit in

V

Wrong extremal quantity — you can fit infinitely many vectors in

V

; dimension is the size of a maximal independent set (equivalently a minimal spanning set). "Maximal independent" means: independent, and you cannot add any vector while staying independent — that maximal independent set is exactly a basis.

"In the exchange lemma proof, we needed some

c_{j} \neq = 0

; that follows because the

w

's are independent."

The error names the wrong hypothesis — here

c_{j}

is the coefficient on

w_{j}

when we write

v_{k + 1} = c_{k + 1} w_{k + 1} + \dots + c_{n} w_{n}

plus earlier

v

's. Some

c_{j} \neq = 0

follows from independence of the incoming set

L = {v_{1}, \dots, v_{m}}

; if all

c_{j} = 0

then

v_{k + 1}

would be a combination of earlier

v

's, contradicting independence of $L$.

B_{1}

spans and

B_{2}

spans, so applying the lemma both ways gives

∣ B_{1} ∣ = ∣ B_{2} ∣

The error uses spanning on both sides; the lemma compares independent-vs-spanning, so you need each basis to play both roles — indep(

B_{1}

)

\leq

span(

B_{2}

) and indep(

B_{2}

)

\leq

span(

B_{1}

"Since

{1, 1 + x, 1 + x + x^{2}}

looks nothing like

{1, x, x^{2}}

, it should span a differently-sized space."

Appearance is not count — both have

3

elements and both span

P_{2}

, so both certify

dim P_{2} = 3

; the theorem forbids the size from changing with a change of basis.

"Rank-nullity says

dim V = rank + nullity

, so dimension depends on the chosen map."

The error ties

dim V

to a map — with rank

= dim (image)

and nullity

= dim (ker)

as defined above,

dim V

is fixed by basis cardinality alone; rank-nullity merely splits that fixed number differently for each map, it never changes the total.

Why questions

Why does the exchange lemma give an inequality (indep

\leq

span) rather than equality?

Because a spanning set can carry redundancy (extra vectors), so it may have strictly more members than the independent set it dominates; equality appears only when both sets are bases.

Why must we apply the lemma twice to get the Dimension Theorem?

One application gives

∣ B_{1} ∣ \leq ∣ B_{2} ∣

; only by swapping roles do we also get

∣ B_{2} ∣ \leq ∣ B_{1} ∣

, and two opposite inequalities are what force equality.

Why is independence of

L

the only place independence is used in the proof?

Independence of

L = {v_{1}, \dots, v_{m}}

guarantees no incoming vector is already built from the swapped-in ones, so a

w

always survives to be kicked out — that keeps the swap process running for all

m

steps.

Why does dimension being well-defined matter at all?

Without it "

dim R^{3} = 3

" would be meaningless, since some odd basis might have

5

vectors; the theorem guarantees the answer is the same no matter which basis you pick.

Why can a spanning set in the lemma be dependent without breaking the argument?

The proof only ever expresses incoming vectors in terms of the current set and swaps — it never needs the

w

's to be uniquely written, so mere spanning suffices.

Why does swapping preserve the spanning property at every step?

Each kicked-out

w_{j}

is shown to be a combination of the vectors that remain plus the incoming

v

, so nothing reachable before becomes unreachable — the set still spans.

Edge cases

What is the dimension of the zero vector space

{0}

It is

0

— the empty set is (vacuously) independent and spans

{0}

, so it is the basis, giving cardinality

0

Is the empty set a valid basis, and of what?

Yes — it is independent (no nontrivial relation exists) and its span is

{0}

, so it is the unique basis of the zero space.

For an infinite-dimensional space, does "independent

\leq

spanning" still say something?

Yes — if

V

has independent sets of every finite size (e.g.

1, x, x^{2}, x^{3}, \dots

in all polynomials), then no finite set can span it, since a finite spanning set of size

n

would cap independent sets at

n

. Concretely, the space of all polynomials has the infinite basis

{1, x, x^{2}, \dots}

and no finite basis at all.

When we say an infinite-dimensional space has "dimension an infinite cardinal," what does that concretely mean?

It just means every basis has infinitely many vectors and any two bases can be matched one-to-one; e.g. all polynomials have basis

{1, x, x^{2}, \dots}

, which pairs off with the counting numbers

0, 1, 2, \dots

— so its dimension is "countably infinite," the size of that list.

For an infinite space, why does a maximal independent set even exist to serve as a basis?

Because you can keep adding independent vectors, and a standard set-theory principle (Zorn's Lemma) guarantees this growing chain reaches a maximal one; that maximal independent set can add nothing more without becoming dependent, so it spans and is a basis.

V

is spanned by

n

vectors and you find

n

independent vectors in

V

, must they form a basis?

Yes — they hit the maximum independent size

n

, and

n

independent vectors in an

n

-dimensional space automatically span, so they are a basis.

Can a single vector

{v}

with

v \neq = 0

be a basis?

Only if

dim V = 1

; then

v \neq = 0

is independent and its scalar multiples fill the whole line, so it spans and is a basis.

If two spaces have the same dimension, are they "the same" space?

Not literally, but they are isomorphic — equal dimension means a coordinate map matches their bases one-to-one, so they behave identically as vector spaces.

Connections

Steinitz Exchange Lemma — the single tool behind nearly every answer here.
Linear Independence — source of the "some coefficient ≠ 0" step and the too-small traps.
Spanning Sets — source of the slots and the too-big traps.
Basis of a Vector Space — the both-at-once object every trap circles around.
Rank-Nullity Theorem — dimension splitting, not dimension changing.
Coordinates and Change of Basis — why different-looking bases still share one count.