Bases 145

The vectors of the foowing set do not

form a basis.

1

0

0

1

,

3

1

,

1

2

,

The set

To understand why they don't form a basis,

have a lk at the foowing equation:

x

2

1

2

1 3 x

1

O

= c

1

y

1

y

2

1

0

0

1

+ c

2

3

1

+ c

3

1

2

+ c

4

where is an arbitrary vector in R

2

.

can be formed in many dierent ways

(using dierent choices for c

1

, c

2

, c

3

, and c

4

).

Because of this, the set does not form “a

minimal set of vectors nded to expre

an arbitrary vector in R

m

.”

y

1

y

2

y

1

y

2

146 Chapter 6 More Vectors

Neither of the two vector sets below is able

to describe the vector , and if they can’t

describe that vector, then there's no way that

they could describe “an arbitrary vector in R

3

.”

Because of this, they're not bases.

,

The set

1

0

0

0

1

0

,

The set

1

0

0

,

0

1

0

1

2

0

0

0

1

,

1

0

0

,

0

1

0

0

0

1

Just because a set of vectors is linearly

independent doesn't mean that it forms a basis.

For instance, the set forms a basis,

while the set does not, even though

they're both linearly independent.

,

1

0

0

0

1

0

x

2

x

3

1

1

O

x

1

x

2

x

3

1

1

O

2

x

1

Bases 147

Linear Independence

= c

1

+ c

2

+ … + c

n

a

11

a

21

a

m1

a

12

a

22

a

m2

a

1n

a

2n

a

mn

0

0

0

c

1

= 0

c

2

= 0

c

n

= 0

where the left side is the zero vector of R

m

.

to the equation

if there’s only one solution

We say that a set of vectors is linearly independent

, … ,

a

11

a

21

a

m1

a

12

a

22

a

m2

a

1n

a

2n

a

mn

,A set of vectors forms a basis if there’s only

y

1

y

2

y

m

where the left side is an arbitrary vector in R

m

. And once again, a basis

is a minimal set of vectors needed to express an arbitrary vector in R

m

.

y

1

y

2

y

m

= c

1

+ c

2

+ … + c

n

a

11

a

21

a

m1

a

12

a

22

a

m2

a

1n

a

2n

a

mn

one solution to the equation

, … ,

a

11

a

21

a

m1

a

12

a

22

a

m2

a

1n

a

2n

a

mn

,

Since bases and linear independence are

confusingly similar, I thought I'd talk a bit

about the dierences betwn the two.

Bases

148 Chapter 6 More Vectors

So...

While linear

independence is about

finding a clear-cut path

back to the origin,

We're

linearly

independent!

They're

bases.

They

are.

Ye p.

Exactly!

Not a lot of

people are

able to grasp

the dierence

betwn the

two that fast!

I must say I'm

impreed!

No big deal!

That's a for

tod—

Ah, wait a

sec!

bases are about finding

clear-cut paths to any

vector in a given space

R

m

?

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