Basis — definition, uniqueness of representation: Question bank

True or false — justify

TF1. Any set that spans

R^{3}

is a basis of

R^{3}

False — spanning gives existence of a representation but if the set is too big it is dependent, so coordinates aren't unique; e.g. four vectors can span

R^{3}

yet never form a basis.

TF2. Any linearly independent set in

V

is a basis of

V

False — independence gives uniqueness but not reach;

{e_{1}, e_{2}}

is independent in

R^{3}

yet cannot build

(0, 0, 1)

, so it fails spanning.

TF3. Every basis of a given space

V

has the same number of vectors.

True — that common size is the invariant dimension

dim V

(see Dimension); no basis can be smaller or larger.

TF4. A single nonzero vector is a basis of the line (1-dimensional subspace) it spans.

True — one nonzero vector is automatically independent and spans its own line, so it's a basis of that 1-D subspace.

TF5. The empty set

{}

is a basis of the zero space

{0}

True — the empty set is vacuously independent and spans

{0}

(the empty sum is

0

), so

dim {0} = 0

TF6. If a set has exactly

dim V

vectors, it is automatically a basis.

False — the right count is necessary but not sufficient;

{(1, 0), (2, 0)}

has two vectors in

R^{2}

but is dependent and spans only a line.

TF7. Coordinates of a vector are the same in every basis.

False — coordinates are addresses relative to a chosen basis; the same arrow

(4, 2)

(4, 2)

in the standard basis but

(3, 1)

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

(see Coordinate Vectors and Change of Basis).

TF8. Reordering the basis vectors changes whether it's a basis.

False — spanning and independence don't care about order; but reordering does permute the coordinate entries, so the address list changes even though the set is still a basis.

TF9. Two different bases can give the same coordinates to the same vector.

True — the standard basis and any basis that happens to send

w

to the same tuple can coincide for that vector, but generically the addresses differ; equality for one vector doesn't make the bases equal.

TF10. If every vector has at least one representation, the set must be a basis.

False — "at least one for all" is exactly spanning; you also need "at most one," which is independence, before you may call it a basis.

Spot the error

SE1. "

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

spans

R^{2}

and lets me write any vector, so it's a basis."

The error is skipping uniqueness:

(2, 3) = 2 e_{1} + 3 e_{2}

and

= e_{1} + 2 e_{2} + (1, 1)

, two addresses, so it is not a basis despite spanning.

SE2. "Since

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

is independent, it's a basis of

R^{3}

It's a basis of the xy-plane subspace, not of

R^{3}

— it fails to span because nothing reaches out of the plane to

(0, 0, 1)

SE3. "In the uniqueness proof we subtract the two representations and immediately conclude

c_{i} = d_{i}

A step is missing: subtracting gives

\sum (c_{i} - d_{i}) v_{i} = 0

, and only then does independence force each coefficient to be zero, yielding

c_{i} = d_{i}

SE4. "

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

is a basis of

P_{2}

because it contains

1, x, x^{2}

The repeated

x^{2}

makes the set dependent (

x^{2} - x^{2} = 0

with nonzero coefficients), so it cannot be a basis even though it spans

P_{2}

SE5. "The vector

(0, 0)

has infinitely many representations in a basis, since

0 = \sum 0 v_{i}

trivially."

In a genuine basis

0

has exactly one representation, the all-zero one — that uniqueness is the definition of independence; infinitely many would mean the set is dependent.

SE6. "

{(2, 0), (0, 3)}

isn't a basis because the entries aren't

1

Entries needn't be

1

; these two are independent and span

R^{2}

, so they form a perfectly valid basis — only the coordinates of a vector will scale.

SE7. "Adding one more independent-looking vector to a basis keeps it a basis, just a bigger one."

You cannot add any vector to a basis and stay independent — the basis already spans, so the new vector is a combination of the old ones, making the enlarged set dependent.

Why questions

WHY1. Why does spanning alone fail to give a coordinate system?

Because it only promises each vector some recipe; without independence a vector may have many recipes, so there's no single well-defined address to call its coordinates.

WHY2. Why does the uniqueness proof turn "two equal sums" into "something

= 0

Because independence is a statement about combinations equal to the zero vector; rewriting the equality as

\sum (c_{i} - d_{i}) v_{i} = 0

is what lets us invoke that definition.

WHY3. Why must a basis have exactly

dim V

vectors — no more, no fewer?

Fewer than

dim V

can't span (too few directions to reach everywhere); more than

dim V

must be dependent (extra directions are redundant), and only the exact count can be both.

WHY4. Why does the standard basis make coordinates equal the vector's entries?

Each

e_{i}

switches on exactly one slot, so the coefficient needed to build slot

i

is precisely that entry — the basis is "invisible" only because of this special alignment.

WHY5. Why is uniqueness of representation the payoff that lets us use matrices?

A unique address turns each abstract vector into one definite coordinate column; a linear map then acts predictably on those columns (see Matrix of a Linear Map).

WHY6. Why can abstract spaces like

P_{2}

have a basis at all, with no arrows to draw?

A basis needs only the vector-space operations (add, scale), not geometry;

{1, x, x^{2}}

is independent and spanning, so

P_{2}

behaves just like

R^{3}

coordinate-wise.

WHY7. Why does applying uniqueness specifically to

w = 0

recover independence?

The obvious representation of

0

is all-zero coefficients; uniqueness says it's the only one, and "

\sum c_{i} v_{i} = 0 \Rightarrow

all

c_{i} = 0

" is precisely the definition of independence.

Edge cases

EC1. Is

{0}

(the single zero vector) ever part of a basis?

Never — the zero vector is dependent by itself (

1 \cdot 0 = 0

with a nonzero coefficient), so any set containing it fails independence.

EC2. What is a basis of the zero space

{0}

, and what is its dimension?

The empty set is its basis and

dim {0} = 0

, since there are no directions to specify and

0

is already the only vector.

EC3. Can an infinite set be a basis?

Yes for infinite-dimensional spaces (e.g.

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

for all polynomials), but every basis of a finite-dimensional

V

is finite with exactly

dim V

vectors.

EC4. In

R^{n}

, is any set of

n

vectors a basis as long as none is the zero vector?

No — they must be independent;

n

nonzero vectors can still be dependent (e.g. two lie on the same line), and then they neither span nor give unique coordinates.

EC5. If two vectors are parallel (one is a scalar multiple of the other), can they both be in a basis?

No — parallel vectors are dependent, so at most one can appear; the other adds a redundant direction that breaks uniqueness.

EC6. Does swapping a vector's sign (

v_{i} \to - v_{i}

) still leave a basis?

Yes —

- v_{i}

is independent from and spans the same space as

v_{i}

; the set stays a basis, though that vector's coordinate flips sign.

Connections

Linear Independence — the uniqueness half every trap here tests.
Span and Spanning Sets — the existence half.
Dimension — why the vector count matters.
Coordinate Vectors and Change of Basis — why addresses depend on the basis.
Subspaces — bases of smaller spaces inside $V$ .
Matrix of a Linear Map — the payoff of unique coordinates.