3.4 Basis and Dimension · Linear Algebra Refresher

A basis is the Goldilocks set: enough vectors to build everything (spanning), not one more than necessary (independent). Dimension is then well defined, and it turns two conditions into one.

Definition · Basis

{v_{1}, \dots, v_{n}}

is a basis for

V

if (i) the vectors are linearly independent and (ii) they span

V

V

has dimension

n

if it has a basis of

n

vectors;

dim {0} = 0

; if no finite basis exists,

V

is infinite-dimensional.

Space	Standard basis	$dim$
$R^{n}$	$e_{1}, \dots, e_{n}$	$n$
$P_{n}$	$1, x, x^{2}, \dots, x^{n - 1}$	$n$
$R^{2 \times 2}$	$E_{11}, E_{12}, E_{21}, E_{22}$	$4$
$C [a, b]$	— none finite —	$\infty$

Theorem 3.4.1 & Corollary 3.4.2

{v_{1}, \dots, v_{n}}

spans

V

, then any set of more than $n$ vectors in

V

is linearly dependent. Consequently, any two bases of $V$ have the same number of elements — so dimension is well defined.

Theorem 3.4.3 · The count does the work

Let

dim V = n > 0

. Then (I) any

n

linearly independent vectors span

V

; (II) any

n

vectors that span

V

are linearly independent.

This halves your workload. Once the number of vectors equals the dimension, verifying independence or spanning is enough to conclude 'basis'.

Theorem 3.4.4

dim V = n > 0

: (i) no set of fewer than

n

vectors can span

V

; (ii) any linearly independent set of fewer than

n

vectors can be extended to a basis; (iii) any spanning set of more than

n

vectors can be pared down to a basis.

Worked example

A non-obvious basis for $P_{3}$

0/5 steps

Show that

{1, x - 1, (x - 1)^{2}}

is a basis for

P_{3}

(polynomials of degree less than 3).

★Key idea

Reading dimension off a description is an exam staple. 'Symmetric

3 \times 3

matrices': 6 free entries ⟹

dim = 6

. 'Vectors in

R^{4}

with

x_{1} + x_{2} + x_{3} + x_{4} = 0

': one constraint on four unknowns ⟹

dim = 3

. Count the free parameters.

Check yourself

V

has dimension

6

. You are handed

6

linearly independent vectors from

V

. What can you conclude?

T / F

A set of 6 vectors can be a basis for

R^{5}

T / F

Any linearly independent set of 3 vectors in a 5-dimensional space can be extended to a basis.

3.4Basis and Dimension

What you must be able to do

A non-obvious basis for $P_{3}$

Check yourself

3.4Basis and Dimension

What you must be able to do

A non-obvious basis for P3​

Check yourself

A non-obvious basis for $P_{3}$